problem solving in linear equation

Word Problems on Linear Equations

Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.

There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.

Steps involved in solving a linear equation word problem: ● Read the problem carefully and note what is given and what is required and what is given. ● Denote the unknown by the variables as x, y, ……. ● Translate the problem to the language of mathematics or mathematical statements. ● Form the linear equation in one variable using the conditions given in the problems. ● Solve the equation for the unknown. ● Verify to be sure whether the answer satisfies the conditions of the problem.

Step-by-step application of linear equations to solve practical word problems:

1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers. 

Solution: Then the other number = x + 9 Let the number be x.  Sum of two numbers = 25 According to question, x + x + 9 = 25 ⇒ 2x + 9 = 25 ⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9)  ⇒ 2x = 16 ⇒ 2x/2 = 16/2 (divide by 2 on both the sides)  ⇒ x = 8 Therefore, x + 9 = 8 + 9 = 17 Therefore, the two numbers are 8 and 17.

2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers?  Solution:   Let the common ratio be x.  Let the common ratio be x.  Their difference = 48 According to the question,  7x - 3x = 48  ⇒ 4x = 48  ⇒ x = 48/4  ⇒ x = 12 Therefore, 7x = 7 × 12 = 84           3x = 3 × 12 = 36  Therefore, the two numbers are 84 and 36.

3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle.  Solution: Let the breadth of the rectangle be x,  Then the length of the rectangle = 2x Perimeter of the rectangle = 72 Therefore, according to the question 2(x + 2x) = 72 ⇒ 2 × 3x = 72 ⇒ 6x = 72  ⇒ x = 72/6 ⇒ x = 12 We know, length of the rectangle = 2x                       = 2 × 12 = 24 Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.

4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages. 

Solution: Let Ron’s present age be x.  Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4.  According to the question;  Ron will be twice as old as Aaron.  Therefore, x + 4 = 2(x - 5 + 4)  ⇒ x + 4 = 2(x - 1)  ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.

5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts.  Solution: Let one part of the number be x Then the other part of the number = x + 10 The ratio of the two numbers is 5 : 3 Therefore, (x + 10)/x = 5/3 ⇒ 3(x + 10) = 5x  ⇒ 3x + 30 = 5x ⇒ 30 = 5x - 3x ⇒ 30 = 2x  ⇒ x = 30/2  ⇒ x = 15 Therefore, x + 10 = 15 + 10 = 25 Therefore, the number = 25 + 15 = 40  The two parts are 15 and 25. 

More solved examples with detailed explanation on the word problems on linear equations.

6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages.  Solution: Let Robert’s age be x years.  Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question,  4x + 5 = 3(x + 5)  ⇒ 4x + 5 = 3x + 15  ⇒ 4x - 3x = 15 - 5  ⇒ x = 10 ⇒ 4x = 4 × 10 = 40  Robert’s present age is 10 years and that of his father’s age = 40 years.  

7. The sum of two consecutive multiples of 5 is 55. Find these multiples.  Solution: Let the first multiple of 5 be x.  Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2  ⇒ x = 25  Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.  

8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles.  Solution: Let the angle be x.  Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51  Therefore, the two complementary angles are 39° and 51°

9. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair.  Solution: The table cost $ 40 more than the chair.  Let us assume the cost of the chair to be x.  Then the cost of the table = $ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x)  Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is $125 and that of each table is $165. 

10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number?  Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2  According to the question,  3/5 ᵗʰ of the number is 4 more than 1/2 of the number.  ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40.  

Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.

●   Equations

What is an Equation?

What is a Linear Equation?

How to Solve Linear Equations?

Solving Linear Equations

Problems on Linear Equations in One Variable

Word Problems on Linear Equations in One Variable

Practice Test on Linear Equations

Practice Test on Word Problems on Linear Equations

●   Equations - Worksheets

Worksheet on Linear Equations

Worksheet on Word Problems on Linear Equation

7th Grade Math Problems   8th Grade Math Practice   From Word Problems on Linear Equations to HOME PAGE

New! Comments

Didn't find what you were looking for? Or want to know more information about Math Only Math . Use this Google Search to find what you need.

• Preschool Activities
• Kindergarten Math
• 1st Grade Math
• 2nd Grade Math
• 3rd Grade Math
• 4th Grade Math
• 5th Grade Math
• 6th Grade Math
• 7th Grade Math
• 8th Grade Math
• 9th Grade Math
• 10th Grade Math
• 11 & 12 Grade Math
• Concepts of Sets
• Probability
• Boolean Algebra
• Math Coloring Pages
• Multiplication Table
• Cool Maths Games
• Math Flash Cards
• Online Math Quiz
• Math Puzzles
• Binary System
• Math Dictionary
• Conversion Chart
• Homework Sheets
• Math Problem Ans
• Free Math Answers
• Printable Math Sheet
• Funny Math Answers
• Employment Test
• Math Patterns
• Link Partners
• Privacy Policy

Recent Articles

Properties of Division | Division of Property Overview|Math Properties

Apr 04, 24 03:22 PM

Long Division | Method | Steps | Examples | Long Division Worksheets

Apr 04, 24 01:09 PM

Division of Two-Digit by a One-Digit Numbers | Dividing Larger Numbers

Apr 04, 24 09:31 AM

Dividing 3-Digit by 1-Digit Number | Long Division |Worksheet Answer

Apr 04, 24 09:30 AM

Months of the Year | List of 12 Months of the Year |Jan, Feb, Mar, Apr

Apr 02, 24 02:08 PM

© and ™ math-only-math.com. All Rights Reserved. 2010 - 2024.

Word Problems Linear Equations

Andymath.com features free videos, notes, and practice problems with answers! Printable pages make math easy. Are you ready to be a mathmagician?

\(\textbf{1)}\) Joe and Steve are saving money. Joe starts with $105 and saves $5 per week. Steve starts with $5 and saves $15 per week. After how many weeks do they have the same amount of money? Show Equations \(y= 5x+105,\,\,\,y=15x+5\) Show Answer 10 weeks ($155)

\(\textbf{2)}\) mike and sarah collect rocks. together they collected 50 rocks. mike collected 10 more rocks than sarah. how many rocks did each of them collect show equations \(m+s=50,\,\,\,m=s+10\) show answer mike collected 30 rocks, sarah collected 20 rocks., \(\textbf{3)}\) in a classroom the ratio of boys to girls is 2:3. there are 25 students in the class. how many are girls show equations \(b+g=50,\,\,\,3b=2g\) show answer 15 girls (10 boys), \(\textbf{4)}\) kyle makes sandals at home. the sandal making tools cost $100 and he spends $10 on materials for each sandal. he sells each sandal for $30. how many sandals does he have to sell to break even show equations \(c=10x+100,\,\,\,r=30x\) show answer 5 sandals ($150), \(\textbf{5)}\) molly is throwing a beach party. she still needs to buy beach towels and beach balls. towels are $3 each and beachballs are $4 each. she bought 10 items in total and it cost $34. how many beach balls did she get show equations show answer 4 beachballs (6 towels), \(\textbf{6)}\) anna volunteers at a pet shelter. they have cats and dogs. there are 36 pets in total at the shelter, and the ratio of dogs to cats is 4:5. how many cats are at the shelter show equations \(c+d=40,\,\,\,5d=4c\) show answer 20 cats (16 dogs), \(\textbf{7)}\) a store sells oranges and apples. oranges cost $1.00 each and apples cost $2.00 each. in the first sale of the day, 15 fruits were sold in total, and the price was $25. how many of each type of frust was sold show equations \(o+a=15,\,\,\,1o+2a=25\) show answer 10 apples and 5 oranges, \(\textbf{8)}\) the ratio of red marbles to green marbles is 2:7. there are 36 marbles in total. how many are red show equations \(r+g=36,\,\,\,7r=2g\) show answer 8 red marbles (28 green marbles), \(\textbf{9)}\) a tennis club charges $100 to join the club and $10 for every hour using the courts. write an equation to express the cost \(c\) in terms of \(h\) hours playing tennis. show equation the equation is \(c=10h+100\), \(\textbf{10)}\) emma and liam are saving money. emma starts with $80 and saves $10 per week. liam starts with $120 and saves $6 per week. after how many weeks will they have the same amount of money show equations \(e = 10x + 80,\,\,\,l = 6x + 120\) show answer 10 weeks ($180 each), \(\textbf{11)}\) mark and lisa collect stamps. together they collected 200 stamps. mark collected 40 more stamps than lisa. how many stamps did each of them collect show equations \(m + l = 200,\,\,\,m = l + 40\) show answer mark collected 120 stamps, lisa collected 80 stamps., \(\textbf{12)}\) in a classroom, the ratio of boys to girls is 3:5. there are 40 students in the class. how many are boys show equations \(b + g = 40,\,\,\,5b = 3g\) show answer 15 boys (25 girls), \(\textbf{13)}\) lisa is selling handmade jewelry. the materials cost $60, and she sells each piece for $20. how many pieces does she have to sell to break even show equations \(c=60,\,\,\,r=20x\) show answer 3 pieces, \(\textbf{14)}\) tom is buying books and notebooks for school. books cost $15 each, and notebooks cost $3 each. he bought 12 items in total, and it cost $120. how many notebooks did he buy show equations \(b + n = 12,\,\,\,15b+3n=120\) show answer 5 notebooks (7 books), \(\textbf{15)}\) emily volunteers at an animal shelter. they have rabbits and guinea pigs. there are 36 animals in total at the shelter, and the ratio of guinea pigs to rabbits is 4:5. how many guinea pigs are at the shelter show equations \(r + g = 36,\,\,\,5g=4r\) show answer 16 guinea pigs (20 rabbits), \(\textbf{16)}\) mike and sarah are going to a theme park. mike’s ticket costs $40, and sarah’s ticket costs $30. they also bought $20 worth of food. how much did they spend in total show equations \(m + s + f = t,\,\,\,m=40,\,\,\,s=30,\,\,\,f=20\) show answer they spent $90 in total., \(\textbf{17)}\) the ratio of red marbles to blue marbles is 2:3. there are 50 marbles in total. how many are blue show equations \(r + b = 50,\,\,\,3r=2b\) show answer 30 blue marbles (20 red marbles), \(\textbf{18)}\) a pizza restaurant charges $12 for a large pizza and $8 for a small pizza. if a customer buys 5 pizzas in total, and it costs $52, how many large pizzas did they buy show equations \(l + s = 5,\,\,\,12l+8s=52\) show answer they bought 3 large pizzas (2 small pizzas)., \(\textbf{19)}\) the area of a rectangle is 48 square meters. if the length is 8 meters, what is the width of the rectangle show equations \(a=l\times w,\,\,\,l=8,\,\,\,a=48\) show answer the width is 6 meters., \(\textbf{20)}\) two numbers have a sum of 50. one number is 10 more than the other. what are the two numbers show equations \(x+y=50,\,\,\,x=y+10\) show answer the numbers are 30 and 20., \(\textbf{21)}\) a store sells jeans for $40 each and t-shirts for $20 each. in the first sale of the day, they sold 8 items in total, and the price was $260. how many of each type of item was sold show equations \(j+t=8,\,\,\,40j+20t=260\) show answer 5 jeans and 3 t-shirts were sold., \(\textbf{22)}\) the ratio of apples to carrots is 3:4. there are 28 fruits in total. how many are apples show equations \(\)a+c=28,\,\,\,4a=3c show answer there are 12 apples and 16 carrots., \(\textbf{23)}\) a phone plan costs $30 per month, and there is an additional charge of $0.10 per minute for calls. write an equation to express the cost \(c\) in terms of \(m\) minutes. show equation the equation is \(\)c=30+0.10m, \(\textbf{24)}\) a triangle has a base of 8 inches and a height of 6 inches. calculate its area. show equations \(a=0.5\times b\times h,\,\,\,b=8,\,\,\,h=6\) show answer the area is 24 square inches., \(\textbf{25)}\) a store sells shirts for $25 each and pants for $45 each. in the first sale of the day, 4 items were sold, and the price was $180. how many of each type of item was sold show equations \(t+p=4,\,\,\,25t+45p=180\) show answer 0 shirts and 4 pants were sold., \(\textbf{26)}\) a garden has a length of 12 feet and a width of 10 feet. calculate its area. show equations \(a=l\times w,\,\,\,l=12,\,\,\,w=10\) show answer the area is 120 square feet., \(\textbf{27)}\) the sum of two consecutive odd numbers is 56. what are the two numbers show equations \(x+y=56,\,\,\,x=y+2\) show answer the numbers are 27 and 29., \(\textbf{28)}\) a toy store sells action figures for $15 each and toy cars for $5 each. in the first sale of the day, 10 items were sold, and the price was $110. how many of each type of item was sold show equations \(a+c=10,\,\,\,15a+5c=110\) show answer 6 action figures and 4 toy cars were sold., \(\textbf{29)}\) a bakery sells pie for $2 each and cookies for $1 each. in the first sale of the day, 14 items were sold, and the price was $25. how many of each type of item was sold show equations \(p+c=14,\,\,\,2p+c=25\) show answer 11 pies and 3 cookies were sold., \(\textbf{for 30-33}\) two car rental companies charge the following values for x miles. car rental a: \(y=3x+150 \,\,\) car rental b: \(y=4x+100\), \(\textbf{30)}\) which rental company has a higher initial fee show answer company a has a higher initial fee, \(\textbf{31)}\) which rental company has a higher mileage fee show answer company b has a higher mileage fee, \(\textbf{32)}\) for how many driven miles is the cost of the two companies the same show answer the companies cost the same if you drive 50 miles., \(\textbf{33)}\) what does the \(3\) mean in the equation for company a show answer for company a, the cost increases by $3 per mile driven., \(\textbf{34)}\) what does the \(100\) mean in the equation for company b show answer for company b, the initial cost (0 miles driven) is $100., \(\textbf{for 35-39}\) andy is going to go for a drive. the formula below tells how many gallons of gas he has in his car after m miles. \(g=12-\frac{m}{18}\), \(\textbf{35)}\) what does the \(12\) in the equation represent show answer andy has \(12\) gallons in his car when he starts his drive., \(\textbf{36)}\) what does the \(18\) in the equation represent show answer it takes \(18\) miles to use up \(1\) gallon of gas., \(\textbf{37)}\) how many miles until he runs out of gas show answer the answer is \(216\) miles, \(\textbf{38)}\) how many gallons of gas does he have after 90 miles show answer the answer is \(7\) gallons, \(\textbf{39)}\) when he has \(3\) gallons remaining, how far has he driven show answer the answer is \(162\) miles, \(\textbf{for 40-42}\) joe sells paintings. each month he makes no commission on the first $5,000 he sells but then makes a 10% commission on the rest., \(\textbf{40)}\) find the equation of how much money x joe needs to sell to earn y dollars per month. show answer the answer is \(y=.1(x-5,000)\), \(\textbf{41)}\) how much does joe need to sell to earn $10,000 in a month. show answer the answer is \($105,000\), \(\textbf{42)}\) how much does joe earn if he sells $45,000 in a month show answer the answer is \($4,000\), see related pages\(\), \(\bullet\text{ word problems- linear equations}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- averages}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- consecutive integers}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- distance, rate and time}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- break even}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- ratios}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- age}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- mixtures and concentration}\) \(\,\,\,\,\,\,\,\,\), linear equations are a type of equation that has a linear relationship between two variables, and they can often be used to solve word problems. in order to solve a word problem involving a linear equation, you will need to identify the variables in the problem and determine the relationship between them. this usually involves setting up an equation (or equations) using the given information and then solving for the unknown variables . linear equations are commonly used in real-life situations to model and analyze relationships between different quantities. for example, you might use a linear equation to model the relationship between the cost of a product and the number of units sold, or the relationship between the distance traveled and the time it takes to travel that distance. linear equations are typically covered in a high school algebra class. these types of problems can be challenging for students who are new to algebra, but they are an important foundation for more advanced math concepts. one common mistake that students make when solving word problems involving linear equations is failing to set up the problem correctly. it’s important to carefully read the problem and identify all of the relevant information, as well as any given equations or formulas that you might need to use. other related topics involving linear equations include graphing and solving systems. understanding linear equations is also useful for applications in fields such as economics, engineering, and physics., about andymath.com, andymath.com is a free math website with the mission of helping students, teachers and tutors find helpful notes, useful sample problems with answers including step by step solutions, and other related materials to supplement classroom learning. if you have any requests for additional content, please contact andy at [email protected] . he will promptly add the content. topics cover elementary math , middle school , algebra , geometry , algebra 2/pre-calculus/trig , calculus and probability/statistics . in the future, i hope to add physics and linear algebra content. visit me on youtube , tiktok , instagram and facebook . andymath content has a unique approach to presenting mathematics. the clear explanations, strong visuals mixed with dry humor regularly get millions of views. we are open to collaborations of all types, please contact andy at [email protected] for all enquiries. to offer financial support, visit my patreon page. let’s help students understand the math way of thinking thank you for visiting. how exciting.

