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Mathematics LibreTexts

2.7: Solve Linear Inequalities

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Learning Objectives

By the end of this section, you will be able to:

  • Graph inequalities on the number line
  • Solve inequalities using the Subtraction and Addition Properties of inequality
  • Solve inequalities using the Division and Multiplication Properties of inequality
  • Solve inequalities that require simplification
  • Translate to an inequality and solve

Before you get started, take this readiness quiz.

  • Translate from algebra to English: \(15>x\). If you missed this problem, review Exercise 1.3.1 .
  • Solve: \(n−9=−42\). If you missed this problem, review Exercise 2.1.7 .
  • Solve: \(−5p=−23\). If you missed this problem, review Exercise 2.2.1 .
  • Solve: \(3a−12=7a−20\). If you missed this problem, review Exercise 2.3.22 .

Graph Inequalities on the Number Line

Do you remember what it means for a number to be a solution to an equation? A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

What about the solution of an inequality? What number would make the inequality \(x > 3\) true? Are you thinking, ‘ x could be 4’? That’s correct, but x could be 5 too, or 20, or even 3.001. Any number greater than 3 is a solution to the inequality \(x > 3\).

We show the solutions to the inequality \(x > 3\) on the number line by shading in all the numbers to the right of 3, to show that all numbers greater than 3 are solutions. Because the number 3 itself is not a solution, we put an open parenthesis at 3. The graph of \(x > 3\) is shown in Figure \(\PageIndex{1}\). Please note that the following convention is used: light blue arrows point in the positive direction and dark blue arrows point in the negative direction.

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than 3 is graphed on the number line, with an open parenthesis at x equals 3, and a red line extending to the right of the parenthesis.

The graph of the inequality \(x \geq 3\) is very much like the graph of \(x > 3\), but now we need to show that 3 is a solution, too. We do that by putting a bracket at \(x = 3\), as shown in Figure \(\PageIndex{2}\).

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than or equal to 3 is graphed on the number line, with an open bracket at x equals 3, and a red line extending to the right of the bracket.

Notice that the open parentheses symbol, (, shows that the endpoint of the inequality is not included. The open bracket symbol, [, shows that the endpoint is included.

Exercise \(\PageIndex{1}\)

Graph on the number line:

  • \(x\leq 1\)
  • \(x>−1\)

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is less than or equal to 1 is graphed on the number line, with an open bracket at x equals 1, and a red line extending to the left of the bracket.

Exercise \(\PageIndex{2}\)

  • \(x\leq −1\)

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is less than or equal to negative 1 is graphed on the number line, with an open bracket at x equals negative 1, and a dark line extending to the left of the bracket.

Exercise \(\PageIndex{3}\)

  • \(x>−2\)
  • \(x<−3\)
  • \(x\geq −1\)

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than negative 2 is graphed on the number line, with an open parenthesis at x equals negative 2, and a dark line extending to the right of the parenthesis.

We can also represent inequalities using interval notation. As we saw above, the inequality \(x>3\) means all numbers greater than 3. There is no upper end to the solution to this inequality. In interval notation , we express \(x>3\) as \((3, \infty)\). The symbol \(\infty\) is read as ‘infinity’. It is not an actual number. Figure \(\PageIndex{3}\) shows both the number line and the interval notation.

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than 3 is graphed on the number line, with an open parenthesis at x equals 3, and a red line extending to the right of the parenthesis. The inequality is also written in interval notation as parenthesis, 3 comma infinity, parenthesis.

The inequality \(x\leq 1\) means all numbers less than or equal to 1. There is no lower end to those numbers. We write \(x\leq 1\) in interval notation as \((-\infty, 1]\). The symbol \(-\infty\) is read as ‘negative infinity’. Figure \(\PageIndex{4}\) shows both the number line and interval notation.

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is less than or equal to 1 is graphed on the number line, with an open bracket at x equals 1, and a red line extending to the left of the bracket. The inequality is also written in interval notation as parenthesis, negative infinity comma 1, bracket.

INEQUALITIES, NUMBER LINES, AND INTERVAL NOTATION

This figure show four number lines, all without tick marks. The inequality x is greater than a is graphed on the first number line, with an open parenthesis at x equals a, and a red line extending to the right of the parenthesis. The inequality is also written in interval notation as parenthesis, a comma infinity, parenthesis. The inequality x is greater than or equal to a is graphed on the second number line, with an open bracket at x equals a, and a red line extending to the right of the bracket. The inequality is also written in interval notation as bracket, a comma infinity, parenthesis. The inequality x is less than a is graphed on the third number line, with an open parenthesis at x equals a, and a red line extending to the left of the parenthesis. The inequality is also written in interval notation as parenthesis, negative infinity comma a, parenthesis. The inequality x is less than or equal to a is graphed on the last number line, with an open bracket at x equals a, and a red line extending to the left of the bracket. The inequality is also written in interval notation as parenthesis, negative infinity comma a, bracket.

Did you notice how the parenthesis or bracket in the interval notation matches the symbol at the endpoint of the arrow? These relationships are shown in Figure \(\PageIndex{5}\).

This figure shows the same four number lines as above, with the same interval notation labels. Below the interval notation for each number line, there is text indicating how the notation on the number lines is similar to the interval notation. The first number line is a graph of x is greater than a, and the interval notation is parenthesis, a comma infinity, parenthesis. The text below reads: “Both have a left parenthesis.” The second number line is a graph of x is greater than or equal to a, and the interval notation is bracket, a comma infinity, parenthesis. The text below reads: “Both have a left bracket.” The third number line is a graph of x is less than a, and the interval notation is parenthesis, negative infinity comma a, parenthesis. The text below reads: “Both have a right parenthesis.” The last number line is a graph of x is less than or equal to a, and the interval notation is parenthesis, negative infinity comma a, bracket. The text below reads: “Both have a right bracket.”

Exercise \(\PageIndex{4}\)

Graph on the number line and write in interval notation.

  • \(x \geq -3\)
  • \(x<2.5\)
  • \(x\leq \frac{3}{5}\)

Exercise \(\PageIndex{5}\)

Graph on the number line and write in interval notation:

  • \(x\leq −1.5\)
  • \(x\geq \frac{3}{4}\)

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than 2 is graphed on the number line, with an open parenthesis at x equals 2, and a dark line extending to the right of the parenthesis. The inequality is also written in interval notation as parenthesis, 2 comma infinity, parenthesis.

Exercise \(\PageIndex{6}\)

  • \(x\leq −4\)
  • \(x\geq 0.5\)
  • \(x<-\frac{2}{3}\)

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is less than or equal to negative 4 is graphed on the number line, with an open bracket at x equals negative 4, and a dark line extending to the left of the bracket. The inequality is also written in interval notation as parenthesis, negative infinity comma negative 4, bracket.

Solve Inequalities using the Subtraction and Addition Properties of Inequality

The Subtraction and Addition Properties of Equality state that if two quantities are equal, when we add or subtract the same amount from both quantities, the results will be equal.

PROPERTIES OF EQUALITY

\[\begin{array} { l l } { \textbf { Subtraction Property of Equality } } & { \textbf { Addition Property of Equality } } \\ { \text { For any numbers } a , b , \text { and } c , } & { \text { For any numbers } a , b , \text { and } c } \\ { \text { if } \qquad \quad a = b , } & { \text { if } \qquad \quad a = b } \\ { \text { then } a - c = b - c . } & { \text { then } a + c = b + c } \end{array}\]

Similar properties hold true for inequalities.

Similarly we could show that the inequality also stays the same for addition.

This leads us to the Subtraction and Addition Properties of Inequality.

PROPERTIES OF INEQUALITY

\[\begin{array} { l l } { \textbf { Subtraction Property of Inequality } } & { \textbf { Addition Property of Inequality } } \\ { \text { For any numbers } a , b , \text { and } c , } & { \text { For any numbers } a , b , \text { and } c } \\ { \text { if }\qquad \quad a < b } & { \text { if } \qquad \quad a < b } \\ { \text { then } a - c < b - c . } & { \text { then } a + c < b + c } \\\\ { \text { if } \qquad \quad a > b } & { \text { if } \qquad \quad a > b } \\ { \text { then } a - c > b - c . } & { \text { then } a + c > b + c } \end{array}\]

We use these properties to solve inequalities, taking the same steps we used to solve equations. Solving the inequality \(x+5>9\), the steps would look like this:

\[\begin{array}{rrll} {} &{x + 5} &{ >} &{9} \\ {\text{Subtract 5 from both sides to isolate }x.} &{x + 5 - 5} &{ >} &{9 - 5} \\{} &{x} &{ >} &{4} \\ \end{array}\]

Any number greater than 4 is a solution to this inequality.

Exercise \(\PageIndex{7}\)

Solve the inequality \(n - \frac{1}{2} \leq \frac{5}{8}\), graph the solution on the number line, and write the solution in interval notation.

Exercise \(\PageIndex{8}\)

Solve the inequality, graph the solution on the number line, and write the solution in interval notation.

\(p - \frac{3}{4} \geq \frac{1}{6}\)

This figure shows the inequality p is greater than or equal to 11/12. Below this inequality is the inequality graphed on a number line ranging from 0 to 4, with tick marks at each integer. There is a bracket at p equals 11/12, and a dark line extends to the right from 11/12. Below the number line is the solution written in interval notation: bracket, 11/12 comma infinity, parenthesis.

Exercise \(\PageIndex{9}\)

\(r - \frac{1}{3} \leq \frac{7}{12}\)

This figure shows the inequality r is less than or equal to 11/12. Below this inequality is the inequality graphed on a number line ranging from 0 to 4, with tick marks at each integer. There is a bracket at r equals 11/12, and a dark line extends to the left from 11/12. Below the number line is the solution written in interval notation: parenthesis, negative infinity comma 11/12, bracket.

Solve Inequalities using the Division and Multiplication Properties of Inequality

The Division and Multiplication Properties of Equality state that if two quantities are equal, when we divide or multiply both quantities by the same amount, the results will also be equal (provided we don’t divide by 0).

\[\begin{array}{ll} {\textbf{Division Property of Equality}} &{\textbf{MUltiplication Property of Equality}} \\ {\text{For any numbers a, b, c, and c} \neq 0} &{\text{For any numbers a, b, c}} \\ {\text{if } \qquad a = b} &{\text{if} \qquad \quad a = b} \\ {\text{then }\quad \frac{a}{c} = \frac{b}{c}} &{\text{then } \quad ac = bc} \end{array}\]

Are there similar properties for inequalities? What happens to an inequality when we divide or multiply both sides by a constant?

Consider some numerical examples.

The inequality signs stayed the same.

Does the inequality stay the same when we divide or multiply by a negative number?

The inequality signs reversed their direction.

When we divide or multiply an inequality by a positive number, the inequality sign stays the same. When we divide or multiply an inequality by a negative number, the inequality sign reverses.

Here are the Division and Multiplication Properties of Inequality for easy reference.

DIVISION AND MULTIPLICATION PROPERTIES OF INEQUALITY

For any real numbers a,b,c

\[\begin{array}{ll} {\text{if } a < b \text{ and } c > 0, \text{ then}} &{\frac{a}{c} < \frac{b}{c} \text{ and } ac < bc} \\ {\text{if } a > b \text{ and } c > 0, \text{ then}} &{\frac{a}{c} > \frac{b}{c} \text{ and } ac > bc} \\ {\text{if } a < b \text{ and } c < 0, \text{ then}} &{\frac{a}{c} > \frac{b}{c} \text{ and } ac > bc} \\ {\text{if } a > b \text{ and } c < 0, \text{ then}} &{\frac{a}{c} < \frac{b}{c} \text{ and } ac < bc} \end{array}\]

When we divide or multiply an inequality by a:

  • positive number, the inequality stays the same .
  • negative number, the inequality reverses .

Exercise \(\PageIndex{10}\)

Solve the inequality \(7y<​​42\), graph the solution on the number line, and write the solution in interval notation.

Exercise \(\PageIndex{11}\)

\(9c>72\)

This figure is a number line ranging from 6 to 10 with tick marks for each integer. The inequality c is greater than 8 is graphed on the number line, with an open parenthesis at c equals 8, and a dark line extending to the right of the parenthesis.

\((8, \infty)\)

Exercise \(\PageIndex{12}\)

\(12d\leq 60\)

\(d\leq 5\)

This figure is a number line ranging from 3 to 7 with tick marks for each integer. The inequality d is less than or equal to 5 is graphed on the number line, with an open bracket at d equals 5, and a dark line extending to the left of the bracket. The inequality is also written in interval notation as parenthesis, negative infinity comma 5, bracket.

\((-\infty, 5]\)

Exercise \(\PageIndex{13}\)

Solve the inequality \(−10a\geq 50\), graph the solution on the number line, and write the solution in interval notation.

Exercise \(\PageIndex{14}\)

Solve each inequality, graph the solution on the number line, and write the solution in interval notation.

\(−8q<32\)

\(q>−4\)

This figure is a number line ranging from negative 6 to negative 3 with tick marks for each integer. The inequality q is greater than negative 4 is graphed on the number line, with an open parenthesis at q equals negative 4, and a dark line extending to the right of the parenthesis. The inequality is also written in interval notation as parenthesis, negative 4 comma infinity, parenthesis.

Exercise \(\PageIndex{15}\)

\(−7r\leq −70\)

This figure is a number line ranging from 9 to 13 with tick marks for each integer. The inequality r is greater than or equal to 10 is graphed on the number line, with an open bracket at r equals 10, and a dark line extending to the right of the bracket. The inequality is also written in interval notation as bracket, 10 comma infinity, parenthesis.

SOLVING INEQUALITIES

\[\begin{array}{l} x > a\text{ has the same meaning as } a < x \end{array}\]

Think about it as “If Xavier is taller than Alex, then Alex is shorter than Xavier.”

Exercise \(\PageIndex{16}\)

Solve the inequality \(-20 < \frac{4}{5}u\), graph the solution on the number line, and write the solution in interval notation.

Exercise \(\PageIndex{17}\)

\(24 \leq \frac{3}{8}m\)

This figure shows the inequality m is greater than or equal to 64. Below this inequality is a number line ranging from 63 to 67 with tick marks for each integer. The inequality m is greater than or equal to 64 is graphed on the number line, with an open bracket at m equals 64, and a dark line extending to the right of the bracket. The inequality is also written in interval notation as bracket, 64 comma infinity, parenthesis.

Exercise \(\PageIndex{18}\)

\(-24 < \frac{4}{3}n\)

This figure shows the inequality n is greater than negative 18. Below this inequality is a number line ranging from negative 20 to negative 16 with tick marks for each integer. The inequality n is greater than negative 18 is graphed on the number line, with an open parenthesis at n equals negative 18, and a dark line extending to the right of the parenthesis. The inequality is also written in interval notation as parenthesis, negative 18 comma infinity, parenthesis.

Exercise \(\PageIndex{19}\)

Solve the inequality \(\frac{t}{-2} \geq 8\), graph the solution on the number line, and write the solution in interval notation.

Exercise \(\PageIndex{20}\)

\(\frac{k}{-12}\leq 15\)

This figure shows the inequality k is greater than or equal to negative 180. Below this inequality is a number line ranging from negative 181 to negative 177 with tick marks for each integer. The inequality k is greater than or equal to negative 180 is graphed on the number line, with an open bracket at n equals negative 180, and a dark line extending to the right of the bracket. The inequality is also written in interval notation as bracket, negative 180 comma infinity, parenthesis.