Solving Linear Equations

Solving linear equations means finding the value of the variable(s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation. Sometimes, we may have to find the values of variables involved in a linear equation. When we are given two or more such linear equations, we can find the values of each variable by solving linear equations. There are a few methods to solve linear equations. Let us discuss each of these methods in detail.

Solving Linear Equations in One Variable

A linear equation in one variable is an equation of degree one and has only one variable term. It is of the form 'ax+b = 0', where 'a' is a non zero number and 'x' is a variable. By solving linear equations in one variable, we get only one solution for the given variable. An example for this is 3x - 6 = 0. The variable 'x' has only one solution, which is calculated as 3x - 6 = 0 3x = 6 x = 6/3 x = 2

For solving linear equations with one variable, simplify the equation such that all the variable terms are brought to one side and the constant value is brought to the other side. If there are any fractional terms then find the LCM ( Least Common Multiple ) and simplify them such that the variable terms are on one side and the constant terms are on the other side. Let us work out a small example to understand this.

4x + 8 = 8x - 10. To find the value of 'x', let us simplify and bring the 'x' terms to one side and the constant terms to another side.

4x - 8x = -10 - 8 -4x = -18 4x = 18 x = 18/4 On simplifying, we get x = 9/2.

Solving Linear Equations by Substitution Method

The substitution method is one of the methods of solving linear equations. In the substitution method , we rearrange the equation such that one of the values is substituted in the second equation. Now that we are left with an equation that has only one variable, we can solve it and find the value of that variable. In the two given equations, any equation can be taken and the value of a variable can be found and substituted in another equation. For solving linear equations using the substitution method, follow the steps mentioned below. Let us understand this with an example of solving the following system of linear equations. x + y = 6 --------------(1) 2x + 4y = 20 -----------(2)

Step 1: Find the value of one of the variables using any one of the equations. In this case, let us find the value of 'x' from equation (1). x + y = 6 ---------(1) x = 6 - y Step 2: Substitute the value of the variable found in step 1 in the second linear equation. Now, let us substitute the value of 'x' in the second equation 2x + 4y = 20.

x = 6 - y Substituting the value of 'x' in 2x + 4y = 20, we get,

2(6 - y) + 4y = 20 12 - 2y + 4y = 20 12 + 2y = 20 2y = 20 - 12 2y = 8 y = 8/2 y = 4 Step 3: Now substitute the value of 'y' in either equation (1) or (2). Let us substitute the value of 'y' in equation (1).

x + y = 6 x + 4 = 6 x = 6 - 4 x = 2 Therefore, by substitution method, the linear equations are solved, and the value of x is 2 and y is 4.

Solving Linear Equations by Elimination Method

The elimination method is another way to solve a system of linear equations. Here we make an attempt to multiply either the 'x' variable term or the 'y' variable term with a constant value such that either the 'x' variable terms or the 'y' variable terms cancel out and gives us the value of the other variable. Let us understand the steps of solving linear equations by elimination method . Consider the given linear equations: 2x + y = 11 ----------- (1) x + 3y = 18 ---------- (2) Step 1: Check whether the terms are arranged in a way such that the 'x' term is followed by a 'y' term and an equal to sign and after the equal to sign the constant term should be present. The given set of linear equations are already arranged in the correct way which is ax+by=c or ax+by-c=0.

Step 2: The next step is to multiply either one or both the equations by a constant value such that it will make either the 'x' terms or the 'y' terms cancel out which would help us find the value of the other variable. Now in equation (2), let us multiply every term by the number 2 to make the coefficients of x the same in both the equations. x + 3y = 18 ---------- (2) Multiplying all the terms in equation (2) by 2, we get,

2(x) + 2(3y) = 2(18). Now equation (2) becomes, 2x + 6y = 36 -----------(2)

Therefore, y = 5. Step 4: Using the value obtained in step 3, find out the value of another variable by substituting the value in any of the equations. Let us substitute the value of 'y' in equation (1). We get, 2x + y = 11 2x + 5 = 11 2x = 11 - 5 2x = 6 x = 6/2 x = 3

Therefore, by solving linear equations, we get the value of x = 3 and y = 5.

Graphical Method of Solving Linear Equations

Another method for solving linear equations is by using the graph. When we are given a system of linear equations, we graph both the equations by finding values for 'y' for different values of 'x' in the coordinate system. Once it is done, we find the point of intersection of these two lines. The (x,y) values at the point of intersection give the solution for these linear equations. Let us take two linear equations and solve them using the graphical method.

x + y = 8 -------(1)

y = x + 2 --------(2)

Let us take some values for 'x' and find the values for 'y' for the equation x + y = 8. This can also be rewritten as y = 8 - x.

Let us take some values for 'x' and find the values for 'y' in the equation y = x + 2.

Plotting these points on the coordinate plane, we get a graph like this.

Now, we find the point of intersection of these lines to find the values of 'x' and 'y'. The two lines intersect at the point (3,5). Therefore, x = 3 and y = 5 by using the graphical method of solving linear equations .

This method is also used to find the optimal solution of linear programming problems. Let us look at one more method of solving linear equations, which is the cross multiplication method.

Cross Multiplication Method of Solving Linear Equations

The cross multiplication method enables us to solve linear equations by picking the coefficients of all the terms ('x' , 'y' and the constant terms) in the format shown below and apply the formula for finding the values of 'x' and 'y'.

Topics Related to Solving Linear Equations

Check the given articles related to solving linear equations.

• Linear Equations
• Application of Linear Equations
• Two-Variable Linear Equations
• Linear Equations and Half Planes
• One Variable Linear Equations and Inequations

Solving Linear Equations Examples

Example 1: Solve the following linear equations by the substitution method.

3x + y = 13 --------- (1) 2x + 3y = 18 -------- (2)

By using the substitution method of solving linear equations, let us take the first equation and find the value of 'y' and substitute it in the second equation.

From equation (1), y = 13-3x. Now, substituting the value of 'y' in equation (2), we get, 2x + 3 (13 - 3x) = 18 2x + 39 - 9x = 18 -7x + 39 = 18 -7x = 18 - 39 -7x = -21 x = -21/-7 x = 3 Now, let us substitute the value of 'x = 3' in equation (1) and find the value of 'y'. 3x + y = 13 ------- (1) 3(3) + y = 13 9 + y = 13 y = 13 - 9 y = 4

Therefore, by the substitution method, the value of x is 3 and y is 4.

Example 2: Using the elimination method of solving linear equations find the values of 'x' and 'y'.

3x + y = 21 ------ (1) 2x + 3y = 28 -------- (2)

By using the elimination method, let us make the 'y' variable to be the same in both the equations (1) and (2). To do this let us multiply all the terms of the first equation by 3. Therefore equation (1) becomes,

3(3x) + 3(y) = 63 9x + 3y = 63 ---------- (3) The second equation is, 2x + 3y = 28 Now let us cancel the 'y' terms and find the value of 'x' by subtracting equation (2) from equation (3). This is done by changing the signs of all the terms in equation (2).

Example 3: Using the cross multiplication method of solving linear equations, solve the following equations.

x + 2y - 16 = 0 --------- (1) 4x - y - 10 = 0 ---------- (2)

Compare the given equation with \(a_{1}\)x + \(b_{1}\)y + \(c_{1}\) = 0, and \(a_{2}\)x+\(b_{2}\)y+\(c_{2}\) = 0. From the given equations,

\(a_{1}\) = 1, \(a_{2}\) = 4, \(b_{1}\) = 2, \(b_{2}\) = -1, \(c_{1}\) = -16, and \(c_{2}\) = -10.

By cross multiplication method,

x = \(b_{1}\)\(c_{2}\) - \(b_{2}\)\(c_{1}\)/\(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\) y = \(c_{1}\)\(a_{2}\) - \(c_{2}\)\(a_{1}\) / \(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\)

Substituting the values in the formula we get,

x = ((2)(-10)) - ((-1)(-16)) / ((1)(-1)) - ((4)(2)) x = (-20-16)/(-1-8) x = -36/-9 x = 36/9 x = 4 y = ((-16)(4)) - ((-10)(1)) / ((1)(-1)) - ((4)(2)) y = (-64 + 10) / (-1 - 8) y = -54 / -9 y = 54/9 y = 6 Therefore, by the cross multiplication method, the value of x is 4 and y is 6.

go to slide go to slide go to slide

Book a Free Trial Class

Practice Questions on Solving Linear Equations

Faqs on solving linear equations, what does it mean by solving linear equations.