Exercise \(\PageIndex{21}\)

\(\frac{u}{-4}\geq -16\)

This figure shows the inequality u is less than or equal to 64. Below this inequality is a number line ranging from 62 to 66 with tick marks for each integer. The inequality u is less than or equal to 64 is graphed on the number line, with an open bracket at u equals 64, and a dark line extending to the left of the bracket. The inequality is also written in interval notation as parenthesis, negative infinity comma 64, bracket.

Solve Inequalities That Require Simplification

Most inequalities will take more than one step to solve. We follow the same steps we used in the general strategy for solving linear equations, but be sure to pay close attention during multiplication or division.

Exercise \(\PageIndex{22}\)

Solve the inequality \(4m\leq 9m+17\), graph the solution on the number line, and write the solution in interval notation.

Exercise \(\PageIndex{23}\)

Solve the inequality \(3q\geq 7q−23\), graph the solution on the number line, and write the solution in interval notation.

This figure shows the inequality q is less than or equal to 23/4. Below this inequality is a number line ranging from 4 to 8 with tick marks for each integer. The inequality q is less than or equal to 23/4 is graphed on the number line, with an open bracket at q equals 23/4 (written in), and a dark line extending to the left of the bracket. The inequality is also written in interval notation as parenthesis, negative infinity comma 23/4, bracket.

Exercise \(\PageIndex{24}\)

Solve the inequality \(6x<10x+19\), graph the solution on the number line, and write the solution in interval notation.

This figure shows the inequality x is greater than negative 19/4. Below this inequality is a number line ranging from negative 7 to negative 3, with tick marks for each integer. The inequality x is greater than negative 19/4 is graphed on the number line, with an open parenthesis at x equals negative 19/4 (written in), and a dark line extending to the right of the parenthesis. The inequality is also written in interval notation as parenthesis, negative 19/4 comma infinity, parenthesis.

Exercise \(\PageIndex{25}\)

Solve the inequality \(8p+3(p−12)>7p−28\) graph the solution on the number line, and write the solution in interval notation.

Exercise \(\PageIndex{26}\)

Solve the inequality \(9y+2(y+6)>5y−24\), graph the solution on the number line, and write the solution in interval notation.

This figure shows the inequality y is greater than negative 6. Below this inequality is a number line ranging from negative 7 to negative 3 with tick marks for each integer. The inequality y is greater than negative 6 is graphed on the number line, with an open parenthesis at y equals negative 6, and a dark line extending to the right of the parenthesis. The inequality is also written in interval notation as parenthesis, negative 6 comma infinity, parenthesis.

Exercise \(\PageIndex{27}\)

Solve the inequality \(6u+8(u−1)>10u+32\), graph the solution on the number line, and write the solution in interval notation.

This figure shows the inequality u is greater than 10. Below this inequality is a number line ranging from 9 to 13 with tick marks for each integer. The inequality u is greater than 10 is graphed on the number line, with an open parenthesis at u equals 10, and a dark line extending to the right of the parenthesis. The inequality is also written in interval notation as parenthesis, 10 comma infinity, parenthesis.

Just like some equations are identities and some are contradictions, inequalities may be identities or contradictions, too. We recognize these forms when we are left with only constants as we solve the inequality. If the result is a true statement, we have an identity. If the result is a false statement, we have a contradiction.

Exercise \(\PageIndex{28}\)

Solve the inequality \(8x−2(5−x)<4(x+9)+6x\), graph the solution on the number line, and write the solution in interval notation.

Exercise \(\PageIndex{29}\)

Solve the inequality \(4b−3(3−b)>5(b−6)+2b\), graph the solution on the number line, and write the solution in interval notation.

This figure shows an inequality that is an identity. Below this inequality is a number line ranging from negative 2 to 2 with tick marks for each integer. The identity is graphed on the number line, with a dark line extending in both directions. The inequality is also written in interval notation as parenthesis, negative infinity comma infinity, parenthesis.

Exercise \(\PageIndex{30}\)

Solve the inequality \(9h−7(2−h)<8(h+11)+8h\), graph the solution on the number line, and write the solution in interval notation.

This figure shows an inequality that is an identity. Below this inequality is a number line ranging from negative 2 to 2 with tick marks for each integer. The identity is graphed on the number line, with a dark line extending in both directions. The inequality is also written in interval notation as parenthesis, negative infinity comma infinity, parenthesis.

Exercise \(\PageIndex{31}\)

Solve the inequality \(\frac{1}{3}a - \frac{1}{8}a > \frac{5}{24}a + \frac{3}{4}\), graph the solution on the number line, and write the solution in interval notation.

Exercise \(\PageIndex{32}\)

Solve the inequality \(\frac{1}{4}x - \frac{1}{12}x > \frac{1}{6}x + \frac{7}{8}\), graph the solution on the number line, and write the solution in interval notation.

This figure shows an inequality that is a contradiction. Below this is a number line ranging from negative 2 to 2 with tick marks for each integer. No inequality is graphed on the number line. Below the number line is the statement: “No solution.”

Exercise \(\PageIndex{33}\)

Solve the inequality \(\frac{2}{5}z - \frac{1}{3}z < \frac{1}{15}z - \frac{3}{5}\), graph the solution on the number line, and write the solution in interval notation.

This figure shows an inequality that is a contradiction. Below this is a number line ranging from negative 2 to 2 with tick marks for each integer. No inequality is graphed on the number line. Below the number line is the statement: “No solution.”

Translate to an Inequality and Solve

To translate English sentences into inequalities, we need to recognize the phrases that indicate the inequality. Some words are easy, like ‘more than’ and ‘less than’. But others are not as obvious.

Think about the phrase ‘at least’ – what does it mean to be ‘at least 21 years old’? It means 21 or more. The phrase ‘at least’ is the same as ‘greater than or equal to’.

Table \(\PageIndex{4}\) ​ shows some common phrases that indicate inequalities.

Exercise \(\PageIndex{34}\)

Translate and solve. Then write the solution in interval notation and graph on the number line.

Twelve times c is no more than 96.

Exercise \(\PageIndex{35}\)

Twenty times y is at most 100

This figure shows the inequality 20y is less than or equal to 100, and then its solution: y is less than or equal to 5. Below this inequality is a number line ranging from 4 to 8 with tick marks for each integer. The inequality y is less than or equal to 5 is graphed on the number line, with an open bracket at y equals 5, and a dark line extending to the left of the bracket. The inequality is also written in interval notation as parenthesis, negative infinity comma 5, bracket.

Exercise \(\PageIndex{36}\)

Nine times z is no less than 135

This figure shows the inequality 9z is greater than or equal to 135, and then its solution: z is greater than or equal to 15. Below this inequality is a number line ranging from 14 to 18 with tick marks for each integer. The inequality z is greater than or equal to 15 is graphed on the number line, with an open bracket at z equals 15, and a dark line extending to the right of the bracket. The inequality is also written in interval notation as bracket, 15 comma infinity, parenthesis.

Exercise \(\PageIndex{37}\)

Thirty less than x is at least 45.

Exercise \(\PageIndex{38}\)

Nineteen less than p is no less than 47

This figure shows the inequality p minus 19 is greater than or equal to 47, and then its solution: p is greater than or equal to 66. Below this inequality is a number line ranging from 65 to 69 with tick marks for each integer. The inequality p is greater than or equal to 66 is graphed on the number line, with an open bracket at p equals 66, and a dark line extending to the right of the bracket. The inequality is also written in interval notation as bracket, 66 comma infinity, parenthesis.

Exercise \(\PageIndex{39}\)

Four more than a is at most 15.

This figure shows the inequality a plus 4 is less than or equal to 15, and then its solution: a is less than or equal to 11. Below this inequality is a number line ranging from 10 to 14 with tick marks for each integer. The inequality a is less than or equal to 11 is graphed on the number line, with an open bracket at a equals 11, and a dark line extending to the left of the bracket. The inequality is also written in interval notation as parenthesis, negative infinity 11, bracket.

Key Concepts

  • Subtraction Property of Inequality For any numbers a, b, and c, if a<b then a−c<b−c and if a>b then a−c>b−c.
  • Addition Property of Inequality For any numbers a, b, and c, if a<b then a+c<b+c and if a>b then a+c>b+c.
  • Division and Multiplication Properties of Inequalit y For any numbers a, b, and c, if a<b and c>0, then ac<bc and ac>bc. if a>b and c>0, then ac>bc and ac>bc. if a<b and c<0, then ac>bc and ac>bc. if a>b and c<0, then ac<bc and ac<bc.

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Solving Linear Inequalities Worksheets with Solutions

Solving linear inequalities is a foundational skill that you will build throughout your studies of elementary and high school algebra. While it isn’t the hardest concept you will encounter in the entire class, there are a few tips and tricks you should get used to. And the best way to do that is by practicing! 

That’s why I put together this collection of solving linear inequalities worksheets! Let’s dig in so that you can master this important algebra concept!

What is a Linear Inequality?

A linear inequality is a type of statement in algebra that compares two linear equations. To make this comparison, we place an inequality symbol in between the two sides of the inequality. For example, \(x+1>2x-1\).

Take a moment to review this inequality symbol list:

  • < (less than)
  • > (greater than)
  • ≤ (less than or equal to)
  • ≥ (greater than or equal to)

Linear inequalities are similar to linear equations, however instead of an equals sign, a linear inequality contains one of the inequality symbols from the above list.

How to Solve a Linear Inequality

The steps that you take to solve a linear inequality are a very similar to those that you would take while  solving an algebraic equation . When solving linear inequalities, you often perform similar operations as you would when  solving linear equations . 

Just like in linear equations, you can add or subtract terms on both sides, and multiply or divide by constants on both sides. However, unlike solving linear equations, linear inequalities have one  very  important rule:

Remember: If you multiply or divide both sides of an inequality by a negative number, the inequality sign must be flipped! 

You will see some examples of this in the solving linear inequalities worksheets linked below!

Another important difference between solving a linear equation and solving a linear inequality is that the solution to a linear inequality will be the  set of all values  that make the inequality true (rather than just a  single value  that makes the equation true).

The solution to a linear inequality will tell you the places on the graph where one line is greater than or less than the other.

For example, consider the coordinate plane below that shows two lines in slope-intercept form. 

the graph of x+1 shown in red, the graph of 2x-1 shown in green

As you can see, the red line is above the green line as long as \(x<2\). After this, the green line is above the red line. So if we wanted to write the solution to \(x+1>2x-1\), we would say that the solution set is \(x<2\), or all x-values less than 2 (note that we do not include 2 in the solution set since that is the point where the lines are equal ).

Solving Algebraically

In order to solve \(x+1>2x-1\) algebraically, we will use the same algebra strategies that we use when solving linear equations. We will collect all x-terms on the left side of the equation, and all non-variable terms on the right side of the equation.

\begin{split} x+1&>2x-1 \\ \\ -x&>-2 \\ \\ x&<2 \end{split}

Since we had to divide both sides by -1, notice that we flipped the sign of the inequality symbol! Therefore, the solution to this inequality is x<2 (which we can confirm using the coordinate plane above).

Another way that we can represent this solution is on a number line . We draw a line to the left of 2 with a hollow circle at 2 to indicate that 2 is  not  part of our solution set. We would use a filled in dot to indicate a value that  is  part of the solution set.

the solution to two different linear inequalities represented on a number line

There are many ways to solve linear inequalities and represent their solutions. In the math worksheets that follow below, you will focus mainly on solving algebraically and representing your answer on a number line!

One-Step Inequalities Math Worksheet

To get started, try the set of one-step inequalities in the worksheet linked below. These problems will help you become familiar with solving a linear equality algebraically, and representing the solution on a number line. 

I have created this worksheet in PDF format for your convenience. Remember to check the answer key provided to make sure that you understand the basics of solving linear inequalities.

Download the PDF worksheet by clicking below!

Multi-Step Inequalities Math Worksheet

Now that you have had some practice with solving one-step inequality problems, use the following worksheet to try your hand at some more complex linear inequality problems. This worksheet contains two-step inequalities as well as some more difficult multi-step linear inequality problems! 

This worksheet is also provided in PDF format for ease of access. Remember to check the answer key to confirm your understanding of solving more complex linear inequalities.

Practice Solving Linear Inequalities

Inequalities may seem scary at first because of the inequality symbol, but as you have seen, they are actually solved in a very similar way to linear equations.

Use the math worksheets linked above to practice getting comfortable with solving a variety of linear inequality problems. Like any math concept, the more you practice the more familiar the concept will start to feel.

I hope these solving linear inequalities worksheets have helped you practice your inequality solving skills, as well as representing the solution set on a number line!

Ready to start applying your skills? Check out this linear inequalities word problems worksheet !

Did you find these solving linear inequalities worksheets helpful? Share this post and subscribe to Math By The Pixel on YouTube for more helpful mathematics content!

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  • 2.5 Solve Linear Inequalities
  • Introduction
  • 1.1 Use the Language of Algebra
  • 1.2 Integers
  • 1.3 Fractions
  • 1.4 Decimals
  • 1.5 Properties of Real Numbers
  • Key Concepts
  • Review Exercises
  • Practice Test
  • 2.1 Use a General Strategy to Solve Linear Equations
  • 2.2 Use a Problem Solving Strategy
  • 2.3 Solve a Formula for a Specific Variable
  • 2.4 Solve Mixture and Uniform Motion Applications
  • 2.6 Solve Compound Inequalities
  • 2.7 Solve Absolute Value Inequalities
  • 3.1 Graph Linear Equations in Two Variables
  • 3.2 Slope of a Line
  • 3.3 Find the Equation of a Line
  • 3.4 Graph Linear Inequalities in Two Variables
  • 3.5 Relations and Functions
  • 3.6 Graphs of Functions
  • 4.1 Solve Systems of Linear Equations with Two Variables
  • 4.2 Solve Applications with Systems of Equations
  • 4.3 Solve Mixture Applications with Systems of Equations
  • 4.4 Solve Systems of Equations with Three Variables
  • 4.5 Solve Systems of Equations Using Matrices
  • 4.6 Solve Systems of Equations Using Determinants
  • 4.7 Graphing Systems of Linear Inequalities
  • 5.1 Add and Subtract Polynomials
  • 5.2 Properties of Exponents and Scientific Notation
  • 5.3 Multiply Polynomials
  • 5.4 Dividing Polynomials
  • Introduction to Factoring
  • 6.1 Greatest Common Factor and Factor by Grouping
  • 6.2 Factor Trinomials
  • 6.3 Factor Special Products
  • 6.4 General Strategy for Factoring Polynomials
  • 6.5 Polynomial Equations
  • 7.1 Multiply and Divide Rational Expressions
  • 7.2 Add and Subtract Rational Expressions
  • 7.3 Simplify Complex Rational Expressions
  • 7.4 Solve Rational Equations
  • 7.5 Solve Applications with Rational Equations
  • 7.6 Solve Rational Inequalities
  • 8.1 Simplify Expressions with Roots
  • 8.2 Simplify Radical Expressions
  • 8.3 Simplify Rational Exponents
  • 8.4 Add, Subtract, and Multiply Radical Expressions
  • 8.5 Divide Radical Expressions
  • 8.6 Solve Radical Equations
  • 8.7 Use Radicals in Functions
  • 8.8 Use the Complex Number System
  • 9.1 Solve Quadratic Equations Using the Square Root Property
  • 9.2 Solve Quadratic Equations by Completing the Square
  • 9.3 Solve Quadratic Equations Using the Quadratic Formula
  • 9.4 Solve Equations in Quadratic Form
  • 9.5 Solve Applications of Quadratic Equations
  • 9.6 Graph Quadratic Functions Using Properties
  • 9.7 Graph Quadratic Functions Using Transformations
  • 9.8 Solve Quadratic Inequalities
  • 10.1 Finding Composite and Inverse Functions
  • 10.2 Evaluate and Graph Exponential Functions
  • 10.3 Evaluate and Graph Logarithmic Functions
  • 10.4 Use the Properties of Logarithms
  • 10.5 Solve Exponential and Logarithmic Equations
  • 11.1 Distance and Midpoint Formulas; Circles
  • 11.2 Parabolas
  • 11.3 Ellipses
  • 11.4 Hyperbolas
  • 11.5 Solve Systems of Nonlinear Equations
  • 12.1 Sequences
  • 12.2 Arithmetic Sequences
  • 12.3 Geometric Sequences and Series
  • 12.4 Binomial Theorem

Learning Objectives

By the end of this section, you will be able to:

  • Graph inequalities on the number line
  • Solve linear inequalities
  • Translate words to an inequality and solve
  • Solve applications with linear inequalities

Be Prepared 2.13

Before you get started, take this readiness quiz.

Translate from algebra to English: 15 > x . 15 > x . If you missed this problem, review Example 1.3 .

Be Prepared 2.14

Translate to an algebraic expression: 15 is less than x . If you missed this problem, review Example 1.8 .

Graph Inequalities on the Number Line

What number would make the inequality x > 3 x > 3 true? Are you thinking, “ x could be four”? That’s correct, but x could be 6, too, or 37, or even 3.001. Any number greater than three is a solution to the inequality x > 3 . x > 3 .

We show all the solutions to the inequality x > 3 x > 3 on the number line by shading in all the numbers to the right of three, to show that all numbers greater than three are solutions. Because the number three itself is not a solution, we put an open parenthesis at three.

We can also represent inequalities using interval notation . There is no upper end to the solution to this inequality. In interval notation, we express x > 3 x > 3 as ( 3 , ∞ ) . ( 3 , ∞ ) . The symbol ∞ ∞ is read as “ infinity .” It is not an actual number.

Figure 2.2 shows both the number line and the interval notation.

We use the left parenthesis symbol, (, to show that the endpoint of the inequality is not included. The left bracket symbol, [, shows that the endpoint is included.

The inequality x ≤ 1 x ≤ 1 means all numbers less than or equal to one. Here we need to show that one is a solution, too. We do that by putting a bracket at x = 1 . x = 1 . We then shade in all the numbers to the left of one, to show that all numbers less than one are solutions. See Figure 2.3 .

There is no lower end to those numbers. We write x ≤ 1 x ≤ 1 in interval notation as ( − ∞ , 1 ] . ( − ∞ , 1 ] . The symbol − ∞ − ∞ is read as “negative infinity.” Figure 2.3 shows both the number line and interval notation.

Inequalities, Number Lines, and Interval Notation

The notation for inequalities on a number line and in interval notation use the same symbols to express the endpoints of intervals.

Example 2.48

Graph each inequality on the number line and write in interval notation.

ⓐ x ≥ −3 x ≥ −3 ⓑ x < 2.5 x < 2.5 ⓒ x ≤ − 3 5 x ≤ − 3 5

Try It 2.95

Graph each inequality on the number line and write in interval notation: ⓐ x > 2 x > 2 ⓑ x ≤ −1.5 x ≤ −1.5 ⓒ x ≥ 3 4 . x ≥ 3 4 .

Try It 2.96

Graph each inequality on the number line and write in interval notation: ⓐ x ≤ −4 x ≤ −4 ⓑ x ≥ 0.5 x ≥ 0.5 ⓒ x < − 2 3 . x < − 2 3 .

What numbers are greater than two but less than five? Are you thinking say, 2.5 , 3 , 3 2 3 , 4 , 4.99 2.5 , 3 , 3 2 3 , 4 , 4.99 ? We can represent all the numbers between two and five with the inequality 2 < x < 5 . 2 < x < 5 . We can show 2 < x < 5 2 < x < 5 on the number line by shading all the numbers between two and five. Again, we use the parentheses to show the numbers two and five are not included. See Figure 2.4 .

Example 2.49

ⓐ −3 < x < 4 −3 < x < 4 ⓑ −6 ≤ x < −1 −6 ≤ x < −1 ⓒ 0 ≤ x ≤ 2.5 0 ≤ x ≤ 2.5

Try It 2.97

Graph each inequality on the number line and write in interval notation:

ⓐ −2 < x < 1 −2 < x < 1 ⓑ −5 ≤ x < −4 −5 ≤ x < −4 ⓒ 1 ≤ x ≤ 4.25 1 ≤ x ≤ 4.25

Try It 2.98

ⓐ −6 < x < 2 −6 < x < 2 ⓑ −3 ≤ x < −1 −3 ≤ x < −1 ⓒ 2.5 ≤ x ≤ 6 2.5 ≤ x ≤ 6

Solve Linear Inequalities

A linear inequality is much like a linear equation—but the equal sign is replaced with an inequality sign. A linear inequality is an inequality in one variable that can be written in one of the forms, a x + b < c , a x + b < c , a x + b ≤ c , a x + b ≤ c , a x + b > c , a x + b > c , or a x + b ≥ c . a x + b ≥ c .

Linear Inequality

A linear inequality is an inequality in one variable that can be written in one of the following forms where a , b , and c are real numbers and a ≠ 0 a ≠ 0 :

When we solved linear equations, we were able to use the properties of equality to add, subtract, multiply, or divide both sides and still keep the equality. Similar properties hold true for inequalities.

We can add or subtract the same quantity from both sides of an inequality and still keep the inequality. For example:

Notice that the inequality sign stayed the same.

This leads us to the Addition and Subtraction Properties of Inequality.

Addition and Subtraction Property of Inequality

For any numbers a , b , and c, if a < b , then a < b , then

For any numbers a , b , and c, if a > b , then a > b , then

We can add or subtract the same quantity from both sides of an inequality and still keep the inequality.

What happens to an inequality when we divide or multiply both sides by a constant?

Let’s first multiply and divide both sides by a positive number.

The inequality signs stayed the same.

Does the inequality stay the same when we divide or multiply by a negative number?

Notice that when we filled in the inequality signs, the inequality signs reversed their direction.

When we divide or multiply an inequality by a positive number, the inequality sign stays the same. When we divide or multiply an inequality by a negative number, the inequality sign reverses.

This gives us the Multiplication and Division Property of Inequality.

Multiplication and Division Property of Inequality

For any numbers a , b , and c ,

When we divide or multiply an inequality by a :

  • positive number, the inequality stays the same.
  • negative number, the inequality reverses.

Sometimes when solving an inequality, as in the next example, the variable ends upon the right. We can rewrite the inequality in reverse to get the variable to the left.

Think about it as “If Xander is taller than Andy, then Andy is shorter than Xander.”

Example 2.50

Solve each inequality. Graph the solution on the number line, and write the solution in interval notation.

ⓐ x − 3 8 ≤ 3 4 x − 3 8 ≤ 3 4 ⓑ 9 y < ​ ​ 54 9 y < ​ ​ 54 ⓒ −15 < 3 5 z −15 < 3 5 z

Try It 2.99

Solve each inequality, graph the solution on the number line, and write the solution in interval notation:

ⓐ p − 3 4 ≥ 1 6 p − 3 4 ≥ 1 6 ⓑ 9 c > 72 9 c > 72 ⓒ 24 ≤ 3 8 m 24 ≤ 3 8 m

Try It 2.100

ⓐ r − 1 3 ≤ 7 12 r − 1 3 ≤ 7 12 ⓑ 12 d ≤ ​ 60 12 d ≤ ​ 60 ⓒ −24 < 4 3 n −24 < 4 3 n

Be careful when you multiply or divide by a negative number—remember to reverse the inequality sign.

Example 2.51

Solve each inequality, graph the solution on the number line, and write the solution in interval notation.

ⓐ −13 m ≥ 65 −13 m ≥ 65 ⓑ n −2 ≥ 8 n −2 ≥ 8

Try It 2.101

ⓐ −8 q < 32 −8 q < 32 ⓑ k −12 ≤ 15 . k −12 ≤ 15 .

Try It 2.102

ⓐ −7 r ≤ ​ − 70 −7 r ≤ ​ − 70 ⓑ u −4 ≥ −16 . u −4 ≥ −16 .

Most inequalities will take more than one step to solve. We follow the same steps we used in the general strategy for solving linear equations, but make sure to pay close attention when we multiply or divide to isolate the variable.

Example 2.52

Solve the inequality 6 y ≤ 11 y + 17 , 6 y ≤ 11 y + 17 , graph the solution on the number line, and write the solution in interval notation.

Try It 2.103

Solve the inequality, graph the solution on the number line, and write the solution in interval notation: 3 q ≥ 7 q − 23 . 3 q ≥ 7 q − 23 .

Try It 2.104

Solve the inequality, graph the solution on the number line, and write the solution in interval notation: 6 x < 10 x + 19 . 6 x < 10 x + 19 .

When solving inequalities, it is usually easiest to collect the variables on the side where the coefficient of the variable is largest. This eliminates negative coefficients and so we don’t have to multiply or divide by a negative—which means we don’t have to remember to reverse the inequality sign.

Example 2.53

Solve the inequality 8 p + 3 ( p − 12 ) > 7 p − 28 , 8 p + 3 ( p − 12 ) > 7 p − 28 , graph the solution on the number line, and write the solution in interval notation.

Try It 2.105

Solve the inequality 9 y + 2 ( y + 6 ) > 5 y − 24 9 y + 2 ( y + 6 ) > 5 y − 24 , graph the solution on the number line, and write the solution in interval notation.

Try It 2.106

Solve the inequality 6 u + 8 ( u − 1 ) > 10 u + 32 6 u + 8 ( u − 1 ) > 10 u + 32 , graph the solution on the number line, and write the solution in interval notation.

Just like some equations are identities and some are contradictions, inequalities may be identities or contradictions, too. We recognize these forms when we are left with only constants as we solve the inequality. If the result is a true statement, we have an identity. If the result is a false statement, we have a contradiction.

Example 2.54

Solve the inequality 8 x − 2 ( 5 − x ) < 4 ( x + 9 ) + 6 x , 8 x − 2 ( 5 − x ) < 4 ( x + 9 ) + 6 x , graph the solution on the number line, and write the solution in interval notation.

Try It 2.107

Solve the inequality 4 b − 3 ( 3 − b ) > 5 ( b − 6 ) + 2 b 4 b − 3 ( 3 − b ) > 5 ( b − 6 ) + 2 b , graph the solution on the number line, and write the solution in interval notation.

Try It 2.108

Solve the inequality 9 h − 7 ( 2 − h ) < 8 ( h + 11 ) + 8 h 9 h − 7 ( 2 − h ) < 8 ( h + 11 ) + 8 h , graph the solution on the number line, and write the solution in interval notation.

We can clear fractions in inequalities much as we did in equations. Again, be careful with the signs when multiplying or dividing by a negative.

Example 2.55

Solve the inequality 1 3 a − 1 8 a > 5 24 a ​ + 3 4 , 1 3 a − 1 8 a > 5 24 a ​ + 3 4 , graph the solution on the number line, and write the solution in interval notation.

Try It 2.109

Solve the inequality 1 4 x − 1 12 x > 1 6 x + 7 8 1 4 x − 1 12 x > 1 6 x + 7 8 , graph the solution on the number line, and write the solution in interval notation.

Try It 2.110

Solve the inequality 2 5 z − 1 3 z < 1 15 z ​ − 3 5 2 5 z − 1 3 z < 1 15 z ​ − 3 5 , graph the solution on the number line, and write the solution in interval notation.

Translate to an Inequality and Solve

To translate English sentences into inequalities, we need to recognize the phrases that indicate the inequality. Some words are easy, like “more than” and “less than.” But others are not as obvious. Table 2.2 shows some common phrases that indicate inequalities.

Example 2.56

Translate and solve. Then graph the solution on the number line, and write the solution in interval notation.

Try It 2.111

Nineteen less than p is no less than 47.

Try It 2.112

Four more than a is at most 15.

Solve Applications with Linear Inequalities

Many real-life situations require us to solve inequalities. The method we will use to solve applications with linear inequalities is very much like the one we used when we solved applications with equations.

We will read the problem and make sure all the words are understood. Next, we will identify what we are looking for and assign a variable to represent it. We will restate the problem in one sentence to make it easy to translate into an inequality. Then, we will solve the inequality.

Sometimes an application requires the solution to be a whole number, but the algebraic solution to the inequality is not a whole number. In that case, we must round the algebraic solution to a whole number. The context of the application will determine whether we round up or down.

Example 2.57

Dawn won a mini-grant of $4,000 to buy tablet computers for her classroom. The tablets she would like to buy cost $254.12 each, including tax and delivery. What is the maximum number of tablets Dawn can buy?

Try It 2.113

Angie has $20 to spend on juice boxes for her son’s preschool picnic. Each pack of juice boxes costs $2.63. What is the maximum number of packs she can buy?

Try It 2.114

Daniel wants to surprise his girlfriend with a birthday party at her favorite restaurant. It will cost $42.75 per person for dinner, including tip and tax. His budget for the party is $500. What is the maximum number of people Daniel can have at the party?

Example 2.58

Taleisha’s phone plan costs her $28.80 a month plus $0.20 per text message. How many text messages can she send/receive and keep her monthly phone bill no more than $50?

Try It 2.115

Sergio and Lizeth have a very tight vacation budget. They plan to rent a car from a company that charges $75 a week plus $0.25 a mile. How many miles can they travel during the week and still keep within their $200 budget?

Try It 2.116

Rameen’s heating bill is $5.42 per month plus $1.08 per therm. How many therms can Rameen use if he wants his heating bill to be a maximum of $87.50.

Profit is the money that remains when the costs have been subtracted from the revenue. In the next example, we will find the number of jobs a small businesswoman needs to do every month in order to make a certain amount of profit.

Example 2.59

Felicity has a calligraphy business. She charges $2.50 per wedding invitation. Her monthly expenses are $650. How many invitations must she write to earn a profit of at least $2,800 per month?

Try It 2.117

Caleb has a pet sitting business. He charges $32 per hour. His monthly expenses are $2,272. How many hours must he work in order to earn a profit of at least $800 per month?

Try It 2.118

Elliot has a landscape maintenance business. His monthly expenses are $1,100. If he charges $60 per job, how many jobs must he do to earn a profit of at least $4,000 a month?

There are many situations in which several quantities contribute to the total expense. We must make sure to account for all the individual expenses when we solve problems like this.

Example 2.60

Malik is planning a six-day summer vacation trip. He has $840 in savings, and he earns $45 per hour for tutoring. The trip will cost him $525 for airfare, $780 for food and sightseeing, and $95 per night for the hotel. How many hours must he tutor to have enough money to pay for the trip?

Try It 2.119

Brenda’s best friend is having a destination wedding and the event will last three days. Brenda has $500 in savings and can earn $15 an hour babysitting. She expects to pay $350 airfare, $375 for food and entertainment and $60 a night for her share of a hotel room. How many hours must she babysit to have enough money to pay for the trip?