An equation that has a degree of 1 is called a linear equation. We can have one variable linear equations , two-variable linear equations , linear equations with three variables, and more depending on the number of variables in it. Solving linear equations means finding the values of all the variables present in the equation. This can be done by substitution method, elimination method, graphical method, and the cross multiplication method . All these methods are different ways of finding the values of the variables.

How to Use the Substitution Method for Solving Linear Equations?

The substitution method of solving equations states that for a given system of linear equations, find the value of either 'x' or 'y' from any of the given equations and then substitute the value found of 'x' or 'y' in another equation so that the other unknown value can be found.

How to Use the Elimination Method for Solving Linear Equations?

In the elimination method of solving linear equations, we multiply a constant or a number with one equation or both the equations such that either the 'x' terms or the 'y' terms are the same. Then we cancel out the same term in both the equations by either adding or subtracting them and find the value of one variable (either 'x' or 'y'). After finding one of the values, we substitute the value in one of the equations and find the other unknown value.

What is the Graphical Method of Solving Linear Equations?

In the graphical method of solving linear equations, we find the value of 'y' from the given equations by putting the values of x as 0, 1, 2, 3, and so on, and plot a graph in the coordinate system for the line for various values of 'x' for both the system of linear equations. We will see that these two lines intersect at a point. This point is the solution for the given system of linear equations. If there is no intersection point between two lines, then we consider them as parallel lines , and if we found that both the lines lie on each other, those are known as coincident lines and have infinitely many solutions.

What are the Steps of Solving Linear Equations that has One Variable?

A linear equation is an equation with degree 1. To solve a linear equation that has one variable we bring the variable to one side and the constant value to the other side. Then, a non-zero number may be added, subtracted, multiplied, or divided on both sides of the equation. For example, a linear equation with one variable will be of the form 'x - 4 = 2'. To find the value of 'x', we add the constant value '4' to both sides of the equation. Therefore, the value of 'x = 6'.

What are the Steps of Solving Linear Equations having Three Variables?

To solve a system of linear equations that has three variables, we take any two equations and variables. We then take another pair of linear equations and also solve for the same variable. Now that, we have two linear equations with two variables, we can use the substitution method or elimination method, or any other method to solve the values of two unknown variables. After finding these two variables, we substitute them in any of the three equations to find the third unknown variable.

What are the 4 Methods of Solving Linear Equations?

The methods for solving linear equations are given below:

• Substitution method
• Elimination method
• Cross multiplication method
• Graphical method

2.2 Linear Equations in One Variable

Learning objectives.

In this section, you will:

• Solve equations in one variable algebraically.
• Solve a rational equation.
• Find a linear equation.
• Given the equations of two lines, determine whether their graphs are parallel or perpendicular.
• Write the equation of a line parallel or perpendicular to a given line.

Caroline is a full-time college student planning a spring break vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays $15.00/hr, and she opened a savings account with an initial deposit of $400 on January 15. She arranged for direct deposit of her payroll checks. If spring break begins March 20 and the trip will cost approximately $2,500, how many hours will she have to work to earn enough to pay for her vacation? If she can only work 4 hours per day, how many days per week will she have to work? How many weeks will it take? In this section, we will investigate problems like this and others, which generate graphs like the line in Figure 1 .

Solving Linear Equations in One Variable

A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form a x + b = 0 a x + b = 0 and are solved using basic algebraic operations.

We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An identity equation is true for all values of the variable. Here is an example of an identity equation.

The solution set consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for x x will make the equation true.

A conditional equation is true for only some values of the variable. For example, if we are to solve the equation 5 x + 2 = 3 x − 6 , 5 x + 2 = 3 x − 6 , we have the following:

The solution set consists of one number: { − 4 } . { − 4 } . It is the only solution and, therefore, we have solved a conditional equation.

An inconsistent equation results in a false statement. For example, if we are to solve 5 x − 15 = 5 ( x − 4 ) , 5 x − 15 = 5 ( x − 4 ) , we have the following:

Indeed, −15 ≠ −20. −15 ≠ −20. There is no solution because this is an inconsistent equation.

Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows.

Linear Equation in One Variable

A linear equation in one variable can be written in the form

where a and b are real numbers, a ≠ 0. a ≠ 0.

Given a linear equation in one variable, use algebra to solve it.

The following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads x = _________, x = _________, if x is the unknown. There is no set order, as the steps used depend on what is given:

• We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero.
• Apply the distributive property as needed: a ( b + c ) = a b + a c . a ( b + c ) = a b + a c .
• Isolate the variable on one side of the equation.
• When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient.

Solving an Equation in One Variable

Solve the following equation: 2 x + 7 = 19. 2 x + 7 = 19.

This equation can be written in the form a x + b = 0 a x + b = 0 by subtracting 19 19 from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations.

The solution is 6.

Solve the linear equation in one variable: 2 x + 1 = −9. 2 x + 1 = −9.

Solving an Equation Algebraically When the Variable Appears on Both Sides

Solve the following equation: 4 ( x −3 ) + 12 = 15 −5 ( x + 6 ) . 4 ( x −3 ) + 12 = 15 −5 ( x + 6 ) .

Apply standard algebraic properties.

This problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line, x = − 5 3 . x = − 5 3 .

Solve the equation in one variable: −2 ( 3 x − 1 ) + x = 14 − x . −2 ( 3 x − 1 ) + x = 14 − x .

Solving a Rational Equation

In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation .

Recall that a rational number is the ratio of two numbers, such as 2 3 2 3 or 7 2 . 7 2 . A rational expression is the ratio, or quotient, of two polynomials. Here are three examples.

Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD).

Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.

Solve the rational equation: 7 2 x − 5 3 x = 22 3 . 7 2 x − 5 3 x = 22 3 .

We have three denominators; 2 x , 3 x , 2 x , 3 x , and 3. The LCD must contain 2 x , 3 x , 2 x , 3 x , and 3. An LCD of 6 x 6 x contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD 6 x . 6 x .

A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as ( x + 1 ) . ( x + 1 ) . Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are x , x , x − 1 , x − 1 , and 3 x − 3. 3 x − 3. First, factor all denominators. We then have x , x , ( x − 1 ) , ( x − 1 ) , and 3 ( x − 1 ) 3 ( x − 1 ) as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of ( x − 1 ) . ( x − 1 ) . The x x in the first denominator is separate from the x x in the ( x − 1 ) ( x − 1 ) denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the x , x , one factor of ( x − 1 ) , ( x − 1 ) , and the 3. Thus, the LCD is the following:

So, both sides of the equation would be multiplied by 3 x ( x − 1 ) . 3 x ( x − 1 ) . Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.

Another example is a problem with two denominators, such as x x and x 2 + 2 x . x 2 + 2 x . Once the second denominator is factored as x 2 + 2 x = x ( x + 2 ) , x 2 + 2 x = x ( x + 2 ) , there is a common factor of x in both denominators and the LCD is x ( x + 2 ) . x ( x + 2 ) .

Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation.

We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign.

Multiply a ( d ) a ( d ) and b ( c ) , b ( c ) , which results in a d = b c . a d = b c .

Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities.

Rational Equations

A rational equation contains at least one rational expression where the variable appears in at least one of the denominators.

Given a rational equation, solve it.

• Factor all denominators in the equation.
• Find and exclude values that set each denominator equal to zero.
• Find the LCD.
• Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left.
• Solve the remaining equation.
• Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator.

Solving a Rational Equation without Factoring

Solve the following rational equation:

We have three denominators: x , x , 2 , 2 , and 2 x . 2 x . No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is 2 x . 2 x . Only one value is excluded from a solution set, 0. Next, multiply the whole equation (both sides of the equal sign) by 2 x . 2 x .

The proposed solution is −1, which is not an excluded value, so the solution set contains one number, −1 , −1 , or { −1 } { −1 } written in set notation.

Solve the rational equation: 2 3 x = 1 4 − 1 6 x . 2 3 x = 1 4 − 1 6 x .

Solving a Rational Equation by Factoring the Denominator

Solve the following rational equation: 1 x = 1 10 − 3 4 x . 1 x = 1 10 − 3 4 x .

First find the common denominator. The three denominators in factored form are x , 10 = 2 ⋅ 5 , x , 10 = 2 ⋅ 5 , and 4 x = 2 ⋅ 2 ⋅ x . 4 x = 2 ⋅ 2 ⋅ x . The smallest expression that is divisible by each one of the denominators is 20 x . 20 x . Only x = 0 x = 0 is an excluded value. Multiply the whole equation by 20 x . 20 x .

The solution is 35 2 . 35 2 .

Solve the rational equation: − 5 2 x + 3 4 x = − 7 4 . − 5 2 x + 3 4 x = − 7 4 .

Solving Rational Equations with a Binomial in the Denominator

Solve the following rational equations and state the excluded values:

• ⓐ 3 x − 6 = 5 x 3 x − 6 = 5 x
• ⓑ x x − 3 = 5 x − 3 − 1 2 x x − 3 = 5 x − 3 − 1 2
• ⓒ x x − 2 = 5 x − 2 − 1 2 x x − 2 = 5 x − 2 − 1 2

The denominators x x and x − 6 x − 6 have nothing in common. Therefore, the LCD is the product x ( x − 6 ) . x ( x − 6 ) . However, for this problem, we can cross-multiply.

The solution is 15. The excluded values are 6 6 and 0. 0.

The LCD is 2 ( x − 3 ) . 2 ( x − 3 ) . Multiply both sides of the equation by 2 ( x − 3 ) . 2 ( x − 3 ) .

The solution is 13 3 . 13 3 . The excluded value is 3. 3.

The least common denominator is 2 ( x − 2 ) . 2 ( x − 2 ) . Multiply both sides of the equation by x ( x − 2 ) . x ( x − 2 ) .

The solution is 4. The excluded value is 2. 2.

Solve − 3 2 x + 1 = 4 3 x + 1 . − 3 2 x + 1 = 4 3 x + 1 . State the excluded values.

Solving a Rational Equation with Factored Denominators and Stating Excluded Values

Solve the rational equation after factoring the denominators: 2 x + 1 − 1 x − 1 = 2 x x 2 − 1 . 2 x + 1 − 1 x − 1 = 2 x x 2 − 1 . State the excluded values.

We must factor the denominator x 2 −1. x 2 −1. We recognize this as the difference of squares, and factor it as ( x − 1 ) ( x + 1 ) . ( x − 1 ) ( x + 1 ) . Thus, the LCD that contains each denominator is ( x − 1 ) ( x + 1 ) . ( x − 1 ) ( x + 1 ) . Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation.

The solution is −3. −3. The excluded values are 1 1 and −1. −1.

Solve the rational equation: 2 x − 2 + 1 x + 1 = 1 x 2 − x − 2 . 2 x − 2 + 1 x + 1 = 1 x 2 − x − 2 .

Finding a Linear Equation

Perhaps the most familiar form of a linear equation is the slope-intercept form, written as y = m x + b , y = m x + b , where m = slope m = slope and b = y -intercept . b = y -intercept . Let us begin with the slope.

The Slope of a Line

The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.

If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper. Some examples are shown in Figure 2 . The lines indicate the following slopes: m = −3 , m = −3 , m = 2 , m = 2 , and m = 1 3 . m = 1 3 .

The slope of a line, m , represents the change in y over the change in x. Given two points, ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) , ( x 2 , y 2 ) , the following formula determines the slope of a line containing these points:

Finding the Slope of a Line Given Two Points

Find the slope of a line that passes through the points ( 2 , −1 ) ( 2 , −1 ) and ( −5 , 3 ) . ( −5 , 3 ) .

We substitute the y- values and the x- values into the formula.

The slope is − 4 7 . − 4 7 .

It does not matter which point is called ( x 1 , y 1 ) ( x 1 , y 1 ) or ( x 2 , y 2 ) . ( x 2 , y 2 ) . As long as we are consistent with the order of the y terms and the order of the x terms in the numerator and denominator, the calculation will yield the same result.

Find the slope of the line that passes through the points ( −2 , 6 ) ( −2 , 6 ) and ( 1 , 4 ) . ( 1 , 4 ) .

Identifying the Slope and y- intercept of a Line Given an Equation

Identify the slope and y- intercept, given the equation y = − 3 4 x − 4. y = − 3 4 x − 4.

As the line is in y = m x + b y = m x + b form, the given line has a slope of m = − 3 4 . m = − 3 4 . The y- intercept is b = −4. b = −4.