Try It 2.120

Josue wants to go on a 10-night road trip with friends next spring. It will cost him $180 for gas, $450 for food, and $49 per night to share a motel room. He has $520 in savings and can earn $30 per driveway shoveling snow. How many driveways must he shovel to have enough money to pay for the trip?

Practice Makes Perfect

In the following exercises, graph each inequality on the number line and write in interval notation.

ⓐ x < −2 x < −2 ⓑ x ≥ −3.5 x ≥ −3.5 ⓒ x ≤ 2 3 x ≤ 2 3

ⓐ x > 3 x > 3 ⓑ x ≤ −0.5 x ≤ −0.5 ⓒ x ≥ 1 3 x ≥ 1 3

ⓐ x ≥ −4 x ≥ −4 ⓑ x < 2.5 x < 2.5 ⓒ x > − 3 2 x > − 3 2

ⓐ x ≤ 5 x ≤ 5 ⓑ x ≥ −1.5 x ≥ −1.5 ⓒ x < − 7 3 x < − 7 3

ⓐ −5 < x < 2 −5 < x < 2 ⓑ −3 ≤ x < 1 −3 ≤ x < 1 ⓒ 0 ≤ x ≤ 1.5 0 ≤ x ≤ 1.5

ⓐ −2 < x < 0 −2 < x < 0 ⓑ −5 ≤ x < −3 −5 ≤ x < −3 ⓒ 0 ≤ x ≤ 3.5 0 ≤ x ≤ 3.5

ⓐ −1 < x < 3 −1 < x < 3 ⓑ −3 < x ≤ −2 −3 < x ≤ −2 ⓒ −1.25 ≤ x ≤ 0 −1.25 ≤ x ≤ 0

ⓐ −4 < x < 2 −4 < x < 2 ⓑ −5 < x ≤ −2 −5 < x ≤ −2 ⓒ −3.75 ≤ x ≤ 0 −3.75 ≤ x ≤ 0

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

ⓐ a + 3 4 ≥ 7 10 a + 3 4 ≥ 7 10 ⓑ 8 x > 72 8 x > 72 ⓒ 20 > 2 5 h 20 > 2 5 h

ⓐ b + 7 8 ≥ 1 6 b + 7 8 ≥ 1 6 ⓑ 6 y < 48 6 y < 48 ⓒ 40 < 5 8 k 40 < 5 8 k

ⓐ f − 13 20 < − 5 12 f − 13 20 < − 5 12 ⓑ 9 t ≥ −27 9 t ≥ −27 ⓒ 7 6 j ≥ 42 7 6 j ≥ 42

ⓐ g − 11 12 < − 5 18 g − 11 12 < − 5 18 ⓑ 7 s < −28 7 s < −28 ⓒ 9 4 g ≤ 36 9 4 g ≤ 36

ⓐ −5 u ≥ 65 −5 u ≥ 65 ⓑ a −3 ≤ 9 a −3 ≤ 9

ⓐ −8 v ≤ 96 −8 v ≤ 96 ⓑ b −10 ≥ 30 b −10 ≥ 30

ⓐ −9 c < 126 −9 c < 126 ⓑ −25 < p −5 −25 < p −5

ⓐ −7 d > 105 −7 d > 105 ⓑ −18 > q −6 −18 > q −6

4 v ≥ 9 v − 40 4 v ≥ 9 v − 40

5 u ≤ 8 u − 21 5 u ≤ 8 u − 21

13 q < 7 q − 29 13 q < 7 q − 29

9 p > 14 p − 18 9 p > 14 p − 18

12 x + 3 ( x + 7 ) > 10 x − 24 12 x + 3 ( x + 7 ) > 10 x − 24

9 y + 5 ( y + 3 ) < 4 y − 35 9 y + 5 ( y + 3 ) < 4 y − 35

6 h − 4 ( h − 1 ) ≤ 7 h − 11 6 h − 4 ( h − 1 ) ≤ 7 h − 11

4 k − ( k − 2 ) ≥ 7 k − 26 4 k − ( k − 2 ) ≥ 7 k − 26

8 m − 2 ( 14 − m ) ≥ ​ 7 ( m − 4 ) + 3 m 8 m − 2 ( 14 − m ) ≥ ​ 7 ( m − 4 ) + 3 m

6 n − 12 ( 3 − n ) ≤ 9 ( n − 4 ) + 9 n 6 n − 12 ( 3 − n ) ≤ 9 ( n − 4 ) + 9 n

3 4 b − 1 3 b < 5 12 b − 1 2 3 4 b − 1 3 b < 5 12 b − 1 2

9 u + 5 ( 2 u − 5 ) ≥ 12 ( u − 1 ) + 7 u 9 u + 5 ( 2 u − 5 ) ≥ 12 ( u − 1 ) + 7 u

2 3 g − 1 2 ( g − 14 ) ≤ 1 6 ( g + 42 ) 2 3 g − 1 2 ( g − 14 ) ≤ 1 6 ( g + 42 )

4 5 h − 2 3 ( h − 9 ) ≥ 1 15 ( 2 h + 90 ) 4 5 h − 2 3 ( h − 9 ) ≥ 1 15 ( 2 h + 90 )

5 6 a − 1 4 a > 7 12 a + 2 3 5 6 a − 1 4 a > 7 12 a + 2 3

12 v + 3 ( 4 v − 1 ) ≤ 19 ( v − 2 ) + 5 v 12 v + 3 ( 4 v − 1 ) ≤ 19 ( v − 2 ) + 5 v

15 k ≤ −40 15 k ≤ −40

35 k ≥ −77 35 k ≥ −77

23 p − 2 ( 6 − 5 p ) > 3 ( 11 p − 4 ) 23 p − 2 ( 6 − 5 p ) > 3 ( 11 p − 4 )

18 q − 4 ( 10 − 3 q ) < 5 ( 6 q − 8 ) 18 q − 4 ( 10 − 3 q ) < 5 ( 6 q − 8 )

− 9 4 x ≥ − 5 12 − 9 4 x ≥ − 5 12

− 21 8 y ≤ − 15 28 − 21 8 y ≤ − 15 28

c + 34 < −99 c + 34 < −99

d + 29 > −61 d + 29 > −61

m 18 ≥ −4 m 18 ≥ −4

n 13 ≤ −6 n 13 ≤ −6

In the following exercises, translate and solve. Then graph the solution on the number line and write the solution in interval notation.

Three more than h is no less than 25.

Six more than k exceeds 25.

Ten less than w is at least 39.

Twelve less than x is no less than 21.

Negative five times r is no more than 95.

Negative two times s is lower than 56.

Nineteen less than b is at most −22 . −22 .

Fifteen less than a is at least −7 . −7 .

In the following exercises, solve.

Alan is loading a pallet with boxes that each weighs 45 pounds. The pallet can safely support no more than 900 pounds. How many boxes can he safely load onto the pallet?

The elevator in Yehire’s apartment building has a sign that says the maximum weight is 2100 pounds. If the average weight of one person is 150 pounds, how many people can safely ride the elevator?

Andre is looking at apartments with three of his friends. They want the monthly rent to be no more than $2,360. If the roommates split the rent evenly among the four of them, what is the maximum rent each will pay?

Arleen got a $20 gift card for the coffee shop. Her favorite iced drink costs $3.79. What is the maximum number of drinks she can buy with the gift card?

Teegan likes to play golf. He has budgeted $60 next month for the driving range. It costs him $10.55 for a bucket of balls each time he goes. What is the maximum number of times he can go to the driving range next month?

Ryan charges his neighbors $17.50 to wash their car. How many cars must he wash next summer if his goal is to earn at least $1,500?

Keshad gets paid $2,400 per month plus 6% of his sales. His brother earns $3,300 per month. For what amount of total sales will Keshad’s monthly pay be higher than his brother’s monthly pay?

Kimuyen needs to earn $4,150 per month in order to pay all her expenses. Her job pays her $3,475 per month plus 4% of her total sales. What is the minimum Kimuyen’s total sales must be in order for her to pay all her expenses?

Andre has been offered an entry-level job. The company offered him $48,000 per year plus 3.5% of his total sales. Andre knows that the average pay for this job is $62,000. What would Andre’s total sales need to be for his pay to be at least as high as the average pay for this job?

Nataly is considering two job offers. The first job would pay her $83,000 per year. The second would pay her $66,500 plus 15% of her total sales. What would her total sales need to be for her salary on the second offer be higher than the first?

Jake’s water bill is $24.80 per month plus $2.20 per ccf (hundred cubic feet) of water. What is the maximum number of ccf Jake can use if he wants his bill to be no more than $60?

Kiyoshi’s phone plan costs $17.50 per month plus $0.15 per text message. What is the maximum number of text messages Kiyoshi can use so the phone bill is no more than $56.60?

Marlon’s TV plan costs $49.99 per month plus $5.49 per first-run movie. How many first-run movies can he watch if he wants to keep his monthly bill to be a maximum of $100?

Kellen wants to rent a banquet room in a restaurant for her cousin’s baby shower. The restaurant charges $350 for the banquet room plus $32.50 per person for lunch. How many people can Kellen have at the shower if she wants the maximum cost to be $1,500?

Moshde runs a hairstyling business from her house. She charges $45 for a haircut and style. Her monthly expenses are $960. She wants to be able to put at least $1,200 per month into her savings account order to open her own salon. How many “cut & styles” must she do to save at least $1,200 per month?

Noe installs and configures software on home computers. He charges $125 per job. His monthly expenses are $1,600. How many jobs must he work in order to make a profit of at least $2,400?

Katherine is a personal chef. She charges $115 per four-person meal. Her monthly expenses are $3,150. How many four-person meals must she sell in order to make a profit of at least $1,900?

Melissa makes necklaces and sells them online. She charges $88 per necklace. Her monthly expenses are $3,745. How many necklaces must she sell if she wants to make a profit of at least $1,650?

Five student government officers want to go to the state convention. It will cost them $110 for registration, $375 for transportation and food, and $42 per person for the hotel. There is $450 budgeted for the convention in the student government savings account. They can earn the rest of the money they need by having a car wash. If they charge $5 per car, how many cars must they wash in order to have enough money to pay for the trip?

Cesar is planning a four-day trip to visit his friend at a college in another state. It will cost him $198 for airfare, $56 for local transportation, and $45 per day for food. He has $189 in savings and can earn $35 for each lawn he mows. How many lawns must he mow to have enough money to pay for the trip?

Alonzo works as a car detailer. He charges $175 per car. He is planning to move out of his parents’ house and rent his first apartment. He will need to pay $120 for application fees, $950 for security deposit, and first and last months’ rent at $1,140 per month. He has $1,810 in savings. How many cars must he detail to have enough money to rent the apartment?

Eun-Kyung works as a tutor and earns $60 per hour. She has $792 in savings. She is planning an anniversary party for her parents. She would like to invite 40 guests. The party will cost her $1,520 for food and drinks and $150 for the photographer. She will also have a favor for each of the guests, and each favor will cost $7.50. How many hours must she tutor to have enough money for the party?

Everyday Math

Maximum load on a stage In 2014, a high school stage collapsed in Fullerton, California, when 250 students got on stage for the finale of a musical production. Two dozen students were injured. The stage could support a maximum of 12,750 pounds. If the average weight of a student is assumed to be 140 pounds, what is the maximum number of students who could safely be on the stage?

Maximum weight on a boat In 2004, a water taxi sank in Baltimore harbor and five people drowned. The water taxi had a maximum capacity of 3,500 pounds (25 people with average weight 140 pounds). The average weight of the 25 people on the water taxi when it sank was 168 pounds per person. What should the maximum number of people of this weight have been?

Wedding budget Adele and Walter found the perfect venue for their wedding reception. The cost is $9850 for up to 100 guests, plus $38 for each additional guest. How many guests can attend if Adele and Walter want the total cost to be no more than $12,500?

Shower budget Penny is planning a baby shower for her daughter-in-law. The restaurant charges $950 for up to 25 guests, plus $31.95 for each additional guest. How many guests can attend if Penny wants the total cost to be no more than $1,500?

Writing Exercises

Explain why it is necessary to reverse the inequality when solving −5 x > 10 . −5 x > 10 .

Explain why it is necessary to reverse the inequality when solving n −3 < 12 . n −3 < 12 .

Find your last month’s phone bill and the hourly salary you are paid at your job. Calculate the number of hours of work it would take you to earn at least enough money to pay your phone bill by writing an appropriate inequality and then solving it. Do you feel this is an appropriate number of hours? Is this the appropriate phone plan for you?

Find out how many units you have left, after this term, to achieve your college goal and estimate the number of units you can take each term in college. Calculate the number of terms it will take you to achieve your college goal by writing an appropriate inequality and then solving it. Is this an acceptable number of terms until you meet your goal? What are some ways you could accelerate this process?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

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Unit 2: Solving equations & inequalities

About this unit, linear equations with variables on both sides.

  • Why we do the same thing to both sides: Variable on both sides (Opens a modal)
  • Intro to equations with variables on both sides (Opens a modal)
  • Equations with variables on both sides: 20-7x=6x-6 (Opens a modal)
  • Equation with variables on both sides: fractions (Opens a modal)
  • Equation with the variable in the denominator (Opens a modal)
  • Equations with variables on both sides Get 3 of 4 questions to level up!
  • Equations with variables on both sides: decimals & fractions Get 3 of 4 questions to level up!

Linear equations with parentheses

  • Equations with parentheses (Opens a modal)
  • Reasoning with linear equations (Opens a modal)
  • Multi-step equations review (Opens a modal)
  • Equations with parentheses Get 3 of 4 questions to level up!
  • Equations with parentheses: decimals & fractions Get 3 of 4 questions to level up!
  • Reasoning with linear equations Get 3 of 4 questions to level up!

Analyzing the number of solutions to linear equations

  • Number of solutions to equations (Opens a modal)
  • Worked example: number of solutions to equations (Opens a modal)
  • Creating an equation with no solutions (Opens a modal)
  • Creating an equation with infinitely many solutions (Opens a modal)
  • Number of solutions to equations Get 3 of 4 questions to level up!
  • Number of solutions to equations challenge Get 3 of 4 questions to level up!

Linear equations with unknown coefficients

  • Linear equations with unknown coefficients (Opens a modal)
  • Why is algebra important to learn? (Opens a modal)
  • Linear equations with unknown coefficients Get 3 of 4 questions to level up!

Multi-step inequalities

  • Inequalities with variables on both sides (Opens a modal)
  • Inequalities with variables on both sides (with parentheses) (Opens a modal)
  • Multi-step inequalities (Opens a modal)
  • Using inequalities to solve problems (Opens a modal)
  • Multi-step linear inequalities Get 3 of 4 questions to level up!
  • Using inequalities to solve problems Get 3 of 4 questions to level up!

Compound inequalities

  • Compound inequalities: OR (Opens a modal)
  • Compound inequalities: AND (Opens a modal)
  • A compound inequality with no solution (Opens a modal)
  • Double inequalities (Opens a modal)
  • Compound inequalities examples (Opens a modal)
  • Compound inequalities review (Opens a modal)
  • Solving equations & inequalities: FAQ (Opens a modal)
  • Compound inequalities Get 3 of 4 questions to level up!

linear inequalities problem solving with solution

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Examples of Solving Harder Linear Inequalities

Intro & Formatting Worked Examples Harder Examples & Word Prob's

Once you'd learned how to solve one-variable linear equations, you were then given word problems. To solve these problems, you'd have to figure out a linear equation that modelled the situation, and then you'd have to solve that equation to find the answer to the word problem.

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Solving Linear Inequalities on MathHelp.com

Solving Inequalities

So, now that you know how to solve linear inequalities — you guessed it! — they give you word problems.

  • The velocity of an object fired directly upward is given by V = 80 − 32 t , where the time t is measured in seconds. When will the velocity be between 32 and 64 feet per second (inclusive)?

This question is asking when the velocity, V , will be between two given values. So I'll take the expression for the velocity,, and put it between the two values they've given me. They've specified that the interval of velocities is inclusive, which means that the interval endpoints are included. Mathematically, this means that the inequality for this model will be an "or equal to" inequality. Because the solution is a bracket (that is, the solution is within an interval), I'll need to set up a three-part (that is, a compound) inequality.

I will set up the compound inequality, and then solve for the range of times t :

32 ≤ 80 − 32 t ≤ 64

32 − 80 ≤ 80 − 80 − 32 t ≤ 64 − 80

−48 ≤ −32 t ≤ −16

−48 / −32 ≥ −32 t / −32 ≥ −16 / −32

1.5 ≥ t ≥ 0.5

Note that, since I had to divide through by a negative, I had to flip the inequality signs.

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Note also that you might (as I do) find the above answer to be more easily understood if it's written the other way around, with "less than" inequalities.

And, because this is a (sort of) real world problem, my working should show the fractions, but my answer should probably be converted to decimal form, because it's more natural to say "one and a half seconds" than it is to say "three-halves seconds". So I convert the last line above to the following:

0.5 ≤ t ≤ 1.5

Looking back at the original question, it did not ask for the value of the variable " t ", but asked for the times when the velocity was between certain values. So the actual answer is:

The velocity will be between 32 and 64 feet per second between 0.5 seconds after launch and 1.5 seconds after launch.

Okay; my answer above was *extremely* verbose and "complete"; you don't likely need to be so extreme. You can probably safely get away with saying something simpler like, "between 0.5 seconds and 1.5 seconds". Just make sure that you do indeed include the approprioate units (in this case, "seconds").

Always remember when doing word problems, that, once you've found the value for the variable, you need to go back and re-read the problem to make sure that you're answering the actual question. The inequality 0.5 ≤  t  ≤ 1.5 did not answer the actual question regarding time. I had to interpret the inequality and express the values in terms of the original question.

  • Solve 5 x + 7 < 3( x  + 1) .

First I'll multiply through on the right-hand side, and then solve as usual:

5 x + 7 < 3( x + 1)

5 x + 7 < 3 x + 3

2 x + 7 < 3

2 x < −4

x < −2

In solving this inequality, I divided through by a positive 2 to get the final answer; as a result (that is, because I did *not* divide through by a minus), I didn't have to flip the inequality sign.

  • You want to invest $30,000 . Part of this will be invested in a stable 5% -simple-interest account. The remainder will be "invested" in your father's business, and he says that he'll pay you back with 7% interest. Your father knows that you're making these investments in order to pay your child's college tuition with the interest income. What is the least you can "invest" with your father, and still (assuming he really pays you back) get at least $1900 in interest?

First, I have to set up equations for this. The interest formula for simple interest is I = Prt , where I is the interest, P is the beginning principal, r is the interest rate expressed as a decimal, and t is the time in years.

Since no time-frame is specified for this problem, I'll assume that t  = 1 ; that is, I'll assume (hope) that he's promising to pay me at the end of one year. I'll let x be the amount that I'm going to "invest" with my father. Then the rest of my money, being however much is left after whatever I give to him, will be represented by "the total, less what I've already given him", so 30000 −  x will be left to invest in the safe account.

Then the interest on the business investment, assuming that I get paid back, will be:

I = ( x )(0.07)(1) = 0.07 x

The interest on the safe investment will be:

(30 000 − x )(0.05)(1) = 1500 − 0.05 x

The total interest is the sum of what is earned on each of the two separate investments, so my expression for the total interest is:

0.07 x + (1500 − 0.05 x ) = 0.02 x + 1500

I need to get at least $1900 ; that is, the sum of the two investments' interest must be greater than, or at least equal to, $1,900 . This allows me to create my inequality:

0.02 x + 1500 ≥ 1900

0.02 x ≥ 400

x ≥ 20 000

That is, I will need to "invest" at least $20,000 with my father in order to get $1,900 in interest income. Since I want to give him as little money as possible, I will give him the minimum amount:

I will invest $20,000 at 7% .

Algebra Tutors

  • An alloy needs to contain between 46% copper and 50% copper. Find the least and greatest amounts of a 60% copper alloy that should be mixed with a 40% copper alloy in order to end up with thirty pounds of an alloy containing an allowable percentage of copper.

This is similar to a mixture word problem , except that this will involve inequality symbols rather than "equals" signs. I'll set it up the same way, though, starting with picking a variable for the unknown that I'm seeking. I will use x to stand for the pounds of 60% copper alloy that I need to use. Then 30 −  x will be the number of pounds, out of total of thirty pounds needed, that will come from the 40% alloy.

Of course, I'll remember to convert the percentages to decimal form for doing the algebra.

How did I get those values in the bottom right-hand box? I multiplied the total number of pounds in the mixture ( 30 ) by the minimum and maximum percentages ( 46% and 50% , respectively). That is, I multiplied across the bottom row, just as I did in the " 60% alloy" row and the " 40% alloy" row, to get the right-hand column's value.

The total amount of copper in the mixture will be the sum of the copper contributed by each of the two alloys that are being put into the mixture. So I'll add the expressions for the amount of copper from each of the alloys, and place the expression for the total amount of copper in the mixture as being between the minimum and the maximum allowable amounts of copper:

13.8 ≤ 0.6 x + (12 − 0.4 x ) ≤ 15

13.8 ≤ 0.2 x + 12 ≤ 15

1.8 ≤ 0.2 x ≤ 3

9 ≤ x ≤ 15

Checking back to my set-up, I see that I chose my variable to stand for the number of pounds that I need to use of the 60% copper alloy. And they'd only asked me for this amount, so I can ignore the other alloy in my answer.

I will need to use between 9 and 15 pounds of the 60% alloy.

Per yoozh, I'm verbose in my answer. You can answer simply as " between 9 and 15 pounds ".

  • Solve 3( x − 2) + 4 ≥ 2(2 x − 3)

First I'll multiply through and simplify; then I'll solve:

3( x − 2) + 4 ≥ 2(2 x − 3)

3 x − 6 + 4 ≥ 4 x − 6

3 x − 2 ≥ 4 x − 6

−2 ≥ x − 6            (*)

Why did I move the 3 x over to the right-hand side (to get to the line marked with a star), instead of moving the 4 x to the left-hand side? Because by moving the smaller term, I was able to avoid having a negative coefficient on the variable, and therefore I was able to avoid having to remember to flip the inequality when I divided through by that negative coefficient. I find it simpler to work this way; I make fewer errors. But it's just a matter of taste; you do what works for you.

Why did I switch the inequality in the last line and put the variable on the left? Because I'm more comfortable with inequalities when the answers are formatted this way. Again, it's only a matter of taste. The form of the answer in the previous line, 4 ≥ x , is perfectly acceptable.

As long as you remember to flip the inequality sign when you multiply or divide through by a negative, you shouldn't have any trouble with solving linear inequalities.

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  • -x+3\gt 2x+1
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  • -17<3+10x\le 33
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  • How do I solve linear inequalities?
  • To solve linear inequalities, isolate the variable on one side of the inequality, keeping track of the sign of the inequality when multiplying or dividing by a negative number, and express the solution as an interval.
  • What is a linear inequality?
  • A linear inequality is a first degree equation with an inequality sign
  • What is a linear expression?
  • A linear expression is a mathematical expression that consists of a constant or a variable raised to the first power, multiplied by a constant coefficient and added or subtracted together.

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Chapter 1: Sets

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Solving Linear Inequalities Word Problems

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We are well versed with equations in multiple variables. Linear Equations represent a point in a single dimension, a line in a two-dimensional, and a plane in a three-dimensional world. Solutions to linear inequalities represent a region of the Cartesian plane. It becomes essential for us to know how to translate real-life problems into linear inequalities. 

Linear Inequalities 

Before defining the linear inequalities formally, let’s see them through a real-life situation and observe why their need arises in the first place. Let’s say Albert went to buy some novels for himself at the book fair. He has a total of Rs 200 with him. The book fair has a special sale policy which offers any book at Rs 70. Now he knows that he may not be able to spend the full amount on the books. Let’s say x is the number of books he bought. This situation can be represented mathematically by the following equation, 

70x < 200

Since he can’t spend all the amount on books, and also the amount spent by him will always be less than Rs 200. The present situation can only be represented by the equation given above. Now let’s study the linear inequalities with a formal description, 

Two real numbers or two algebraic expressions which are related by symbols such as ‘>’, ‘<‘, ‘≥’ and’≤’ form the inequalities. Linear inequalities are formed by linear equations which are connected with these symbols. These inequalities can be further classified into two parts:  Strict Inequalities: The inequalities with the symbols such as ‘>’ or ‘<‘. Slack Inequalities: The inequalities with the symbols such as ‘≥’ or ‘≤’.

Rules of Solving Linear Inequalities:

There are certain rules which we should keep in mind while solving linear inequalities. 

  • Equal numbers can be added or subtracted from both sides of the inequality without affecting its sign.
  • Both sides of Inequality can be divided or multiplied by any positive number but when they are multiplied or divided by a negative number, the sign of the linear inequality is reversed.

Now with this brief introduction to linear inequalities, let’s see some word problems on this concept. 

Sample Problems

Question 1: Considering the problem given in the beginning. Albert went to buy some novels for himself at the book fair. He has a total of Rs 200 with him. The book fair has a special sale policy which offers any book at Rs 70. Now he knows that he may not be able to spend the full amount on the books. Let’s say x is the number of books he bought. Represent this situation mathematically and graphically. 

Solution: 

We know that Albert cannot buy books for all the money he has. So, let’s say the number of books he buys is “x”. Then,  70x < 200 ⇒ x <  To plot the graph of this inequality, put x = 0.  0 <   Thus, x = 0 satisfies the inequality. So, the graph for the following inequality will look like, 

Question 2: Consider the performance of the strikers of the football club Real Madrid in the last 3 matches. Ronaldo and Benzema together scored less than 9 goals in the last three matches. It is also known that Ronaldo scored three more goals than Benzema. What can be the possible number of goals Ronaldo scored? 

Let’s say the number of goals scored by Benzema and Ronaldo are y and x respectively.  x = y + 3 …..(1) x + y < 9  …..(2)  Substituting the value of x from equation (1) in equation (2).  y + 3 + y < 9  ⇒2y  < 6  ⇒y < 3 Possible values of y: 0,1,2  Possible values of x: 3,4,5

Question 3: A classroom can fit at least 9 tables with an area of a one-meter square. We know that the perimeter of the classroom is 12m. Find the bounds on the length and breadth of the classroom. 

It can fit 9 tables, that means the area of the classroom is atleast 9m 2 . Let’s say the length of the classroom is x and breadth is y meters.  2(x + y) = 12 {Perimeter of the classroom} ⇒ x + y = 6  Area of the rectangle is given by,  xy > 9  ⇒x(6 – x) > 9  ⇒6x – x 2 > 9  ⇒ 0 > x 2 – 6x + 9  ⇒ 0 > (x – 3) 2 ⇒ 0 > x – 3  ⇒ x < 3  Thus, length of the classroom must be less than 3 m.  So, then the breadth of the classroom will be greater than 3 m. 

Question 4: Formulate the linear inequality for the following situation and plot its graph. 

Let’s say Aman and Akhil went to a stationery shop. Aman bought 3 notebooks and Akhil bought 4 books. Let’s say cost of each notebook was “x” and each book was “y”. The total expenditure was less than Rs 500. 

Cost of each notebook was “x” and for each book, it was “y”. Then the inequality can be described as, 3x + 4y < 500 Putting (x,y) → (0,0)  3(0) + 4(0) < 500 Origin satisfies the inequality. Thus, the graph of its solutions will look like, x.

Question 5: Formulate the linear inequality for the following situation and plot its graph. 

A music store sells its guitars at five times their cost price. Find the shopkeeper’s minimum cost price if his profit is more than Rs 3000. 

Let’s say the selling price of the guitar is y, and the cost price is x. y – x > 3000 ….(1) It is also given that,  y = 5x ….(2)  Substituting the value of y from equation (2) to equation (1).  5x – x > 3000  ⇒ 4x > 3000  ⇒  x >  ⇒ x> 750  Thus, the cost price must be greater than Rs 750. 

Question 6: The length of the rectangle is 4 times its breadth. The perimeter of the rectangle is less than 20. Formulate a linear inequality in two variables for the given situation, plot its graph and calculate the bounds for both length and breadth. 

Let’s say the length is “x” and breadth is “y”.  Perimeter = 2(x + y) < 20 ….(1) ⇒ x + y < 10  Given : x = 4y  Substituting the value of x in equation (1).  x + y < 10 ⇒ 5y < 10  ⇒ y < 2 So, x < 8 and y < 2. 

Question 7: Rahul and Rinkesh play in the same football team. In the previous game, Rahul scored 2 more goals than Rinkesh, but together they scored less than 8 goals. Solve the linear inequality and plot this on a graph.

The equations obtained from the given information in the question, Suppose Rahul scored x Number of goals and Rinkesh scored y number of goals, Equations obtained will be, x = y+2 ⇢ (1) x+y< 8 ⇢ (2) Solving both the equations, y+2 + y < 8 2y < 6 y< 3 Putting this value in equation (2), x< 5

Question 8: In a class of 100 students, there are more girls than boys, Can it be concluded that how many girls would be there?

Let’s suppose that B is denoted for boys and G is denoted for girls. Now, since Girls present in the class are more than boys, it can be written in equation form as, G > B  The total number of students present in class is 100 (given), It can be written as, G+ B= 100 B = 100- G Substitute G> B in the equation formed, G> 100 – G 2G > 100 G> 50 Hence, it is fixed that the number of girls has to be more than 50 in class, it can be 60, 65, etc. Basically any number greater than 50 and less than 100.

Question 9: In the previous question, is it possible for the number of girls to be exactly 50 or exactly 100? If No, then why?

No, It is not possible for the Number of girls to be exactly 50 since while solving, it was obtained that, G> 50  In any case if G= 50 is a possibility, from equation G+ B= 100, B = 50 will be obtained. This simply means that the number of boys is equal to the number of girls which contradicts to what is given in the question. No, it is not possible for G to be exactly 100 as well, as this proves that there are 0 boys in the class.

Question 10: Solve the linear inequality and plot the graph for the same,

7x+ 8y < 30

The linear inequality is given as, 7x+ 8y< 30  At x= 0, y= 30/8= 3.75 At y= 0, x= 30/7= 4.28 These values are the intercepts. The graph for the above shall look like,  Putting x= y/2, that is, y= 2x in the linear inequality, 7x + 16x < 30 x = 1.304 y = 2.609

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NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities

Ncert solutions class 11 maths linear inequalities – free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 5.

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities are curated by the highly qualified expert teachers at BYJU’S. These NCERT Solutions of Maths help students in solving problems efficiently. The solutions for NCERT textbook questions contain steps with appropriate formulas and examples. All these NCERT solutions are written as per the latest update of the CBSE Syllabus 2023-24 to help students in scoring full marks.