The y -intercept is the point at which the line crosses the y- axis. On the y- axis, x = 0. x = 0. We can always identify the y- intercept when the line is in slope-intercept form, as it will always equal b. Or, just substitute x = 0 x = 0 and solve for y.

The Point-Slope Formula

Given the slope and one point on a line, we can find the equation of the line using the point-slope formula.

This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.

Given one point and the slope, the point-slope formula will lead to the equation of a line:

Finding the Equation of a Line Given the Slope and One Point

Write the equation of the line with slope m = −3 m = −3 and passing through the point ( 4 , 8 ) . ( 4 , 8 ) . Write the final equation in slope-intercept form.

Using the point-slope formula, substitute −3 −3 for m and the point ( 4 , 8 ) ( 4 , 8 ) for ( x 1 , y 1 ) . ( x 1 , y 1 ) .

Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.

Given m = 4 , m = 4 , find the equation of the line in slope-intercept form passing through the point ( 2 , 5 ) . ( 2 , 5 ) .

Finding the Equation of a Line Passing Through Two Given Points

Find the equation of the line passing through the points ( 3 , 4 ) ( 3 , 4 ) and ( 0 , −3 ) . ( 0 , −3 ) . Write the final equation in slope-intercept form.

First, we calculate the slope using the slope formula and two points.

Next, we use the point-slope formula with the slope of 7 3 , 7 3 , and either point. Let’s pick the point ( 3 , 4 ) ( 3 , 4 ) for ( x 1 , y 1 ) . ( x 1 , y 1 ) .

In slope-intercept form, the equation is written as y = 7 3 x − 3. y = 7 3 x − 3.

To prove that either point can be used, let us use the second point ( 0 , −3 ) ( 0 , −3 ) and see if we get the same equation.

We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.

Standard Form of a Line

Another way that we can represent the equation of a line is in standard form . Standard form is given as

where A , A , B , B , and C C are integers. The x- and y- terms are on one side of the equal sign and the constant term is on the other side.

Finding the Equation of a Line and Writing It in Standard Form

Find the equation of the line with m = −6 m = −6 and passing through the point ( 1 4 , −2 ) . ( 1 4 , −2 ) . Write the equation in standard form.

We begin using the point-slope formula.

From here, we multiply through by 2, as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right.

This equation is now written in standard form.

Find the equation of the line in standard form with slope m = − 1 3 m = − 1 3 and passing through the point ( 1 , 1 3 ) . ( 1 , 1 3 ) .

Vertical and Horizontal Lines

The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as

where c is a constant. The slope of a vertical line is undefined, and regardless of the y- value of any point on the line, the x- coordinate of the point will be c .

Suppose that we want to find the equation of a line containing the following points: ( −3 , −5 ) , ( −3 , 1 ) , ( −3 , 3 ) , ( −3 , −5 ) , ( −3 , 1 ) , ( −3 , 3 ) , and ( −3 , 5 ) . ( −3 , 5 ) . First, we will find the slope.

Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x- coordinates are the same and we find a vertical line through x = −3. x = −3. See Figure 3 .

The equation of a horizontal line is given as

where c is a constant. The slope of a horizontal line is zero, and for any x- value of a point on the line, the y- coordinate will be c .

Suppose we want to find the equation of a line that contains the following set of points: ( −2 , −2 ) , ( 0 , −2 ) , ( 3 , −2 ) , ( −2 , −2 ) , ( 0 , −2 ) , ( 3 , −2 ) , and ( 5 , −2 ) . ( 5 , −2 ) . We can use the point-slope formula. First, we find the slope using any two points on the line.

Use any point for ( x 1 , y 1 ) ( x 1 , y 1 ) in the formula, or use the y -intercept.

The graph is a horizontal line through y = −2. y = −2. Notice that all of the y- coordinates are the same. See Figure 3 .

Finding the Equation of a Line Passing Through the Given Points

Find the equation of the line passing through the given points: ( 1 , −3 ) ( 1 , −3 ) and ( 1 , 4 ) . ( 1 , 4 ) .

The x- coordinate of both points is 1. Therefore, we have a vertical line, x = 1. x = 1.

Find the equation of the line passing through ( −5 , 2 ) ( −5 , 2 ) and ( 2 , 2 ) . ( 2 , 2 ) .

Determining Whether Graphs of Lines are Parallel or Perpendicular

Parallel lines have the same slope and different y- intercepts. Lines that are parallel to each other will never intersect. For example, Figure 4 shows the graphs of various lines with the same slope, m = 2. m = 2.

All of the lines shown in the graph are parallel because they have the same slope and different y- intercepts.

Lines that are perpendicular intersect to form a 90° 90° -angle. The slope of one line is the negative reciprocal of the other. We can show that two lines are perpendicular if the product of the two slopes is −1 : m 1 ⋅ m 2 = −1. −1 : m 1 ⋅ m 2 = −1. For example, Figure 5 shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of − 1 3 . − 1 3 .

Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither

Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: 3 y = − 4 x + 3 3 y = − 4 x + 3 and 3 x − 4 y = 8. 3 x − 4 y = 8.

The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.

First equation:

Second equation:

See the graph of both lines in Figure 6

From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.

The slopes are negative reciprocals of each other, confirming that the lines are perpendicular.

Graph the two lines and determine whether they are parallel, perpendicular, or neither: 2 y − x = 10 2 y − x = 10 and 2 y = x + 4. 2 y = x + 4.

Writing the Equations of Lines Parallel or Perpendicular to a Given Line

As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line.

Given an equation for a line, write the equation of a line parallel or perpendicular to it.

• Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form.
• Use the slope and the given point with the point-slope formula.
• Simplify the line to slope-intercept form and compare the equation to the given line.

Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point

Write the equation of line parallel to a 5 x + 3 y = 1 5 x + 3 y = 1 and passing through the point ( 3 , 5 ) . ( 3 , 5 ) .

First, we will write the equation in slope-intercept form to find the slope.

The slope is m = − 5 3 . m = − 5 3 . The y- intercept is 1 3 , 1 3 , but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope. The one exception is that if the y- intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula.

The equation of the line is y = − 5 3 x + 10. y = − 5 3 x + 10. See Figure 7 .

Find the equation of the line parallel to 5 x = 7 + y 5 x = 7 + y and passing through the point ( −1 , −2 ) . ( −1 , −2 ) .

Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

Find the equation of the line perpendicular to 5 x − 3 y + 4 = 0 5 x − 3 y + 4 = 0 and passing through the point ( − 4 , 1 ) . ( − 4 , 1 ) .

The first step is to write the equation in slope-intercept form.

We see that the slope is m = 5 3 . m = 5 3 . This means that the slope of the line perpendicular to the given line is the negative reciprocal, or − 3 5 . − 3 5 . Next, we use the point-slope formula with this new slope and the given point.

Access these online resources for additional instruction and practice with linear equations.

• Solving rational equations
• Equation of a line given two points
• Finding the equation of a line perpendicular to another line through a given point
• Finding the equation of a line parallel to another line through a given point

2.2 Section Exercises

What does it mean when we say that two lines are parallel?

What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)?

How do we recognize when an equation, for example y = 4 x + 3 , y = 4 x + 3 , will be a straight line (linear) when graphed?

What does it mean when we say that a linear equation is inconsistent?

When solving the following equation:

2 x − 5 = 4 x + 1 2 x − 5 = 4 x + 1

explain why we must exclude x = 5 x = 5 and x = −1 x = −1 as possible solutions from the solution set.

For the following exercises, solve the equation for x . x .

7 x + 2 = 3 x − 9 7 x + 2 = 3 x − 9

4 x − 3 = 5 4 x − 3 = 5

3 ( x + 2 ) − 12 = 5 ( x + 1 ) 3 ( x + 2 ) − 12 = 5 ( x + 1 )

12 − 5 ( x + 3 ) = 2 x − 5 12 − 5 ( x + 3 ) = 2 x − 5

1 2 − 1 3 x = 4 3 1 2 − 1 3 x = 4 3

x 3 − 3 4 = 2 x + 3 12 x 3 − 3 4 = 2 x + 3 12

2 3 x + 1 2 = 31 6 2 3 x + 1 2 = 31 6

3 ( 2 x − 1 ) + x = 5 x + 3 3 ( 2 x − 1 ) + x = 5 x + 3

2 x 3 − 3 4 = x 6 + 21 4 2 x 3 − 3 4 = x 6 + 21 4

x + 2 4 − x − 1 3 = 2 x + 2 4 − x − 1 3 = 2

For the following exercises, solve each rational equation for x . x . State all x -values that are excluded from the solution set.

3 x − 1 3 = 1 6 3 x − 1 3 = 1 6

2 − 3 x + 4 = x + 2 x + 4 2 − 3 x + 4 = x + 2 x + 4

3 x − 2 = 1 x − 1 + 7 ( x − 1 ) ( x − 2 ) 3 x − 2 = 1 x − 1 + 7 ( x − 1 ) ( x − 2 )

3 x x − 1 + 2 = 3 x − 1 3 x x − 1 + 2 = 3 x − 1

5 x + 1 + 1 x − 3 = − 6 x 2 − 2 x − 3 5 x + 1 + 1 x − 3 = − 6 x 2 − 2 x − 3

1 x = 1 5 + 3 2 x 1 x = 1 5 + 3 2 x

For the following exercises, find the equation of the line using the point-slope formula. Write all the final equations using the slope-intercept form.

( 0 , 3 ) ( 0 , 3 ) with a slope of 2 3 2 3

( 1 , 2 ) ( 1 , 2 ) with a slope of − 4 5 − 4 5

x -intercept is 1, and ( −2 , 6 ) ( −2 , 6 )

y -intercept is 2, and ( 4 , −1 ) ( 4 , −1 )

( −3 , 10 ) ( −3 , 10 ) and ( 5 , −6 ) ( 5 , −6 )

( 1 , 3 )  and   ( 5 , 5 ) ( 1 , 3 )  and   ( 5 , 5 )

parallel to y = 2 x + 5 y = 2 x + 5 and passes through the point ( 4 , 3 ) ( 4 , 3 )

perpendicular to 3 y = x − 4 3 y = x − 4 and passes through the point ( −2 , 1 ) ( −2 , 1 ) .

For the following exercises, find the equation of the line using the given information.

( − 2 , 0 ) ( − 2 , 0 ) and ( −2 , 5 ) ( −2 , 5 )

( 1 , 7 ) ( 1 , 7 ) and ( 3 , 7 ) ( 3 , 7 )

The slope is undefined and it passes through the point ( 2 , 3 ) . ( 2 , 3 ) .

The slope equals zero and it passes through the point ( 1 , −4 ) . ( 1 , −4 ) .

The slope is 3 4 3 4 and it passes through the point ( 1 , 4 ) ( 1 , 4 ) .

( –1 , 3 ) ( –1 , 3 ) and ( 4 , –5 ) ( 4 , –5 )

For the following exercises, graph the pair of equations on the same axes, and state whether they are parallel, perpendicular, or neither.

y = 2 x + 7 y = − 1 2 x − 4 y = 2 x + 7 y = − 1 2 x − 4

3 x − 2 y = 5 6 y − 9 x = 6 3 x − 2 y = 5 6 y − 9 x = 6

y = 3 x + 1 4 y = 3 x + 2 y = 3 x + 1 4 y = 3 x + 2

x = 4 y = −3 x = 4 y = −3

For the following exercises, find the slope of the line that passes through the given points.

( 5 , 4 ) ( 5 , 4 ) and ( 7 , 9 ) ( 7 , 9 )

( −3 , 2 ) ( −3 , 2 ) and ( 4 , −7 ) ( 4 , −7 )

( −5 , 4 ) ( −5 , 4 ) and ( 2 , 4 ) ( 2 , 4 )

( −1 , −2 ) ( −1 , −2 ) and ( 3 , 4 ) ( 3 , 4 )

( 3 , −2 ) ( 3 , −2 ) and ( 3 , −2 ) ( 3 , −2 )

For the following exercises, find the slope of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular.

( −1 , 3 )  and   ( 5 , 1 ) ( −2 , 3 )  and   ( 0 , 9 ) ( −1 , 3 )  and   ( 5 , 1 ) ( −2 , 3 )  and   ( 0 , 9 )

( 2 , 5 )  and   ( 5 , 9 ) ( −1 , −1 )  and   ( 2 , 3 ) ( 2 , 5 )  and   ( 5 , 9 ) ( −1 , −1 )  and   ( 2 , 3 )

For the following exercises, express the equations in slope intercept form (rounding each number to the thousandths place). Enter this into a graphing calculator as Y1, then adjust the ymin and ymax values for your window to include where the y -intercept occurs. State your ymin and ymax values.