The PDF of Class 11 Maths NCERT Solutions for Chapter 6 Linear Inequalities, serves as a detailed study material for students. BYJU’S designs all the concepts and solutions in simple words so that students can easily learn the chapter in less time. Also, the solutions of BYJU’S are accurate, and students need not worry about the quality and correctness of the solutions. The knowledge of fundamental concepts helps in gaining knowledge and remembering more effectively for their upcoming exams. Linear Inequalities are used in many real-life applications such as income and expenditure problems, to find the proportion of the amount spent on various things. Two types of linear inequalities are explained in NCERT Solutions for Class 11 Maths of Chapter 6, i.e., linear inequalities in one variable and linear inequalities in two variables.

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NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities

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Access answers of NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities

Exercise 6.1 Page No: 122

1. Solve 24x < 100, when

(i) x is a natural number.

(ii) x is an integer.

(i) Given that 24x < 100

Now we have to divide the inequality by 24 then we get x < 25/6

Now when x is a natural integer then

It is clear that the only natural number less than 25/6 are 1, 2, 3, 4.

Thus, 1, 2, 3, 4 will be the solution of the given inequality when x is a natural number.

Hence {1, 2, 3, 4} is the solution set.

(ii) Given that 24x < 100

now when x is an integer then

It is clear that the integer number less than 25/6 are…-1, 0, 1, 2, 3, 4.

Thus, solution of 24 x < 100 are…,-1, 0, 1, 2, 3, 4, when x is an integer.

Hence {…, -1, 0, 1, 2, 3, 4} is the solution set.

2. Solve – 12x > 30, when

(i) Given that, – 12x > 30

Now, by dividing the inequality by -12 on both sides we get, x < -5/2

When x is a natural integer then

It is clear that there is no natural number less than -2/5 because -5/2 is a negative number and natural numbers are positive numbers.

Therefore there would be no solution of the given inequality when x is a natural number.

(ii) Given that, – 12x > 30

Now by dividing the inequality by -12 on both sides we get, x < -5/2

When x is an integer then

It is clear that the integer number less than -5/2 are…, -5, -4, – 3

Thus, solution of – 12x > 30 is …,-5, -4, -3, when x is an integer.

Therefore the solution set is {…, -5, -4, -3}

3. Solve 5x – 3 < 7, when

(i) x is an integer

(ii) x is a real number

(i) Given that, 5x – 3 < 7

Now by adding 3 on both sides, we get,

5x – 3 + 3 < 7 + 3

The above inequality becomes

Again, by dividing both sides by 5 we get,

5x/5 < 10/5

When x is an integer, then

It is clear that that the integer number less than 2 are…, -2, -1, 0, 1.

Thus, solution of 5x – 3 < 7 is …,-2, -1, 0, 1, when x is an integer.

Therefore the solution set is {…, -2, -1, 0, 1}

(ii) Given that, 5x – 3 < 7

Above inequality becomes

Again, by dividing both sides by 5, we get,

When x is a real number, then

It is clear that the solutions of 5x – 3 < 7 will be given by x < 2 which states that all the real numbers that are less than 2.

Hence the solution set is x ∈ (-∞, 2)

4. Solve 3x + 8 >2, when

(i) x is an integer .

(ii) x is a real number.

(i) Given that, 3x + 8 > 2

Now by subtracting 8 from both sides, we get,

3x + 8 – 8 > 2 – 8

The above inequality becomes,

3x > – 6

Again by dividing both sides by 3, we get,

3x/3 > -6/3

Hence x > -2

It is clear that the integer numbers greater than -2 are -1, 0, 1, 2,…

Thus, solution of 3x + 8 > 2 is -1, 0, 1, 2,… when x is an integer.

Hence the solution set is {-1, 0, 1, 2,…}

(ii) Given that, 3x + 8 > 2

Now by subtracting 8 from both sides we get,

Again, by dividing both sides by 3, we get,

When x is a real number.

It is clear that the solutions of 3x + 8 >2 will be given by x > -2 which means all the real numbers that are greater than -2.

Therefore the solution set is x ∈ (-2, ∞)

Solve the inequalities in Exercises 5 to 16 for real x.

5. 4x + 3 < 5x + 7

Given that, 4x + 3 < 5x + 7

Now by subtracting 7 from both the sides, we get

4x + 3 – 7 < 5x + 7 – 7

4x – 4 < 5x

Again, by subtracting 4x from both the sides,

4x – 4 – 4x < 5x – 4x

∴The solutions of the given inequality are defined by all the real numbers greater than -4.

The required solution set is (-4, ∞)

6. 3x – 7 > 5x – 1

Given that,

3x – 7 > 5x – 1

Now, by adding 7 to both the sides, we get

3x – 7 +7 > 5x – 1 + 7

3x > 5x + 6

Again, by subtracting 5x from both the sides,

3x – 5x > 5x + 6 – 5x

Dividing both sides by -2 to simplify, we get

-2x/-2 < 6/-2

∴ The solutions of the given inequality are defined by all the real numbers less than -3.

Hence the required solution set is (-∞, -3)

7. 3(x – 1) ≤ 2 (x – 3)

Given that, 3(x – 1) ≤ 2 (x – 3)

By multiplying, the above inequality can be written as

3x – 3 ≤ 2x – 6

Now, by adding 3 to both the sides, we get

3x – 3+ 3 ≤ 2x – 6+ 3

3x ≤ 2x – 3

Again, by subtracting 2x from both the sides,

3x – 2x ≤ 2x – 3 – 2x

Therefore, the solutions of the given inequality are defined by all the real numbers less than or equal to -3.

Hence, the required solution set is (-∞, -3]

8. 3 (2 – x) ≥ 2 (1 – x)

Given that, 3 (2 – x) ≥ 2 (1 – x)

By multiplying, we get

6 – 3x ≥ 2 – 2x

Now, by adding 2x to both the sides,

6 – 3x + 2x ≥ 2 – 2x + 2x

Again, by subtracting 6 from both the sides, we get

6 – x – 6 ≥ 2 – 6

Multiplying throughout inequality by negative sign, we get

∴ The solutions of the given inequality are defined by all the real numbers greater than or equal to 4.

Hence the required solution set is (- ∞, 4]

9. x + x/2 + x/3 < 11

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 1

The solutions of the given inequality are defined by all the real numbers less than 6.

Hence the solution set is (-∞, 6)

10. x/3 > x/2 + 1

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 3

– x/6 > 1

∴ The solutions of the given inequality are defined by all the real numbers less than – 6.

Hence, the required solution set is (-∞, -6)

11. 3(x – 2)/5 ≤ 5 (2 – x)/3

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 4

Now by cross-multiplying the denominators, we get

9(x- 2) ≤ 25 (2 – x)

9x – 18 ≤ 50 – 25x

Now adding 25x both the sides,

9x – 18 + 25x ≤ 50 – 25x + 25x

34x – 18 ≤ 50

Adding 25x both the sides,

34x – 18 + 18 ≤ 50 + 18

Dividing both sides by 34,

34x/34 ≤ 68/34

The solutions of the given inequality are defined by all the real numbers less than or equal to 2.

Required solution set is (-∞, 2]

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 5

∴ The solutions of the given inequality are defined by all the real numbers less than or equal to 120.

Thus, (-∞, 120] is the required solution set.

13. 2 (2x + 3) – 10 < 6 (x – 2)

2 (2x + 3) – 10 < 6 (x – 2)

4x + 6 – 10 < 6x – 12

On simplifying, we get

4x – 4 < 6x – 12

4x – 6x < -12 + 4

-2x < -8

Dividing by 2, we get;

Multiply by “-1” and change the sign.

∴ The solutions of the given inequality are defined by all the real numbers greater than 4.

Hence, the required solution set is (4, ∞).

14. 37 – (3x + 5) ≥ 9x – 8 (x – 3)

Given that, 37 – (3x + 5) ≥ 9x – 8 (x – 3)

= 37 – 3x – 5 ≥ 9x – 8x + 24

= 32 – 3x ≥ x + 24

On rearranging,

= 32 – 24 ≥ x + 3x

All the real numbers of x which are less than or equal to 2 are the solutions of the given inequality

Hence, (-∞, 2] will be the solution for the given inequality

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 8

= 15x < 4 (4x – 1)

= 15x < 16x – 4

All the real numbers of x which are greater than 4 are the solutions of the given inequality

Hence, (4, ∞) will be the solution for the given inequality

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 10

= 20 (2x – 1) ≥ 3 (19x – 18)

= 40x – 20 ≥ 57x – 54

= – 20 + 54 ≥ 57x – 40x

∴ All the real numbers of x which are less than or equal to 2 are the solutions of the given inequality

Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line.

17. 3x – 2 < 2x + 1

3x – 2 < 2x + 1

Solving the given inequality, we get

= 3x – 2x < 1 + 2

Now, the graphical representation of the solution is as follows:

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 13

18. 5x – 3 ≥ 3x – 5

5x – 3 ≥ 3x – 5

On rearranging, we get

= 5x – 3x ≥ -5 + 3

On simplifying,

Now, dividing by 2 on both sides, we get

The graphical representation of the solution is as follows:

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 14

19. 3 (1 – x) < 2 (x + 4)

3 (1 – x) < 2 (x + 4)

Multiplying, we get

= 3 – 3x < 2x + 8

= 3 – 8 < 2x + 3x

= – 5 < 5x

Now by dividing 5 on both sides, we get

-5/5 < 5x/5

= – 1 < x

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 15

On computing we get

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 19

= 15x ≥ 2 (4x – 1)

= 15x ≥ 8x -2

= 15x -8x ≥ 8x -2 -8x

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 21

21. Ravi obtained 70 and 75 marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.

Let us assume that x is the marks obtained by Ravi in his third unit test.

According to the question, all the students should have an average of at least 60 marks

(70 + 75 + x)/3 ≥ 60

= 145 + x ≥ 180

= x ≥ 180 – 145

Hence, all the students must obtain 35 marks in order to have an average of at least 60 marks

22. To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in the first four examinations are 87, 92, 94 and 95, find the minimum marks that Sunita must obtain in the fifth examination to get Grade ‘A’ in the course.

Let us assume Sunita scored x marks in her fifth examination

Now, according to the question, in order to receive A grade in the course, she must obtain an average of 90 marks or more in her five examinations

(87 + 92 + 94 + 95 + x)/5 ≥ 90

= (368 + x)/5 ≥ 90

= 368 + x ≥ 450

= x ≥ 450 – 368

Hence, she must obtain 82 or more marks in her fifth examination

23. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Let us assume x to be the smaller of the two consecutive odd positive integers.

∴ The other integer is = x + 2

It is also given in the question that both the integers are smaller than 10.

∴ x + 2 < 10

x < 8 … (i)

Also, it is given in the question that the sum of two integers is more than 11.

∴ x + (x + 2) > 11

2x + 2 > 11

x > 4.5 … (ii)

Thus, from (i) and (ii), we have,

x is an odd integer and it can take values 5 and 7.

Hence, possible pairs are (5, 7) and (7, 9)

24. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.

Let us assume x is the smaller of the two consecutive even positive integers.

∴ The other integer = x + 2

It is also given in the question that both the integers are larger than 5.

∴ x > 5   ….(i)

Also, it is given in the question that the sum of two integers is less than 23.

∴ x + (x + 2) < 23

2x + 2 < 23

x < 21/2

x < 10.5   … (ii)

Thus, from (i) and (ii) we have x is an even number and it can take values 6, 8 and 10.

Hence, possible pairs are (6, 8), (8, 10) and (10, 12).

25. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

Let us assume the length of the shortest side of the triangle to be x cm.

∴ According to the question, the length of the longest side = 3x cm

And, length of third side = (3x – 2) cm

As, the least perimeter of the triangle = 61 cm

Thus, x + 3x + (3x – 2) cm ≥ 61 cm

= 7x – 2 ≥ 61

Now dividing by 7, we get

= 7x/7 ≥ 63/7

Hence, the minimum length of the shortest side will be 9 cm.

26. A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second?

Let us assume the length of the shortest piece to be x cm

∴ According to the question, length of the second piece = (x + 3) cm

And, length of third piece = 2x cm

As all the three lengths are to be cut from a single piece of board having a length of 91 cm

∴ x + (x + 3) + 2x ≤ 91 cm

= 4x + 3 ≤ 91

= 4x/4 ≤ 88/4

= x ≤ 22 … (i)

Also, it is given in the question that, the third piece is at least 5 cm longer than the second piece.

∴ 2x ≥ (x+3) + 5

x ≥ 8 … (ii)

Thus, from equation (i) and (ii), we have:

Hence, it is clear that the length of the shortest board is greater than or equal to 8 cm and less than or equal to 22 cm.

Exercise 6.2 Page No: 127

Solve the following inequalities graphically in two-dimensional plane:

1. x + y < 5

Given x + y < 5

Now, draw a dotted line x + y = 5 in the graph (∵ x + y = 5 is excluded in the given question)

Now, consider x + y < 5

Select a point (0, 0)

⇒ 0 + 0 < 5

⇒ 0 < 5 (this is true)

∴ Solution region of the given inequality is below the line x + y = 5. (i.e., origin is included in the region)

The graph is as follows:

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 22

2. 2x + y ≥ 6

Given 2x + y ≥ 6

Now, draw a solid line 2x + y = 6 in the graph (∵2x + y = 6 is included in the given question)

Now, consider 2x + y ≥6

⇒ 2 × (0) + 0 ≥ 6

⇒ 0 ≥ 6 (this is false)

∴ Solution region of the given inequality is above the line 2x + y = 6. (away from the origin)

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 23

3. 3x + 4y ≤ 12

Given 3x + 4y ≤ 12

Now, draw a solid line 3x + 4y = 12 in the graph (∵3x + 4y = 12 is included in the given question)

Now, consider 3x + 4y ≤ 12

⇒ 3 × (0) + 4 × (0) ≤ 12

⇒ 0 ≤ 12 (this is true)

∴ Solution region of the given inequality is below the line 3x + 4y = 12. (i.e., origin is included in the region)

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 24

4. y + 8 ≥ 2x

Given y + 8 ≥ 2x

Now, draw a solid line y + 8 = 2x in the graph (∵y + 8 = 2x is included in the given question)

Now, consider y + 8 ≥ 2x

⇒ (0) + 8 ≥ 2 × (0)

⇒ 0≤ 8 (this is true)

∴ Solution region of the given inequality is above the line y + 8 = 2x. (i.e., origin is included in the region)

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 25

5. x – y ≤ 2

Given x – y ≤ 2

Now, draw a solid line x – y = 2 in the graph (∵ x – y = 2 is included in the given question).

Now, consider x – y ≤ 2

⇒ (0) – (0) ≤ 2

⇒ 0 ≤ 2 (this is true)

∴ Solution region of the given inequality is above the line x – y = 2. (i.e., origin is included in the region)

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 26

6. 2x – 3y > 6

Given 2x – 3y > 6

Now draw a dotted line 2x – 3y = 6 in the graph (∵2x – 3y = 6 is excluded in the given question)

Now Consider 2x – 3y > 6

⇒ 2 × (0) – 3 × (0) > 6

⇒ 0 > 6 (this is false)

∴ Solution region of the given inequality is below the line 2x – 3y > 6. (Away from the origin)

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 27

7. – 3x + 2y ≥ – 6

Given – 3x + 2y ≥ – 6

Now, draw a solid line – 3x + 2y = – 6 in the graph (∵– 3x + 2y = – 6 is included in the given question).

Now, consider – 3x + 2y ≥ – 6

⇒ – 3 × (0) + 2 × (0) ≥ – 6

⇒ 0 ≥ – 6 (this is true)

∴ Solution region of the given inequality is above the line – 3x + 2y ≥ – 6. (i.e., origin is included in the region)

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 28

8. y – 5x < 30

Given y – 5x < 30

Now, draw a dotted line 3y – 5x = 30 in the graph (∵3y – 5x = 30 is excluded in the given question)

Now, consider 3y – 5x < 30

⇒ 3 × (0) – 5 × (0) < 30

⇒ 0 < 30 (this is true)

∴ Solution region of the given inequality is below the line 3y – 5x < 30. (i.e., origin is included in the region)

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 29

9. y < – 2

Given y < – 2

Now, draw a dotted line y = – 2 in the graph (∵ y = – 2 is excluded in the given question)

Now, consider y < – 2

⇒ 0 < – 2 (this is false)

∴ Solution region of the given inequality is below the line y < – 2. (i.e., away from the origin)

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 30

10. x > – 3

Given x > – 3

Now, draw a dotted line x = – 3 in the graph (∵x = – 3 is excluded in the given question)

Now, consider x > – 3

⇒ 0 > – 3

⇒ 0 > – 3 (this is true)

∴ Solution region of the given inequality is right to the line x > – 3. (i.e., origin is included in the region)

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 31

Exercise 6.3 Page No: 129

Solve the following system of inequalities graphically:

1. x ≥ 3, y ≥ 2

Given x ≥ 3……… (i)

y ≥ 2…………… (ii)

Since x ≥ 3 means for any value of y the equation will be unaffected, so similarly for y ≥ 2, for any value of x the equation will be unaffected.

Now putting x = 0 in (i)

0 ≥ 3 which is not true

Putting y = 0 in (ii)

0 ≥ 2 which is not true again

This implies the origin doesn’t satisfy in the given inequalities. The region to be included will be on the right side of the two equalities drawn on the graphs.

The shaded region is the desired region.

ncert-sol-c11-maths-chapter-6-ex-3-img-1

2. 3x + 2y ≤ 12, x ≥ 1, y ≥ 2

Given 3x+ 2y ≤ 12

Solving for the value of x and y by putting x = 0 and y = 0 one by one, we get

y = 6 and x = 4

So the points are (0, 6) and (4, 0)

Now checking for (0, 0)

0 ≤ 12 which is also true.

Hence, the origin lies in the plane and the required area is toward the left of the equation.

Now checking for x ≥ 1, the value of x would be unaffected by any value of y.

The origin would not lie on the plane.

⇒ 0 ≥ 1 which is not true

The required area to be included would be on the left of the graph x ≥1

Similarly, for y ≥ 2

Value of y will be unaffected by any value of x in the given equality. Also, the origin doesn’t satisfy the given inequality.

⇒ 0 ≥ 2 which is not true. Hence origin is not included in the solution of the inequality.

The region to be included in the solution would be towards the left of the equality y≥ 2

The shaded region in the graph will give the answer to the required inequalities as it is the region which is covered by all the given three inequalities at the same time satisfying all the given conditions.

ncert-sol-c11-maths-chapter-6-ex-3-img-2

3. 2x + y ≥ 6, 3x + 4y ≤ 12

Given 2x + y ≥ 6…………… (i)

3x + 4y ≤ 12 ……………. (ii)

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 6 and x = 3

So the point for the (0, 6) and (3, 0)

0 ≥ 6 which is not true, hence the origin does not lie in the solution of the equality. The required region is on the right side of the graph.

Checking for 3x + 4y ≤ 12,

Putting value of x= 0 and y = 0 one by one in equation,

We get y = 3, x = 4

The points are (0, 3), (4, 0)

Now, checking for origin (0, 0)

0 ≤ 12 which is true,

So the origin lies in solution of the equation.

The region on the right of the equation is the region required.

The solution is the region which is common to the graphs of both the inequalities.

The shaded region is the required region.

ncert-sol-c11-maths-chapter-6-ex-3-img-3

4. x + y ≥ 4, 2x – y < 0

Given x + y ≥ 4

y = 4 and x = 4

The points for the line are (0, 4) and (4, 0)

Checking for the origin (0, 0)

This is not true,

So the origin would not lie in the solution area. The required region would be on the right of line’s graph.

2x – y < 0

y= 0 and x = 0

Putting x = 1 we get y = 2

So, the points for the given inequality are (0, 0) and (1, 2)

Now that the origin lies on the given equation, we will check for (4, 0) point to check which side of the line’s graph will be included in the solution.

⇒ 8 < 0 which is not true, hence the required region would be on the left side of the line 2x-y < 0

The shaded region is the required solution of the inequalities.

ncert-sol-c11-maths-chapter-6-ex-3-img-4

5. 2x – y >1, x – 2y < – 1

Given 2x – y >1……………… (i)

y = -1 and x = 1/2 = 0.5

The points are (0,-1) and (0.5, 0)

Checking for the origin, putting (0, 0)

0 >1, which is false

Hence the origin does not lie in the solution region. The required region would be on the right of the line`s graph.

x – 2y < – 1………… (ii)

y = ½ = 0.5 and x = -1

The required points are (0, 0.5) and (-1, 0)

Now checking for the origin, (0, 0)

0 < -1 which is false.

Hence, the origin does not lie in the solution area; the required area would be on the left side of the line’s graph.

∴ The shaded area is the required solution of the given inequalities.

ncert-sol-c11-maths-chapter-6-ex-3-img-5

6. x + y ≤ 6, x + y ≥ 4

Given x + y ≤ 6,

y = 6 and x = 6

The required points are (0, 6) and (6, 0)

Checking further for origin (0, 0)

We get 0 ≤ 6, this is true.

Hence the origin would be included in the area of the line’s graph. So, the required solution of the equation would be on the left side of the line graph which will be including origin.

The required points are (0, 4) and (4, 0)

0 ≥ 4 which is false

So, the origin would not be included in the required area. The solution area will be above the line graph or the area on the right of line graph.

Hence, the shaded area in the graph is required graph area.

ncert-sol-c11-maths-chapter-6-ex-3-img-6

7. 2x + y ≥ 8, x + 2y ≥ 10

Given 2x + y ≥ 8

y = 8 and x = 4

The required points are (0, 8) and (4, 0)

Checking if the origin is included in the line`s graph (0, 0)

0 ≥ 8, which is false.

Hence, the origin is not included in the solution area and the required area would be the area to the right of the line’s graph.

x + 2y ≥ 10

y = 5 and x = 10

The required points are (0, 5) and (10, 0)

0 ≥ 10 which is false,

Hence the origin would not lie in the required solution area. The required area would be to the left of the line graph.

The shaded area in the graph is the required solution of the given inequalities.

ncert-sol-c11-maths-chapter-6-ex-3-img-7

8. x + y ≤ 9, y > x, x ≥ 0

Given x + y ≤ 9,

y = 9 and x = 9

The required points are (0, 9) and (9, 0)

Checking if the origin is included in the line’s graph (0, 0)

Which is true. So, the required area would be including the origin and hence, will lie on the left side of the line`s graph.

Solving for y = x

We get x= 0, y = 0, so the origin lies on the line’s graph.

The other points would be (0, 0) and (2, 2)

Checking for (9, 0) in y > x,

We get 0 > 9 which is false, since the area would not include the area below the line’s graph and hence, would be on the left side of the line.

We have x ≥ 0

The area of the required line’s graph would be on the right side of the line’s graph.

Therefore, the shaded area is the required solution of the given inequalities.

ncert-sol-c11-maths-chapter-6-ex-3-img-8

9. 5x + 4y ≤ 20, x ≥ 1, y ≥ 2

Given 5x + 4y ≤ 20,

y = 5 and x= 4

The required points are (0, 5) and (4, 0)

Checking if the origin lies in the solution area (0, 0)

Which is true, hence the origin would lie in the solution area. The required area of the line’s graph is on the left side of the graph.

We have x ≥ 1,

For all the values of y, x would be 1,

The required points would be (1, 0), (1, 2) and so on.

Checking for origin (0, 0)

0 ≥ 1, which is not true.

So, the origin would not lie in the required area. The required area on the graph will be on the right side of the line’s graph.

Consider y ≥ 2

Similarly for all the values of x, y would be 2.

The required points would be (0, 2), (1, 2) and so on.

0 ≥ 2, this is not true.

Hence, the required area would be on the right side of the line’s graph.

The shaded area on the graph shows the required solution of the given inequalities.

ncert-sol-c11-maths-chapter-6-ex-3-img-9

10. 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0

Given 3x + 4y ≤ 60,

y = 15 and x = 20

The required points are (0, 15) and (20, 0)

Checking if the origin lies in the required solution area (0, 0)

0 ≤ 60, this is true.

Hence the origin would lie in the solution area of the line’s graph.

The required solution area would be on the left of the line’s graph.

We have x + 3y ≤ 30,

y = 10 and x = 30

The required points are (0, 10) and (30, 0).

0 ≤ 30, this is true.

Hence the origin lies in the solution area which is given by the left side of the line’s graph.

Consider x ≥ 0,

The given inequalities imply the solution lies in the first quadrant only.

Hence the solution of the inequalities is given by the shaded region in the graph.

ncert-sol-c11-maths-chapter-6-ex-3-img-10

11. 2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6

Given 2x + y ≥ 4,

y = 4 and x =2

The required points are (0, 4) and (2, 0)

0 ≥ 4, this is not true

Hence the origin doesn’t lie in the solution area of the line’s graph. The solution area would be given by the right side of the line’s graph.

y = 3 and x = 3

The required points are (0, 3) and (3, 0)

0 ≤ 3, this is true.

Hence the solution area would include the origin and hence, would be on the left side of the line’s graph.

2x – 3y ≤ 6

y = -2 and x = 3

The required points are (0, -2), (3, 0).

0 ≤ 6 this is true

So the origin lies in the solution area and the area would be on the left of the line’s graph.

Hence, the shaded area in the graph is the required solution area for the given inequalities.

ncert-sol-c11-maths-chapter-6-ex-3-img-11

12. x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0 , y ≥ 1

Given, x – 2y ≤ 3

Putting value of x = 0 and y = 0 in the equation one by one, we get value of

y = -3/2 = -1.5 and x = 3

The required points are (0, -1.5) and (3, 0)

Hence, the solution area would be on the left of the line’s graph

3x + 4y ≥ 12,

y = 3 and x = 4

The required points are (0, 3) and (4, 0)

0 ≥ 12, this is not true.

So, the solution area would include the origin and the required solution area would be on the right side of the line’s graph.

We have x ≥ 0,

For all the values of y, the value of x would be same in the given inequality, which would be the region above the x axis on the graph.

Consider, y ≥ 1

For all the values of x, the value of y would be same in the given inequality.

The solution area of the line would not include origin as 0 ≥ 1 is not true.

The solution area would be on the left side of the line’s graph.

The shaded area in the graph is the required solution area which satisfies all the given inequalities at the same time.

ncert-sol-c11-maths-chapter-6-ex-3-img-12

13. 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0

Given, 4x + 3y ≤ 60,

y = 20 and x = 15

The required points are (0, 20) and (15, 0).

Hence the origin would lie in the solution area. The required area would be on the left of the line’s graph.

We have y ≥ 2x,

y = 0 and x = 0

Hence the line would pass through origin.

To check which side would be included in the line’s graph solution area, we would check for point (15, 0)

⇒ 0 ≥ 15, this is not true. So the required solution area would be to the left of the line’s graph.

Consider, x ≥ 3,

For any value of y, the value of x would be same.

Also the origin (0, 0) doesn’t satisfiy the inequality as 0 ≥ 3.

So, the origin doesn’t lie in the solution area. Hence, the required solution area would be on the right of the line’s graph.

We have x, y ≥ 0

Since it is given both x and y are greater than 0

∴ the solution area would be in the first I st quadrant only.

The shaded area in the graph shows the solution area for the given inequalities.

ncert-sol-c11-maths-chapter-6-ex-3-img-13

14. 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0

Given, 3x + 2y ≤ 150

y = 75 and x = 50

The required points are (0, 75) and (50, 0).

0 ≤ 150, this is true.

Hence, the solution area for the line would be on the left side of the line’s graph, which would be including the origin too.

We have x + 4y ≤ 80,

y = 20 and x = 80

The required points are (0, 20) and (80, 0).

0 ≤ 80, this is also true. So, the origin lies in the solution area.

The required solution area would be toward the left of the line’s graph.

Given x ≤ 15,

For all the values of y, x would be same.

0 ≤ 15, this is true. So, the origin would be included in the solution area. The required solution area would be towards the left of the line’s graph.

Consider y ≥ 0, x ≥ 0

Since x and y are greater than 0, the solution would lie in the 1 st quadrant.

The shaded area in the graph satisfies all the given inequalities, and hence is the solution area for given inequalities.

ncert-sol-c11-maths-chapter-6-ex-3-img-14

15. x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0

Given, x + 2y ≤ 10,

The required points are (0, 5) and (10, 0).

0 ≤ 10, this is true.

Hence, the solution area would be toward origin including the same. The solution area would be toward the left of the line’s graph.

We have x + y ≥ 1,

y = 1 and x = 1

The required points are (0, 1) and (1, 0)

0 ≥ 1, this is not true.

Hence, the origin would not be included in the solution area. The required solution area would be toward the right of the line’s graph.

Consider x – y ≤ 0,

Hence, the origin would lie on the line.

To check which side of the line graph would be included in the solution area, we would check for the (10, 0)

10 ≤ 0, which is not true. Hence, the solution area would be on the left side of the line’s graph.

Again, we have x ≥ 0, y ≥ 0

Since both x and y are greater than 0, the solution area would be in the 1 st quadrant.

Hence, the solution area for the given inequalities would be the shaded area of the graph satisfying all the given inequalities.

ncert-sol-c11-maths-chapter-6-ex-3-img-15

Miscellaneous Exercise Page No: 129

Solve the inequalities in Exercises 1 to 6

1. 2 ≤ 3x – 4 ≤ 5

According to the question,

The inequality given is,

2 ≤ 3x – 4 ≤ 5

⇒ 2 ≤ 3x – 4 ≤ 5

⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4

⇒ 6 ≤ 3x ≤ 9

⇒ 6/3 ≤ 3x/3 ≤ 9/3

⇒ 2 ≤ x ≤ 3

Hence, all real numbers x greater than or equal to 2, but less than or equal to 3 are solutions of given equality.

2. 6 ≤ –3 (2x – 4) < 12

The inequality given is,

6 ≤ –3 (2x – 4) < 12

⇒ 6 ≤ -3 (2x – 4) < 12

Dividing the inequality by 3, we get.

⇒ 2 ≤ – (2x – 4) < 4

Multiplying the inequality by -1,

⇒ -2 ≥ 2x – 4 > -4 [multiplying the inequality with -1 changes the inequality sign.]

⇒ -2 + 4 ≥ 2x – 4 + 4 > -4 + 4

⇒ 2 ≥ 2x > 0

Dividing the inequality by 2,

⇒ 0 < x ≤ 1

Hence, all real numbers x greater than 0, but less than or equal to 1 are solutions of given equality.