0.537 x − 2.19 y = 100 0.537 x − 2.19 y = 100

4,500 x − 200 y = 9,528 4,500 x − 200 y = 9,528

200 − 30 y x = 70 200 − 30 y x = 70

Starting with the point-slope formula y − y 1 = m ( x − x 1 ) , y − y 1 = m ( x − x 1 ) , solve this expression for x x in terms of x 1 , y , y 1 , x 1 , y , y 1 , and m m .

Starting with the standard form of an equation A x + B y = C A x + B y = C solve this expression for y y in terms of A , B , C A , B , C and x x . Then put the expression in slope-intercept form.

Use the above derived formula to put the following standard equation in slope intercept form: 7 x − 5 y = 25. 7 x − 5 y = 25.

Given that the following coordinates are the vertices of a rectangle, prove that this truly is a rectangle by showing the slopes of the sides that meet are perpendicular.

( – 1 , 1 ) , ( 2 , 0 ) , ( 3 , 3 ) ( – 1 , 1 ) , ( 2 , 0 ) , ( 3 , 3 ) and ( 0 , 4 ) ( 0 , 4 )

Find the slopes of the diagonals in the previous exercise. Are they perpendicular?

Real-World Applications

The slope for a wheelchair ramp for a home has to be 1 12 . 1 12 . If the vertical distance from the ground to the door bottom is 2.5 ft, find the distance the ramp has to extend from the home in order to comply with the needed slope.

If the profit equation for a small business selling x x number of item one and y y number of item two is p = 3 x + 4 y , p = 3 x + 4 y , find the y y value when p = $ 453 and   x = 75. p = $ 453 and   x = 75.

For the following exercises, use this scenario: The cost of renting a car is $45/wk plus $0.25/mi traveled during that week. An equation to represent the cost would be y = 45 + .25 x , y = 45 + .25 x , where x x is the number of miles traveled.

What is your cost if you travel 50 mi?

If your cost were $ 63.75 , $ 63.75 , how many miles were you charged for traveling?

Suppose you have a maximum of $100 to spend for the car rental. What would be the maximum number of miles you could travel?

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: College Algebra 2e
• Publication date: Dec 21, 2021
• Location: Houston, Texas
• Book URL: https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/college-algebra-2e/pages/2-2-linear-equations-in-one-variable

© Jan 9, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

NEW CHAPTER 6 Linear Equations and Inequalities in One Variable

6.1.1 solving linear equations in one variable, solving linear equations, learning objectives.

• Determine whether or not an equation in one variable is defined as a linear equation
• Use the addition and subtraction property to solve a linear equation
• Use the multiplication and division property to solve a linear equation
• Equality : The state of two or more entities having the same value.
• Equation : a statement that two expressions are equal.
• Solution : a number or numbers that make an equation true.
• Equivalent equations: two or more equations that have identical solutions.

Linear Equations

An equation with just one variable is said to be  linear when the highest power on the variable is [latex]1[/latex]. Remember that [latex]{x}^{1}[/latex] is equivalent to [latex]x[/latex], so any equation that can be simplified to [latex]ax+b=c[/latex] (where [latex]a,b,c[/latex] are real numbers) is a linear equation in one variable . For example, the equation [latex]2x+3=7, \;5x=7,\; -\frac{3}{4}x+\frac{7}{3}=\frac{6}{5}[/latex] are all linear equations because the highest power of [latex]x[/latex] is [latex]1[/latex]. On the other hand, the equation [latex]{x}^3+2{x}^2=4x-2[/latex] is NOT linear because the highest power of [latex]x[/latex] is [latex]3[/latex], not [latex]1[/latex].

Determine whether the equation is linear:

a) [latex]3x-7=6x+2[/latex]

b) [latex]{x}^{3}-5{x}^{2}=5x+6[/latex]

c) [latex]x(x+6)=8[/latex]

d) [latex]x-\frac{3}{8} =\frac{1}{2}[/latex]

Determine whether the equation [latex]6x-7=8{x}^{2}+5[/latex] is linear.

[latex]6x-7=8{x}^{2}+5[/latex] is NOT linear since the highest power of [latex]x[/latex] is [latex]2[/latex], not [latex]1[/latex].

Determine whether the equation [latex]\frac{4}{5}x+6=\frac{-7}{3}x[/latex] is linear.

[latex]\frac{4}{5}x+6=\frac{-7}{3}x[/latex] is linear since the highest power of [latex]x[/latex] is [latex]1[/latex].

The Addition and Subtraction Property of Equality

When an equation contains a variable such as [latex]x[/latex], this variable is considered an unknown value. In many cases, we can find the possible values for [latex]x[/latex] that would make the equation true. We can  solve  the equation for [latex]x[/latex].

For some equations like [latex]x + 3 =5[/latex] it is easy to guess the solution: the only possible value of [latex]x[/latex] is [latex]2[/latex], because [latex]2 + 3 = 5[/latex]. However, it becomes useful to have a process for finding solutions for unknowns as problems become more complex.

In order to solve an equation, we need to  isolate the variable . Isolating the variable means rewriting an equivalent equation in which the variable is on one side of the equation and everything else is on the other side of the equation. The variable is an unknown quantity whose value we are trying to find. We have a solution when we reorganize the equation into the form [latex]x[/latex] = a constant.

The second we add or subtract a “weight” to one side of the scale, it will become unbalanced. In order to right that balance, we must add or subtract the same “weight” to the other side of the sale.

To solve this equation for [latex]x[/latex], we need to isolate [latex]x[/latex] by removing the [latex]+3[/latex] from the left side of the equation. We do this this by “undoing” the addition of [latex]+3[/latex] by subtracting [latex]3[/latex].  In other words we use the inverse operation of addition, which is subtraction. But if we subtract [latex]3[/latex] from the left side of the equation, we must also subtract [latex]3[/latex] from the right side to keep the equation balanced.

[latex]x+3\;\;\;\;\;\;=5 \\ x+3\color{blue}{-3}=5\color{blue}{-3} \\ x\;\;\;\;\;\;\;\;\;\;\;\;\,=2[/latex]

We can verify that [latex]x=2[/latex] is indeed the solution by substituting [latex]2[/latex] in for [latex]x[/latex] in the original equation: [latex]x+3=5 \\ 2+3=5 \\ 5=5[/latex] Since this is true, [latex]x=2[/latex] is the solution of [latex]x+3=5[/latex].

If the equation is [latex]x-5=9[/latex], we isolate the variable by adding [latex]5[/latex] to both sides of the equation. This is because, to isolate [latex]x[/latex] we must “undo” subtract [latex]5[/latex] by adding [latex]5[/latex]: 

[latex]x-5=9 \\ x-5\color{blue}{+5}=9\color{blue}{+5} \\ x\;\;\;\;\;\;\;\;\;\;\;\;\,=14[/latex]

Again, we can verify the solution: [latex]x-5\;\;\;\;\;\;=9 \\ 14 - 5 = 9 \\ 9=9[/latex] True.

When we add or subtract the same term to both sides of an equation, we get an equivalent equation . Equivalent equations are two or more equations that have identical solutions. This leads us to the the Addition and Subtraction Property of Equality .

Addition & subtraction Property of Equality

For all real numbers [latex]a,b[/latex], and [latex]c[/latex], if [latex]a=b[/latex], then [latex]a+c=b+c[/latex] and [latex]a-c=b-c[/latex].

Adding or subtracting the same term to both sides of an equation will result in an equivalent equation .

Solve: [latex]x+11=-3[/latex]

Solution: To isolate [latex]x[/latex], we undo the addition of [latex]11[/latex] by using the Subtraction Property of Equality.

Since [latex]x=-14[/latex] makes [latex]x+11=-3[/latex] a true statement, we know that it is a solution to the equation.

https://ohm.lumenlearning.com/multiembedq.php?id=141721&theme=oea&iframe_resize_id=mom200

Solve: [latex]m - 4=-5[/latex]

https://ohm.lumenlearning.com/multiembedq.php?id=141723&theme=oea&iframe_resize_id=mom27

Linear Equations Containing Fractions

It is not uncommon to encounter equations that contain fractions; therefore, in the following examples, we will demonstrate how to use the addition property of equality to solve an equation with fractions.

Solve: [latex]n-\frac{3}{8} =\frac{1}{2}[/latex]

https://ohm.lumenlearning.com/multiembedq.php?id=141732&theme=oea&iframe_resize_id=mom25

Watch this video for more examples of solving equations that include fractions and require addition or subtraction.

Linear Equations Containing Decimals

Decimals will be encountered any time money or metric measurements are used.

Solve [latex]a - 3.7=4.3[/latex]

https://ohm.lumenlearning.com/multiembedq.php?id=141736&theme=oea&iframe_resize_id=mom29

The Multiplication and Division Properties of Equality

Just as we can add or subtract the same exact quantity on both sides of an equation, we can also multiply or divide both sides of an equation by the same quantity to write an equivalent equation. This makes sense because multiplication is just repeated addition and division is multiplication by a reciprocal.

For example, to solve the equation [latex]3x=15[/latex] we need to isolate the variable [latex]x[/latex] which is multiplied by [latex]3[/latex]. To “undo” this multiplication, we divide both sides of the equation by [latex]3[/latex]:  [latex]\frac{3x}\color{blue}{{3}}=\frac{15}\color{blue}{{3}}[/latex]. Since [latex]\frac{3}{3}=1[/latex] and [latex]1x=x[/latex], this simplifies to [latex]x=5[/latex] and we have our solution.

On the other hand, in the equation [latex]\frac{1}{2}x=-5[/latex] the variable [latex]x[/latex] is multiplied by [latex]\frac{1}{2}[/latex]. To isolate the variable we need to turn [latex]\frac{1}{2}[/latex] into [latex]1[/latex] by multiplying by the reciprocal [latex]2[/latex]:     [latex]\color{blue}{2}\left(\frac{1}{2}x\right)=\color{blue}{2}(-5)[/latex]. This simplifies to [latex]x=-10[/latex] and we have our solution.

This characteristic of equations is generalized in the M ultiplication & Division Property of Equality .

multiplication & Division Property of Equality

For all real numbers [latex]a,b,c[/latex], and [latex]c\ne 0[/latex], if [latex]a=b[/latex], then [latex]ac=bc[/latex] and [latex]\Large\frac{a}{c}\normalsize =\Large\frac{b}{c}[/latex].

Multiplying or dividing the same non-zero* term to both sides of an equation will result in an equivalent equation .

*Technically we could multiply both sides of an equation by zero but that would wipe out our entire equation to [latex]0=0[/latex]. However, we can never divide by zero as that is undefined.

When the equation involves multiplication or division, we can “undo” these operations by using the inverse operation to isolate the variable.

Solve [latex]3x=24[/latex]. When you are done, check your solution.

Divide both sides of the equation by [latex]3[/latex] to isolate the variable (this is will give you a coefficient of  [latex]1[/latex]). Dividing by [latex]3[/latex] is the same as multiplying by [latex] \frac{1}{3}[/latex].

[latex]\begin{array}{r}\underline{3x}=\underline{24}\\3\,\,\,\,\,\,\,\,\,\,\,\,3\,\\x=8\,\,\,\end{array}[/latex]

Check by substituting your solution, [latex]8[/latex], for the variable in the original equation.

[latex]\begin{array}{r}3x=24 \\ 3\cdot8=24 \\ 24=24\end{array}[/latex]

The solution is correct!

[latex]x=8[/latex]

Solve: [latex]\frac{a}{-7} =-42[/latex]

Solution Here [latex]a[/latex] is divided by [latex]-7[/latex]. We can multiply both sides by [latex]-7[/latex] to isolate [latex]a[/latex].

https://ohm.lumenlearning.com/multiembedq.php?id=141868&theme=oea&iframe_resize_id=mom21

Another way to think about solving an equation when the operation is multiplication or division is that we want to multiply the coefficient by the multiplicative inverse (reciprocal) in order to change the coefficient to [latex]1[/latex].

In the following example, we change the coefficient to  [latex]1[/latex] by multiplying by the multiplicative inverse of [latex]\frac{1}{2}[/latex].

Solve [latex] \frac{1}{2}x=8[/latex] for [latex]x[/latex].

We can multiply both sides by the reciprocal of [latex]\frac{1}{2}[/latex], which is [latex]\frac{2}{1}[/latex].

[latex]\begin{array}{r}\color{blue}{\left(\frac{2}{1}\right)}\frac{1}{2 }{ x }=\color{blue}{\left(\frac{2}{1}\right)}8\\(1)x= 16\,\,\,\,\\{ x }=16\,\,\,\,\,\end{array}[/latex]

The video shows examples of how to use the Multiplication and Division Property of Equality to solve one-step equations with integers and fractions.