3. – 3 ≤ 4 – 7x/2 ≤ 18

– 3 ≤ 4 – 7x/2 ≤ 18

⇒ – 3 – 4 ≤ 4 – 7x/2 – 4 ≤ 18 – 4

⇒ – 7 ≤ – 7x/2 ≤ 18 – 14

Multiplying the inequality by -2,

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 47

⇒ 14 ≥ 7x ≥ -28

⇒ -28 ≤ 7x ≤ 14

Dividing the inequality by 7,

⇒ -4 ≤ x ≤ 2

Hence, all real numbers x greater than or equal to -4, but less than or equal to 2 are solutions of given equality.

x ∈ [-4, 2]

4. – 15 ≤ 3(x – 2)/5 ≤ 0

– 15 ≤ 3(x – 2)/5 ≤ 0

⇒ – 15 < 3(x – 2)/5 ≤ 0

Multiplying the inequality by 5,

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 48

⇒ -75 < 3(x – 2) ≤ 0

Dividing the inequality by 3, we get,

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 49

⇒ -25 < x – 2 ≤ 0

⇒ – 25 + 2 < x – 2 + 2 ≤ 0 + 2

⇒ – 23 < x ≤ 2

Hence, all real numbers x greater than -23, but less than or equal to 2 are solutions of given equality.

x ∈ (-23, 2]

5. – 12 < 4 – 3x/ (-5) ≤ 2

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 50

Hence, all real numbers x greater than -80/3, but less than or equal to -10/3 are solutions of given equality.

x ∈ (-80/3, -10/3]

6. 7 ≤ (3x + 11)/2 ≤ 11

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 51

⇒ 14 ≤ 3x + 11 ≤ 22

⇒ 14 – 11 ≤ 3x + 11 – 11 ≤ 22 – 11

⇒ 3 ≤ 3x ≤ 11

⇒ 1 ≤ x ≤ 11/3

x ∈ [1, 11/3]

Solve the inequalities in Exercises 7 to 11 and represent the solution graphically on number line.

7. 5x + 1 > – 24, 5x – 1 < 24

The inequalities given are,

5x + 1 > -24 and 5x – 1 < 24

5x + 1 > -24

⇒ 5x > -24 – 1

⇒ 5x > -25

⇒ x > -5 ……… (i)

5x – 1 < 24

⇒ 5x < 24 + 1

⇒ 5x < 25

⇒ x < 5 ……….(ii)

From equations (i) and (ii),

We can infer that the solution of given inequalities is (-5, 5).

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 52

8. 2 (x – 1) < x + 5, 3 (x + 2) > 2 – x

2 (x – 1) < x + 5 and 3 (x + 2) > 2 – x

2 (x – 1) < x + 5

⇒ 2x – 2 < x + 5

⇒ 2x – x < 5 + 2

⇒ x < 7 ……… (i)

3 (x + 2) > 2 – x

⇒ 3x + 6 > 2 – x

⇒ 3x + x > 2 – 6

⇒ 4x > -4

⇒ x > -1 ………. (ii)

We can infer that the solution of given inequalities is (-1, 7).

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 53

9. 3x – 7 > 2(x – 6), 6 – x > 11 – 2x

3x – 7 > 2(x – 6) and 6 – x > 11 – 2x

3x – 7 > 2(x – 6)

⇒ 3x – 7 > 2x – 12

⇒ 3x – 2x > 7 – 12

⇒ x > -5 ………… (i)

6 – x > 11 – 2x

⇒ 2x – x > 11 – 6

⇒ x > 5 ……….(ii)

We can infer that the solution of given inequalities is (5, ∞).

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 54

10. 5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47

5(2x – 7) – 3(2x + 3) ≤ 0 and 2x + 19 ≤ 6x + 47

5(2x – 7) – 3(2x + 3) ≤ 0

⇒ 10x – 35 – 6x – 9 ≤ 0

⇒ 4x – 44 ≤ 0

⇒ x ≤ 11 ……(i)

2x + 19 ≤ 6x +47

⇒ 6x – 2x ≥ 19 – 47

⇒ x ≥ -7 ……….(ii)

We can infer that the solution of given inequalities is (-7, 11).

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 55

11. A solution is to be kept between 68° F and 77° F. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by F = (9/5) C + 32?

The solution has to be kept between 68° F and 77° F

So, we get, 68° < F < 77°

Substituting,

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 56

⇒ 20 < C < 25

Hence, we get,

The range of temperature in degree Celsius is between 20° C to 25° C.

12. A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4%, but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?

8% of solution of boric acid = 640 litres

Let the amount of 2% boric acid solution added = x litres

Then we have,

Total mixture = x + 640 litres

We know that,

The resulting mixture has to be more than 4% but less than 6% boric acid.

∴ 2% of x + 8% of 640 > 4% of (x + 640) and

2% of x + 8% of 640 < 6% of (x + 640)

2% of x + 8% of 640 > 4% of (x + 640)

⇒ (2/100) × x + (8/100) × 640 > (4/100) × (x + 640)

⇒ 2x + 5120 > 4x + 2560

⇒ 5120 – 2560 > 4x – 2x

⇒ 2560 > 2x

⇒ x < 1280 ….(i)

⇒ (2/100) × x + (8/100) × 640 < (6/100) × (x + 640)

⇒ 2x + 5120 < 6x + 3840

⇒ 6x – 2x > 5120 – 3840

⇒ 4x > 1280

⇒ x > 320 ……….(i)

From (i) and (ii)

320 < x < 1280

Therefore, the number of litres of 2% of boric acid solution that has to be added will be more than 320 litres but less than 1280 litres.

13. How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

45% of solution of acid = 1125 litres

Let the amount of water added = x litres

Resulting mixture = x + 1125 litres

The resulting mixture has to be more than 25% but less than 30% acid content.

Amount of acid in resulting mixture = 45% of 1125 litres.

∴ 45% of 1125 < 30% of (x + 1125) and 45% of 1125 > 25% of (x + 1125)

45% of 1125 < 30% of (x + 1125)

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 57

⇒ 45 × 1125 < 30x + 30 × 1125

⇒ (45 – 30) × 1125 < 30x

⇒ 15 × 1125 < 30x

⇒ x > 562.5 ………..(i)

45% of 1125 > 25% of (x + 1125)

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 58

⇒ 45 × 1125 > 25x + 25 × 1125

⇒ (45 – 25) × 1125 > 25x

⇒ 25x < 20 × 1125

⇒ x < 900 …..(ii)

∴ 562.5 < x < 900

Therefore, the number of litres of water that has to be added will have to be more than 562.5 litres but less than 900 litres.

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 59

Chronological age = CA = 12 years

IQ for age group of 12 is 80 ≤ IQ ≤ 140.

We get that,

80 ≤ IQ ≤ 140

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Image 60

⇒ 9.6 ≤ MA ≤ 16.8

∴ Range of mental age of the group of 12 year-old-children is 9.6 ≤ MA ≤ 16.8

NCERT Solutions for Class 11 Maths Chapter 6 – Linear Inequalities

The Chapter Linear Inequalities belongs to the unit Algebra of Class 11 Maths CBSE Syllabus for 2023-24, and there exist 3 exercises and a miscellaneous exercise in this chapter, which help the students understand the concepts related to the chapter in detail. The topics covered in Chapter 6 Linear Inequalities of NCERT Solutions for Class 11 include:

6.1 Introduction

Two algebraic expressions or real numbers related by any of the symbols ≤, ≥, <, and > form an inequality. For example: px + qy > 0, 3a –19b < 0. Here, students will be able to know how to solve word problems by converting them into inequalities.

6.2 Inequalities

This topic is well explained with real-life scenarios which can be transformed to linear inequalities. Also, enough practice problems are provided along with solved examples.

6.3 Algebraic Solutions of Linear Inequalities in 1 Variable and their Graphical Representation

In this exercise, students can learn the meaning of a solution of linear inequalities and the graphical representation of these solutions. Besides, the methods of finding the solutions for linear inequalities have been explained with examples.

6.4 Graphical Solution of Linear Inequalities in Two Variables

After this section, students can understand the representation of the solution of linear inequalities in two variables on the Cartesian plane. Also, they can identify the solution region for the given inequalities.

6.5 Solution of System of Linear Inequalities in Two Variables

The solution of the system of linear inequalities in two variables using graphical methods is explained with many examples to help the students to understand the concept in a better way.

Exercise 6.1 Solutions 26 Questions

Exercise 6.2 Solutions 10 Questions

Exercise 6.3 Solutions 15 Questions

Miscellaneous Exercise On Chapter 6 Solutions 14 Questions

The summary of the concepts involved in the NCERT Solutions for Class 11 Maths Chapter 6 of BYJU’S is given below:

  • Two real numbers or two algebraic expressions related by the symbols <, >, ≤ or ≥ form an inequality.
  • Equal numbers may be added to (or subtracted from) both sides of an inequality.
  • Both sides of an inequality can be multiplied (or divided) by the same positive number. But, when both sides are multiplied (or divided) by a negative number, then the inequality is reversed.
  • The values of x, which make an inequality a true statement, are called solutions of the inequality.
  • To represent x < a (or x > a) on a number line, put a circle on the number a and dark line to the left (or right) of the number a.
  • To represent x ≤ a (or x ≥ a) on a number line, put a dark circle on the number a and darken the line to the left (or right) of the number x.
  • If inequality is having ≤ or ≥ symbol, then the points on the line are also included in the solutions of the inequality, and the graph of the inequality lies to the left (below) or right (above) of the graph of the equality represented by a dark line that satisfies an arbitrary point in that part.
  • If an inequality is having < or > symbol, then the points on the line are not included in the solutions of the inequality, and the graph of the inequality lies to the left (below) or right (above) of the graph of the corresponding equality represented by a dotted line that satisfies an arbitrary point in that part.
  • The solution region of a system of inequalities is the region which satisfies all the given inequalities in the system simultaneously.

Key Features of NCERT Solutions for Class 11 Maths Chapter 6 – Linear Inequalities

The solutions for NCERT questions provided by BYJU’S for Class 11 Maths Chapter 6 have covered the below concepts. These solutions are prepared meticulously to avoid mistakes so that students are assured of getting full marks after practising them.

  • Meaning of Linear inequalities.
  • Algebraic solutions of linear inequalities in one variable and their representation on the number line.
  • Graphical solution of linear inequalities in two variables.
  • Graphical method of finding a solution of systems of linear inequalities in two variables.

Disclaimer – 

Dropped Topics – 

6.4 Graphical Solution of Linear Inequalities in Two Variables 6.5 Solution of System of Linear Inequalities in Two Variables Last three points in the Summary

Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 6

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  4. Solving Inequalities (video lessons, examples, solutions)

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  6. Lesson Video: Solving Systems of Linear Inequalities

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  1. Solving Linear Inequalities

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  3. 2.8 Linear Inequalities in Two Variables

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  5. SOLVING LINEAR INEQUALITIES

  6. 9.4 Linear Inequalities in Two Variables

COMMENTS

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    Home Bookshelves Algebra Elementary Algebra 1e (OpenStax) 2: Solving Linear Equations and Inequalities

  2. Algebra

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  3. Solving and graphing linear inequalities (video)

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    A linear inequality is a type of statement in algebra that compares two linear equations. To make this comparison, we place an inequality symbol in between the two sides of the inequality. For example, x + 1 > 2x − 1. Linear inequalities are similar to linear equations, however instead of an equals sign, a linear inequality contains one of ...

  6. Linear Inequalities

    To solve inequalities, we can follow the following steps: Step 1: We simplify the inequality if possible. This includes removing grouping signs such as parentheses, combining like terms, and removing fractions. Step 2: Solve for the variable.

  7. 2.5 Solve Linear Inequalities

    Solve applications with linear inequalities Be Prepared 2.13 Before you get started, take this readiness quiz. Translate from algebra to English: 15 > x. 15 > x. If you missed this problem, review Example 1.3. Be Prepared 2.14 Translate to an algebraic expression: 15 is less than x. If you missed this problem, review Example 1.8.

  8. Solving Linear Inequalities

    Solving Linear Inequalities Most of the rules or techniques involved in solving multi-step equations should easily translate to solving inequalities. The only big difference is how the inequality symbol switches direction when a negative number is multiplied or divided to both sides of an equation.

  9. Solving equations & inequalities

    Unit test About this unit There are lots of strategies we can use to solve equations. Let's explore some different ways to solve equations and inequalities. We'll also see what it takes for an equation to have no solution, or infinite solutions. Linear equations with variables on both sides Learn

  10. Linear Inequalities Questions

    Linear inequalities questions and solutions help the students to solve various problems on inequalities in mathematics. In this article, you will get solved questions on linear inequalities, and additional questions for practice. You will find detailed explanations to the solutions of linear inequalities in one variable and two variables.

  11. Harder linear inequalities & Word problems

    Mathematically, this means that the inequality for this model will be an "or equal to" inequality. Because the solution is a bracket (that is, the solution is within an interval), I'll need to set up a three-part (that is, a compound) inequality. I will set up the compound inequality, and then solve for the range of times t:

  12. Linear Inequalities Calculator

    To solve linear inequalities, isolate the variable on one side of the inequality, keeping track of the sign of the inequality when multiplying or dividing by a negative number, and express the solution as an interval. What is a linear inequality? A linear inequality is a first degree equation with an inequality sign What is a linear expression?

  13. Linear Inequalities

    When we solve linear inequality then we get an ordered pair. So basically, in a system, the solution to all inequalities and the graph of the linear inequality is the graph displaying all solutions of the system. Let us see an example to understand it. Example: Graph the Linear inequality: 2x - y >1, x - 2y < - 1

  14. Algebra

    Section 2.11 : Linear Inequalities. Back to Problem List. 6. Solve the following inequality and give the solution in both inequality and interval notation. 2 < 1 6 − 1 2x ≤ 4 2 < 1 6 − 1 2 x ≤ 4. Show All Steps Hide All Steps.

  15. Solving Linear Inequalities Word Problems

    Solutions to linear inequalities represent a region of the Cartesian plane. It becomes essential for us to know how to translate real-life problems into linear inequalities. Linear Inequalities Before defining the linear inequalities formally, let's see them through a real-life situation and observe why their need arises in the first place.

  16. NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities

    1. Solve 24x < 100, when (i) x is a natural number. (ii) x is an integer. Solution: (i) Given that 24x < 100 Now we have to divide the inequality by 24 then we get x < 25/6 Now when x is a natural integer then It is clear that the only natural number less than 25/6 are 1, 2, 3, 4.

  17. An Oblivious Ellipsoid Algorithm for Solving a System of (In)Feasible

    The ellipsoid algorithm is a fundamental algorithm for computing a solution to the system of m linear inequalities in n variables (P):A⊤x≤u when its set of solutions has positive volume. However, when (P) is infeasible, the ellipsoid algorithm has no mechanism for proving that (P) is infeasible.This is in contrast to the other two fundamental algorithms for tackling (P), namely, the ...

  18. Algebra

    We know that the process of solving a linear inequality is pretty much the same process as solving a linear equation. We do basic algebraic manipulations to get all the \(z\)'s on one side of the inequality and the numbers on the other side. Just remember that what you do to one side of the inequality you have to do to the other side as well.

  19. Algebra

    Section 2.11 : Linear Inequalities. For problems 1 - 10 solve each of the following inequalities. Give the solution in both inequality and interval notations. If −7 < x ≤ 6 − 7 < x ≤ 6 determine a and b for the inequality : a < 3x +8 ≤ b a < 3 x + 8 ≤ b. If −3 ≤ x ≤ −1 − 3 ≤ x ≤ − 1 determine a and b for the ...