Solve: [latex]4x=-28[/latex]

To solve this equation, we use the Division Property of Equality to divide both sides by [latex]4[/latex].

Since this is a true statement, [latex]x=-7[/latex] is a solution to [latex]4x=-28[/latex].

https://ohm.lumenlearning.com/multiembedq.php?id=141857&theme=oea&iframe_resize_id=mom2

As we solve equations that require several steps, it is not unusual to end up with an equation that looks like the one in the next example, with a negative sign in front of the variable.  

Solve: [latex]-r=2[/latex]

Solution: Remember [latex]-r[/latex] is equivalent to [latex]-1r[/latex].

https://ohm.lumenlearning.com/multiembedq.php?id=141865&theme=oea&iframe_resize_id=mom22

The equations [latex]x=4[/latex] and [latex]4=x[/latex] are equivalent. Both say that [latex]x[/latex] is equal to [latex]4[/latex]. This is an example of The Reflection Property of Equality .

The Reflection Property of Equality

If [latex]a=b[/latex], then [latex]b=a[/latex], for all real numbers [latex]a[/latex] and [latex]b[/latex].

This implies that it does not matter which side of the equation the variable term ends up on.

The next video includes examples of using the division and multiplication properties to solve equations with the variable on the right side of the equal sign.

Two-Step Linear Equations

If the equation is in the form [latex]ax+b=c[/latex], where [latex]x[/latex] is the variable, we can solve the equation as before. First we must isolate the [latex]x-[/latex]term by “undoing” the addition or subtraction. Then we isolate the variable by “undoing” the multiplication or division.

1. Solve: [latex]4x+6=-14[/latex]

In this equation, the variable is only on the left side. It makes sense to call the left side the variable side. Therefore, the right side will be the constant side.

2. Solve: [latex]2y - 7=15[/latex]

Solution: Notice that the variable is only on the left side of the equation, so this will be the variable side, and the right side will be the constant side. Since the left side is the variable side, the [latex]7[/latex] is out of place. It is subtracted from the [latex]2y[/latex], so to “undo” subtraction, add [latex]7[/latex] to both sides.

Solve [latex]3y+2=11[/latex]

Subtract [latex]2[/latex] from both sides of the equation to get the term with the variable by itself.

[latex] \displaystyle \begin{array}{r}3y+2\,\,\,=\,\,11\\\underline{\,\,\,\,\,\,\,\color{blue}{-2}\,\,\,\,\,\,\,\,\color{blue}{-2}}\\3y\,\,\,\,=\,\,\,\,\,9\end{array}[/latex]

Divide both sides of the equation by [latex]3[/latex] to get a coefficient of [latex]1[/latex] for the variable.

[latex]\begin{array}{r}\,\,\,\,\,\,\underline{3y}\,\,\,\,=\,\,\,\,\,\underline{9}\\3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9\\\,\,\,\,\,\,\,\,\,\,y\,\,\,\,=\,\,\,\,3\end{array}[/latex]

[latex]y=3[/latex]

https://ohm.lumenlearning.com/multiembedq.php?id=142131&theme=oea&iframe_resize_id=mom2

In the following video, we show examples of solving two step linear equations.

Remember to check the solution of an algebraic equation by substituting the value of the variable into the original equation.

Privacy Policy

Solving Linear Equations: Rich Tasks

You cannot beat a good rich task! For me, a rich task is one that both stimulates and challenges students of all ages and abilities. Here is a selection of some of my favourites.

Median Rich Tasks and Purposeful Practise keyboard_arrow_up Back to Top

There are no words to express how much I love Don Steward and his amazing resources on his Median Blog . They allow for purposeful practise, to enable students to gain fluency in key skills. But they also have more than a sprinkling of richness, that leads students merrily along the path towards hypotheses and generalisations. Amazing.

Forming and Solving

Solving practice, nrich keyboard_arrow_up back to top.

NRICH is simply one of the best websites for rich maths problems in the whole wide world.

Solver Title

Generating PDF...

• Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Number Line Mean, Median & Mode
• Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi (Product) Notation Induction Logical Sets Word Problems
• Pre Calculus Equations Inequalities Scientific Calculator Scientific Notation Arithmetics Complex Numbers Polar/Cartesian Simultaneous Equations System of Inequalities Polynomials Rationales Functions Arithmetic & Comp. Coordinate Geometry Plane Geometry Solid Geometry Conic Sections Trigonometry
• Calculus Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Fourier Transform
• Functions Line Equations Functions Arithmetic & Comp. Conic Sections Transformation
• Linear Algebra Matrices Vectors
• Trigonometry Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify
• Statistics Mean Geometric Mean Quadratic Mean Average Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge Standard Normal Distribution
• Physics Mechanics
• Chemistry Chemical Reactions Chemical Properties
• Finance Simple Interest Compound Interest Present Value Future Value
• Economics Point of Diminishing Return
• Conversions Roman Numerals Radical to Exponent Exponent to Radical To Fraction To Decimal To Mixed Number To Improper Fraction Radians to Degrees Degrees to Radians Hexadecimal Scientific Notation Distance Weight Time Volume
• Pre Algebra
• One-Step Addition
• One-Step Subtraction
• One-Step Multiplication
• One-Step Division
• One-Step Decimals
• Two-Step Integers
• Two-Step Add/Subtract
• Two-Step Multiply/Divide
• Two-Step Fractions
• Two-Step Decimals
• Multi-Step Integers
• Multi-Step with Parentheses
• Multi-Step Rational
• Multi-Step Fractions
• Multi-Step Decimals
• Solve by Factoring
• Completing the Square
• Quadratic Formula
• Biquadratic
• Logarithmic
• Exponential
• Rational Roots
• Floor/Ceiling
• Equation Given Roots
• Newton Raphson
• Substitution
• Elimination
• Cramer's Rule
• Gaussian Elimination
• System of Inequalities
• Perfect Squares
• Difference of Squares
• Difference of Cubes
• Sum of Cubes
• Polynomials
• Distributive Property
• FOIL method
• Perfect Cubes
• Binomial Expansion
• Negative Rule
• Product Rule
• Quotient Rule
• Expand Power Rule
• Fraction Exponent
• Exponent Rules
• Exponential Form
• Logarithmic Form
• Absolute Value
• Rational Number
• Powers of i
• Partial Fractions
• Is Polynomial
• Leading Coefficient
• Leading Term
• Standard Form
• Complete the Square
• Synthetic Division
• Linear Factors
• Rationalize Denominator
• Rationalize Numerator
• Identify Type
• Convergence
• Interval Notation
• Pi (Product) Notation
• Boolean Algebra
• Truth Table
• Mutual Exclusive
• Cardinality
• Caretesian Product
• Age Problems
• Distance Problems
• Cost Problems
• Investment Problems
• Number Problems
• Percent Problems
• Addition/Subtraction
• Multiplication/Division
• Dice Problems
• Coin Problems
• Card Problems
• Pre Calculus
• Linear Algebra
• Trigonometry
• Conversions

Most Used Actions

Number line.

• x^4-5x^2+4=0
• \sqrt{x-1}-x=-7
• \left|3x+1\right|=4
• \log _2(x+1)=\log _3(27)
• 3^x=9^{x+5}
• What is the completing square method?
• Completing the square method is a technique for find the solutions of a quadratic equation of the form ax^2 + bx + c = 0. This method involves completing the square of the quadratic expression to the form (x + d)^2 = e, where d and e are constants.
• What is the golden rule for solving equations?
• The golden rule for solving equations is to keep both sides of the equation balanced so that they are always equal.
• How do you simplify equations?
• To simplify equations, combine like terms, remove parethesis, use the order of operations.
• How do you solve linear equations?
• To solve a linear equation, get the variable on one side of the equation by using inverse operations.

equation-calculator

• High School Math Solutions – Radical Equation Calculator Radical equations are equations involving radicals of any order. We will show examples of square roots; higher...

Please add a message.

Message received. Thanks for the feedback.

Python Numerical Methods

This notebook contains an excerpt from the Python Programming and Numerical Methods - A Guide for Engineers and Scientists , the content is also available at Berkeley Python Numerical Methods .

The copyright of the book belongs to Elsevier. We also have this interactive book online for a better learning experience. The code is released under the MIT license . If you find this content useful, please consider supporting the work on Elsevier or Amazon !

< 14.4 Solutions to Systems of Linear Equations | Contents | 14.6 Matrix Inversion >

Solve Systems of Linear Equations in Python ¶

Though we discussed various methods to solve the systems of linear equations, it is actually very easy to do it in Python. In this section, we will use Python to solve the systems of equations. The easiest way to get a solution is via the solve function in Numpy.

TRY IT! Use numpy.linalg.solve to solve the following equations.

We can see we get the same results as that in the previous section when we calculated by hand. Under the hood, the solver is actually doing a LU decomposition to get the results. You can check the help of the function, it needs the input matrix to be square and of full-rank, i.e., all rows (or, equivalently, columns) must be linearly independent.

TRY IT! Try to solve the above equations using the matrix inversion approach.

We can also get the \(L\) and \(U\) matrices used in the LU decomposition using the scipy package.

TRY IT! Get the \(L\) and \(U\) for the above matrix A.

We can see the \(L\) and \(U\) we get are different from the ones we got in the last section by hand. You will also see there is a permutation matrix \(P\) that returned by the lu function. This permutation matrix record how do we change the order of the equations for easier calculation purposes (for example, if first element in first row is zero, it can not be the pivot equation, since you can not turn the first elements in other rows to zero. Therefore, we need to switch the order of the equations to get a new pivot equation). If you multiply \(P\) with \(A\) , you will see that this permutation matrix reverse the order of the equations for this case.

TRY IT! Multiply \(P\) and \(A\) and see what’s the effect of the permutation matrix on \(A\) .

Please ensure that your password is at least 8 characters and contains each of the following:

• a special character: @$#!%*?&

Help | Advanced Search

Mathematics > Probability

Title: well-posedness of the obstacle problem for stochastic nonlinear diffusion equations: an entropy formulation.

Abstract: In this paper, we establish the existence, uniqueness and stability results for the obstacle problem associated with a degenerate nonlinear diffusion equation perturbed by conservative gradient noise. Our approach revolves round introducing a new entropy formulation for stochastic variational inequalities. As a consequence, we obtain a novel well-posedness result for the obstacle problem of deterministic porous medium equations with nonlinear reaction terms.

Submission history

Access paper:.

• Other Formats

References & Citations

• Google Scholar
• Semantic Scholar

BibTeX formatted citation

Bibliographic and Citation Tools

Code, data and media associated with this article, recommenders and search tools.

• Institution

arXivLabs: experimental projects with community collaborators

arXivLabs is a framework that allows collaborators to develop and share new arXiv features directly on our website.

Both individuals and organizations that work with arXivLabs have embraced and accepted our values of openness, community, excellence, and user data privacy. arXiv is committed to these values and only works with partners that adhere to them.

Have an idea for a project that will add value for arXiv's community? Learn more about arXivLabs .

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons
• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

1.1: Solving Linear Equations and Inequalities

• Last updated
• Save as PDF
• Page ID 48347

Learning Objectives

• Verify linear solutions.
• Use the properties of equality to solve basic linear equations.
• Clear fractions from equations.
• Identify linear inequalities and check solutions.
• Solve linear inequalities and express the solutions graphically on a number line and in interval notation.

Prerequisite Skills

Before you get started, take this prerequisite quiz.

1. Simplify \(2-6(4-7)^2\) without using a calculator.

If you missed this problem, review here . (Note that this will open a different textbook in a new window.)

2. Evaluate \(6x−4\) when \(x=−2\).

3. Evaluate \(-5x^2−x+9\) when \(x=-3\).

4. Simplify \(7x−1−4x+5\).

\(3x+4\) 

Solving Basic Linear Equations

An equation 129 is a statement indicating that two algebraic expressions are equal. A linear equation with one variable 130 , \(x\), is an equation that can be written in the standard form \(ax + b = 0\) where \(a\) and \(b\) are real numbers and \(a ≠ 0\). For example

\(3 x - 12 = 0\)

A solution 131 to a linear equation is any value that can replace the variable to produce a true statement. The variable in the linear equation \(3x − 12 = 0\) is \(x\) and the solution is \(x = 4\). To verify this, substitute the value \(4\) in for \(x\) and check that you obtain a true statement.

\(\begin{aligned} 3 x - 12 & = 0 \\ 3 ( \color{Cerulean}{4}\color{Black}{ )} - 12 & = 0 \\ 12 - 12 & = 0 \\ 0 & = 0 \:\: \color{Cerulean}{✓} \end{aligned}\)

Alternatively, when an equation is equal to a constant, we may verify a solution by substituting the value in for the variable and showing that the result is equal to that constant. In this sense, we say that solutions “satisfy the equation.”

Example \(\PageIndex{1}\):

Is \(a=2\) a solution to \(−10a+5=−25\)?

Recall that when evaluating expressions, it is a good practice to first replace all variables with parentheses, and then substitute the appropriate values. By making use of parentheses, we avoid some common errors when working the order of operations.

\(\begin{align*} - 10 a + 5 &= -25 \\ - 10 ( \color{Cerulean}{2} \color{Black}{ ) +} 5 & = -25 \\ -20 + 5 & = -25 \\ -15 &\neq 25\:\: \color{red}{✗}\end{align*}\)  

No, \(a=2\) does not satisfy the equation and is therefore not a solution.

Developing techniques for solving various algebraic equations is one of our main goals in algebra. This section reviews the basic techniques used for solving linear equations with one variable. We begin by defining equivalent equations 132 as equations with the same solution set.

\(\left. \begin{aligned} 3 x - 5 & = 16 \\ 3 x & = 21 \\ x & = 7 \end{aligned} \right\} \quad \color{Cerulean}{Equivalent \:equations}\)

Here we can see that the three linear equations are equivalent because they share the same solution set, namely, \(\{7\}\). To obtain equivalent equations, use the following properties of equality 133 . Given algebraic expressions \(A\) and \(B\), where \(c\) is a nonzero number:

Table 1.1.1

Multiplying or dividing both sides of an equation by \(0\) is carefully avoided. Dividing by \(0\) is undefined and multiplying both sides by \(0\) results in the equation \(0 = 0\).

We solve algebraic equations by isolating the variable with a coefficient of 1. If given a linear equation of the form \(ax + b = c\), then we can solve it in two steps. First, use the appropriate equality property of addition or subtraction to isolate the variable term. Next, isolate the variable using the equality property of multiplication or division. Checking the solution in the following examples is left to the reader.

Example \(\PageIndex{2}\):

Solve: \(7x − 2 = 19\).

\(\begin{aligned} 7 x - 2 & = 19 \\ 7 x - 2 \color{Cerulean}{+ 2} & = 19 \color{Cerulean}{+ 2} & & \color{Cerulean}{Add\: 2\: to\: both\: sides.} \\ 7 x & = 21 \\ \frac { 7 x } { \color{Cerulean}{7} } & = \frac { 21 } { \color{Cerulean}{7} } & & \color{Cerulean}{Divide\: both\: sides\: by\: 7.} \\ x & = 3 \end{aligned}\)

The solution is \(3\).

Example \(\PageIndex{3}\):

Solve: \(56 = 8 + 12y\).

When no sign precedes the term, it is understood to be positive. In other words, think of this as \(56 = +8 + 12y\). Therefore, we begin by subtracting \(8\) on both sides of the equal sign.

\(\begin{aligned} 56 \color{Cerulean}{- 8} & = 8 + 12 y \color{Cerulean}{- 8} \\ 48 & = 12 y \\ \frac { 48 } { \color{Cerulean}{12} } & = \frac { 12 y } { \color{Cerulean}{12} } \\ 4 & = y \end{aligned}\)

It does not matter on which side we choose to isolate the variable because the symmetric property 134 states that \(4 = y\) is equivalent to \(y = 4\).

The solution is \(4\).

Example \(\PageIndex{4}\):

Solve: \(\frac { 5 } { 3 } x + 2 = - 8\).

Isolate the variable term using the addition property of equality, and then multiply both sides of the equation by the reciprocal of the coefficient \(\frac{5}{3}\) .

\begin{aligned} \frac { 5 } { 3 } x + 2 & = - 8 \\ \frac { 5 } { 3 } x + 2 \color{Cerulean}{- 2} & = - 8 \color{Cerulean}{- 2}\quad \color{Cerulean}{Subtract\: 2\: on\: both\: sides.} \\ \frac { 5 } { 3 } x & = - 10 \\ \color{Cerulean}{\frac { 3 } { 5 }} \color{Black}{ \cdot} \frac { 5 } { 3 } x & = \color{Cerulean}{\frac { 3 } { \cancel{5} }} \color{Black}{\cdot} ( \overset{-2}{\cancel{-10}} )\quad \color{Cerulean}{Multiply \:both \:sides\: by\: \frac{3}{5}.} \\ 1x & = 3 \cdot ( - 2 ) \\ x & = - 6 \end{aligned}

The solution is \(−6\).

In summary, to retain equivalent equations, we must perform the same operation on both sides of the equation.

Exercise \(\PageIndex{1}\)

Solve: \(\frac { 2 } { 3 } x + \frac { 1 } { 2 } = - \frac { 5 } { 6 }\).

Video Solution: www.youtube.com/v/cQwqXs9AD6M

General Guidelines for Solving Linear Equations

Typically linear equations are not given in standard form, and so solving them requires additional steps. When solving linear equations, the goal is to determine what value, if any, will produce a true statement when substituted in the original equation. Do this by isolating the variable using the following steps:

• Step 1: Simplify both sides of the equation using the order of operations and combine all like terms on the same side of the equal sign.
• Step 2a: Add or subtract as needed to isolate the variable.
• Step 2b: Divide or multiply as needed to isolate the variable.
• Step 3: Check to see if the answer solves the original equation.

We will often encounter linear equations where the expressions on each side of the equal sign can be simplified. If this is the case, then it is best to simplify each side first before solving. Normally this involves combining same-side like terms.

At this point in our study of algebra the use of the properties of equality should seem routine. Therefore, displaying these steps in this text, usually in blue, becomes optional.

Example \(\PageIndex{5}\):

Solve: \(- 4 a + 2 - a = 1\).

First combine the like terms on the left side of the equal sign.

\(\begin{aligned} - 4 a + 2 - a = 1 & \quad \color{Cerulean}{ Combine\: same-side\: like\: terms.} \\ - 5 a + 2 = 1 & \quad\color{Cerulean} { Subtract\: 2\: on\: both\: sides.} \\ - 5 a = - 1 & \quad\color{Cerulean} { Divide\: both\: sides\: by\: - 5.} \\ a = \frac { - 1 } { - 5 } = \frac { 1 } { 5 } \end{aligned}\)

Always use the original equation to check to see if the solution is correct.

\(\begin{aligned} - 4 a + 2 - a & = - 4 \left( \color{OliveGreen}{\frac { 1 } { 5 }} \right) + 2 - \color{OliveGreen}{\frac { 1 } { 5 }} \\ & = - \frac { 4 } { 5 } + \frac { 2 } { 1 } \cdot \color{Cerulean}{\frac { 5 } { 5 }}\color{Black}{ -} \frac { 1 } { 5 } \\ & = \frac { - 4 + 10 + 1 } { 5 } \\ & = \frac { 5 } { 5 } = 1 \:\:\color{Cerulean}{✓} \end{aligned}\)

The solution is \(\frac{1}{5}\) .

Given a linear equation in the form \(ax + b = cx + d\), we begin the solving process by combining like terms on opposite sides of the equal sign. To do this, use the addition or subtraction property of equality to place like terms on the same side so that they can be combined. In the examples that remain, the check is left to the reader.

Example \(\PageIndex{6}\):

Solve: \(−2y − 3 = 5y + 11\).

Subtract \(5y\) on both sides so that we can combine the terms involving y on the left side.

\(\begin{array} { c } { - 2 y - 3 \color{Cerulean}{- 5 y}\color{Black}{ =} 5 y + 11 \color{Cerulean}{- 5 y} } \\ { - 7 y - 3 = 11 } \end{array}\)

From here, solve using the techniques developed previously.

\(\begin{aligned} - 7 y - 3 & = 11 \quad\color{Cerulean}{Add\: 3\: to\: both\: sides.} \\ - 7 y & = 14 \\ y & = \frac { 14 } { - 7 } \quad\color{Cerulean}{Divide\: both\: sides\: by\: -7.} \\ y & = - 2 \end{aligned}\)

The solution is \(−2\).

Solving will often require the application of the distributive property.

Example \(\PageIndex{7}\):

Solve: \(- \frac { 1 } { 2 } ( 10 x - 2 ) + 3 = 7 ( 1 - 2 x )\).

Simplify the linear expressions on either side of the equal sign first.

\(\begin{aligned} - \frac { 1 } { 2 } ( 10 x - 2 ) + 3 = 7 ( 1 - 2 x ) & \quad\color{Cerulean} { Distribute } \\ - 5 x + 1 + 3 = 7 - 14 x & \quad\color{Cerulean} { Combine\: same-side\: like\: terms. } \\ - 5 x + 4 = 7 - 14 x & \quad\color{Cerulean} { Combine\: opposite-side\: like\: terms. } \\ 9 x = 3 & \quad\color{Cerulean} { Solve. } \\ x = \frac { 3 } { 9 } = \frac { 1 } { 3 } \end{aligned}\)

The solution is \(\frac{1}{3}\) .

Example \(\PageIndex{8}\):

Solve: \(5(3−a)−2(5−2a)=3\).

Begin by applying the distributive property.

\(\begin{aligned} 5 ( 3 - a ) - 2 ( 5 - 2 a ) & = 3 \\ 15 - 5 a - 10 + 4 a & = 3 \\ 5 - a & = 3 \\ - a & = - 2 \end{aligned}\)

Here we point out that \(−a\) is equivalent to \(−1a\); therefore, we choose to divide both sides of the equation by \(−1\).

\(\begin{array} { c } { - a = - 2 } \\ { \frac { - 1 a } { \color{Cerulean}{- 1} }\color{Black}{ =} \frac { - 2 } { \color{Cerulean}{- 1} } } \\ { a = 2 } \end{array}\)

Alternatively, we can multiply both sides of \(−a=−2\) by negative one and achieve the same result.

\(\begin{aligned} - a & = - 2 \\ \color{Cerulean}{( - 1 )}\color{Black}{ (} - a ) & = \color{Cerulean}{( - 1 )}\color{Black}{ (} - 2 ) \\ a & = 2 \end{aligned}\)

The solution is \(2\).

Exercise \(\PageIndex{2}\)

Solve: \(6 - 3 ( 4 x - 1 ) = 4 x - 7\).

Video Solution: www.youtube.com/v/NAIAZrFjU-o

The coefficients of linear equations may be any real number, even decimals and fractions. When this is the case it is possible to use the multiplication property of equality to clear the fractional coefficients and obtain integer coefficients in a single step. If given fractional coefficients, then multiply both sides of the equation by the least common multiple of the denominators (LCD).

Example \(\PageIndex{9}\):

Solve: \(\frac { 1 } { 3 } x + \frac { 1 } { 5 } = \frac { 1 } { 5 } x - 1\).

Clear the fractions by multiplying both sides by the least common multiple of the given denominators. In this case, it is the \(LCD (3, 5) = 15\).

\(\begin{aligned} \color{Cerulean}{15}\color{Black}{ \cdot} \left( \frac { 1 } { 3 } x + \frac { 1 } { 5 } \right) & = \color{Cerulean}{15}\color{Black}{ \cdot} \left( \frac { 1 } { 5 } x - 1 \right) \quad \color{Cerulean}{Multiply\: both\: sides\: by\: 15.} \\ \color{Cerulean}{15}\color{Black}{ \cdot} \frac { 1 } { 3 } x + \color{Cerulean}{15}\color{Black}{ \cdot} \frac { 1 } { 5 } & = \color{Cerulean}{15}\color{Black}{ \cdot} \frac { 1 } { 5 } x - \color{Cerulean}{15}\color{Black}{ \cdot} 1\quad\color{Cerulean}{Simplify.} \\ 5 x + 3 & = 3 x - 15\quad\quad\quad\color{Cerulean}{Solve.} \\ 2 x & = - 18 \\ x & = \frac { - 18 } { 2 } = - 9 \end{aligned}\)

The solution is \(−9\).

It is important to know that this technique only works for equations. Do not try to clear fractions when simplifying expressions. As a reminder:

Table 1.1.2

We simplify expressions and solve equations. If you multiply an expression by \(6\), you will change the problem. However, if you multiply both sides of an equation by \(6\), you obtain an equivalent equation.

Table 1.1.3

Applications Involving Linear Equations

Algebra simplifies the process of solving real-world problems. This is done by using letters to represent unknowns, restating problems in the form of equations, and by offering systematic techniques for solving those equations. To solve problems using algebra, first translate the wording of the problem into mathematical statements that describe the relationships between the given information and the unknowns. Usually, this translation to mathematical statements is the difficult step in the process. The key to the translation is to carefully read the problem and identify certain key words and phrases.

Table 1.1.4

When translating sentences into mathematical statements, be sure to read the sentence several times and parse out the key words and phrases. It is important to first identify the variable, “ let x represent… ” and state in words what the unknown quantity is. This step not only makes our work more readable, but also forces us to think about what we are looking for.

Example \(\PageIndex{10}\):

When \(6\) is subtracted from twice the sum of a number and \(8\) the result is \(5\). Represent this as an algebraic equation and find the number.

Let n represent the unknown number.

To understand why we included the parentheses in the set up, you must study the structure of the following two sentences and their translations:

Table 1.1.5

The key was to focus on the phrase “ twice the sum ,” this prompted us to group the sum within parentheses and then multiply by \(2\). After translating the sentence into a mathematical statement we then solve.

\(\begin{aligned} 2 ( n + 8 ) - 6 & = 5 \\ 2 n + 16 - 6 & = 5 \\ 2 n + 10 & = 5 \\ 2 n & = - 5 \\ n & = \frac { - 5 } { 2 } \end{aligned}\)

\(\begin{aligned} 2 ( n + 8 ) - 6 & = 2 \left( \color{Cerulean}{- \frac { 5 } { 2 }}\color{Black}{ +} 8 \right) - 6 \\ & = 2 \left( \frac { 11 } { 2 } \right) - 6 \\ & = 11 - 6 \\ & = 5 \quad\color{Cerulean}{✓}\end{aligned}\)       

The number is \(−\frac{5}{2}\).

General guidelines for setting up and solving word problems follow.

• Step 1: Read the problem several times, identify the key words and phrases, and organize the given information.
• Step 2: Identify the variables by assigning a letter or expression to the unknown quantities.
• Step 3: Translate and set up an algebraic equation that models the problem.
• Step 4: Solve the resulting algebraic equation.
• Step 5: Finally, answer the question in sentence form and make sure it makes sense (check it).

For now, set up all of your equations using only one variable. Avoid two variables by looking for a relationship between the unknowns.

Linear Inequalities

A linear inequality 138 is a mathematical statement that relates a linear expression as either less than or greater than another. The following are some examples of linear inequalities, all of which are solved in this section:

A solution to a linear inequality 139 is a real number that will produce a true statement when substituted for the variable.

Example \(\PageIndex{11}\):

Are \(x=−4\) and \(x=6\) solutions to \(5x+7<22\)?

Substitute the values in for \(x\), simplify, and check to see if we obtain a true statement.

Table 1.1.6

\(x=−4\) is a solution and \(x=6\) is not

Linear inequalities have either infinitely many solutions or no solution. If there are infinitely many solutions, we graph the solution set on a number line and/or express the solution using interval notation.

Expressing Solutions to Linear Inequalities

What number would make the inequality \(x>3\) true? Are you thinking, "\(x\) could be four"? That’s correct, but \(x\) could be 6, too, or 137, or even 3.0001. Any number greater than three is a solution to the inequality \(x>3\). We show all the solutions to the inequality \(x>3\) 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 \(x>3\) as \((3,\infty)\). The symbol \(\infty\) is read as “ infinity .” It is not an actual number. Figure \(\PageIndex{2}\) 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, [, would show that the endpoint is included.

The inequality \(x\leq 1\) 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 \(x=1\). We then shade in all the numbers to the left of one, to show that all numbers less than one are solutions (Figure \(\PageIndex{3}\)). There is no lower end to those numbers. We write \(x\leq 1x\leq 1 \)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.

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. Notice that \(\infty\) and \(-\infty\) always use parentheses in interval notation, never brackets.

Example \(\PageIndex{12}\)

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

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

Exercise \(\PageIndex{3}\)

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

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

Solving Linear Inequalities

All but one of the techniques learned for solving linear equations apply to solving linear inequalities. You may add or subtract any real number to both sides of an inequality, and you may multiply or divide both sides by any positive real number to create equivalent inequalities. For example:

\[\begin{align*} 10 &> - 5 \\[4pt] 10{\color{Cerulean}{-7}}\,&{\color{Black}{>}} -5{\color{Cerulean}{-7}} & & {\color{Cerulean}{Subtract\: 7\: on\: both\: sides.}}\\[4pt]  3 &> - 12 & & \color{Cerulean}{✓}\quad\color{Cerulean}{True.} \\[20pt] 10 &>-5\\[4pt] \frac{10}{\color{Cerulean}{5}}\,&\color{Black}{>}\frac{-5}{\color{Cerulean}{5}} & & \color{Cerulean}{Divide\: both\: sides\: by\: 5.}\\[4pt] 2 &>-1 & & \color{Cerulean}{✓\:\:True} \end{align*} \]

Subtracting \(7\) from each side and dividing each side by positive \(5\) results in an inequality that is true.

Example \(\PageIndex{13}\):

Solve and graph the solution set: \(5x+7<22\).

\(\begin{array} { c } { 5 x + 7 < 22 } \\ { 5 x + 7 \color{Cerulean}{- 7}\color{Black}{ < 22}\color{Cerulean}{ - 7} } \\ { 5 x < 15 } \\ { \frac { 5 x } {\color{Cerulean}{ 5} } < \frac { 15 } { \color{Cerulean}{5} } } \\ { x < 3 } \end{array}\)

It is helpful to take a minute and choose a few values in and out of the solution set, substitute them into the original inequality, and then verify the results. As indicated, you should expect \(x=0\) to solve the original inequality and that \(x=5\) should not.

Table 1.1.7

Checking in this manner gives us a good indication that we have solved the inequality correctly.

We can express this solution in two ways: using set notation and interval notation.

\(\begin{array} { r } { \{ x | x < 3 \} } &\color{Cerulean}{Set\: notation} \\ { ( - \infty , 3 ) } &\color{Cerulean}{Interval\: notation} \end{array}\)

In this text we will choose to present answers using interval notation.

\((−∞,  3) \)

When working with linear inequalities, a different rule applies when multiplying or dividing by a negative number. To illustrate the problem, consider the true statement \(10 > −5\) and divide both sides by \(−5\).

\(\begin{array} { l } { 10 > - 5 } \\ { \frac { 10 } { \color{Cerulean}{- 5} } \color{Black}{>} \frac { - 5 } { \color{Cerulean}{- 5} } } \quad \color{Cerulean}{Divide\: both\: sides\: by\: -5.} \\ { - 2 \color{red}{>}\color{Black}{ 1} \quad \color{red}{✗} \color{Cerulean}{ False } } \end{array}\)

Dividing by \(−5\) results in a false statement. To retain a true statement, the inequality must be reversed.

\(\begin{array} { l } { 10 \color{OliveGreen}{>}\color{Black}{ - 5} } \\ { \frac { 10 } { \color{Cerulean}{- 5} } \color{Black}{<} \frac { - 5 } { \color{Cerulean}{- 5} } } \quad \color{Cerulean}{Reverse\: the\: inequality.} \\ { - 2 \color{OliveGreen}{<}\color{Black}{ 1} \quad \color{Cerulean}{✓} \color{Cerulean}{ True } } \end{array}\)

The same problem occurs when multiplying by a negative number. This leads to the following new rule: when multiplying or dividing by a negative number, reverse the inequality . It is easy to forget to do this so take special care to watch for negative coefficients. In general, given algebraic expressions \(A\) and \(B\), where \(c\) is a positive nonzero real number, we have the following properties of inequalities 140 :

Table 1.1.8

We use these properties to obtain an equivalent inequality 141 , one with the same solution set, where the variable is isolated. The process is similar to solving linear equations.

Example \(\PageIndex{14}\):

Solve and graph the solution set: \(−2(x+8)+6≥20\).

\(\begin{aligned} - 2 ( x + 8 ) + 6 & \geq 20 \quad\color{Cerulean}{Distribute.} \\ - 2 x - 16 + 6 & \geq 20 \quad\color{Cerulean}{Combine\: like\: terms.} \\ - 2 x - 10 & \geq 20 \quad\color{Cerulean}{Solve\: for\: x.} \\ - 2 x & \geq 30 \quad\color{Cerulean}{Divide\: both\: sides\: by\: -2.} \\ \frac { - 2 x } { \color{Cerulean}{- 2} } & \color{OliveGreen}{\leq} \frac { \color{Black}{30} } { \color{Cerulean}{- 2} } \quad\color{Cerulean}{Reverse\: the\: inequality.} \\ x & \leq - 15 \end{aligned}\)

Interval notation \((−∞,  −15] \)

Example \(\PageIndex{15}\):

Solve and graph the solution set: \(−2(4x−5)<9−2(x−2)\).

\(\begin{array} { c } { - 2 ( 4 x - 5 ) < 9 - 2 ( x - 2 ) } \\ { - 8 x + 10 < 9 - 2 x + 4 } \\ { - 8 x + 10 < 13 - 2 x } \\ { - 6 x < 3 } \\ { \frac { - 6 x } { \color{Cerulean}{- 6} } \color{OliveGreen}{>} \frac { \color{Black}{3} } { \color{Cerulean}{- 6} } }\color{Cerulean}{Reverse\:the\:inequality.} \\ { x > - \frac { 1 } { 2 } } \end{array}\)

Interval notation \((−\frac{1}{2}, ∞)\)

Example \(\PageIndex{16}\):

Solve and graph the solution set: \(\frac{1}{2}x−2≥\frac{1}{2}(\frac{7}{4}x−9)+1\).

\(\begin{array} { c } { \frac { 1 } { 2 } x - 2 \geq \frac { 1 } { 2 } \left( \frac { 7 } { 4 } x - 9 \right) + 1 } \\ { \frac { 1 } { 2 } x - 2 \geq \frac { 7 } { 8 } x - \frac { 9 } { 2 } + 1 } \\ { \frac { 1 } { 2 } x - \frac { 7 } { 8 } x \geq - \frac { 7 } { 2 } + 2 } \\ { - \frac { 3 } { 8 } x \geq - \frac { 3 } { 2 } } \\ { \left( \color{Cerulean}{- \frac { 8 } { 3 }} \right) \left(\color{Black}{ - \frac { 3 } { 8 } x} \right) \leq \left( \color{Cerulean}{- \frac { 8 } { 3 }} \right) \left( \color{Black}{-} \frac { 3 } { 2 } \right) \quad \color{Cerulean} { Reverse\: the\: inequality. } } \\ { x \leq 4 } \end{array}\)

Interval notation: \((−∞,  4]\)

Exercise \(\PageIndex{4}\)

Solve and graph the solution set: \(10 - 5 ( 2 x + 3 ) \leq 25\)

\([ - 3 , \infty )\);

Video Solution: www.youtube.com/v/COLLNtwYFm8

Translation of Linear Inequalities

Some of the key words and phrases that indicate inequalities are summarized below:

Table 1.1.9

Key Takeaways

• Solving general linear equations involves isolating the variable, with coefficient \(1\), on one side of the equal sign. To do this, first use the appropriate equality property of addition or subtraction to isolate the variable term on one side of the equal sign. Next, isolate the variable using the equality property of multiplication or division. Finally, check to verify that your solution solves the original equation.
• If solving a linear equation leads to a true statement like \(0 = 0\), then the equation is an identity and the solution set consists of all real numbers, \(ℝ\).
• If solving a linear equation leads to a false statement like \(0 = 5\), then the equation is a contradiction and there is no solution, \(Ø\).
• Clear fractions by multiplying both sides of an equation by the least common multiple of all the denominators. Distribute and multiply all terms by the LCD to obtain an equivalent equation with integer coefficients.
• Simplify the process of solving real-world problems by creating mathematical models that describe the relationship between unknowns. Use algebra to solve the resulting equations.
• Inequalities typically have infinitely many solutions. The solutions are presented graphically on a number line or using interval notation or both.
• All but one of the rules for solving linear inequalities are the same as solving linear equations. If you divide or multiply an inequality by a negative number, reverse the inequality to obtain an equivalent inequality.

129 Statement indicating that two algebraic expressions are equal.

130 An equation that can be written in the standard form \(ax + b = 0\), where \(a\) and \(b\) are real numbers and \(a ≠ 0\).

131 Any value that can replace the variable in an equation to produce a true statement.

132 Equations with the same solution set.

133 Properties that allow us to obtain equivalent equations by adding, subtracting, multiplying, and dividing both sides of an equation by nonzero real numbers.

134 Allows you to solve for the variable on either side of the equal sign, because \(x = 5\) is equivalent to \(5 = x\).

135 Equations that are true for particular values.

136 An equation that is true for all possible values.

137 An equation that is never true and has no solution.