Solving Equations

What is an equation.

An equation says that two things are equal. It will have an equals sign "=" like this:

That equations says:

what is on the left (x − 2)  equals  what is on the right (4)

So an equation is like a statement " this equals that "

What is a Solution?

A Solution is a value we can put in place of a variable (such as x ) that makes the equation true .

Example: x − 2 = 4

When we put 6 in place of x we get:

which is true

So x = 6 is a solution.

How about other values for x ?

  • For x=5 we get "5−2=4" which is not true , so x=5 is not a solution .
  • For x=9 we get "9−2=4" which is not true , so x=9 is not a solution .

In this case x = 6 is the only solution.

You might like to practice solving some animated equations .

More Than One Solution

There can be more than one solution.

Example: (x−3)(x−2) = 0

When x is 3 we get:

(3−3)(3−2) = 0 × 1 = 0

And when x is 2 we get:

(2−3)(2−2) = (−1) × 0 = 0

which is also true

So the solutions are:

x = 3 , or x = 2

When we gather all solutions together it is called a Solution Set

The above solution set is: {2, 3}

Solutions Everywhere!

Some equations are true for all allowed values and are then called Identities

Example: sin(−θ) = −sin(θ) is one of the Trigonometric Identities

Let's try θ = 30°:

sin(−30°) = −0.5 and

−sin(30°) = −0.5

So it is true for θ = 30°

Let's try θ = 90°:

sin(−90°) = −1 and

−sin(90°) = −1

So it is also true for θ = 90°

Is it true for all values of θ ? Try some values for yourself!

How to Solve an Equation

There is no "one perfect way" to solve all equations.

A Useful Goal

But we often get success when our goal is to end up with:

x = something

In other words, we want to move everything except "x" (or whatever name the variable has) over to the right hand side.

Example: Solve 3x−6 = 9

Now we have x = something ,

and a short calculation reveals that x = 5

Like a Puzzle

In fact, solving an equation is just like solving a puzzle. And like puzzles, there are things we can (and cannot) do.

Here are some things we can do:

  • Add or Subtract the same value from both sides
  • Clear out any fractions by Multiplying every term by the bottom parts
  • Divide every term by the same nonzero value
  • Combine Like Terms
  • Expanding (the opposite of factoring) may also help
  • Recognizing a pattern, such as the difference of squares
  • Sometimes we can apply a function to both sides (e.g. square both sides)

Example: Solve √(x/2) = 3

And the more "tricks" and techniques you learn the better you will get.

Special Equations

There are special ways of solving some types of equations. Learn how to ...

  • solve Quadratic Equations
  • solve Radical Equations
  • solve Equations with Sine, Cosine and Tangent

Check Your Solutions

You should always check that your "solution" really is a solution.

How To Check

Take the solution(s) and put them in the original equation to see if they really work.

Example: solve for x:

2x x − 3 + 3 = 6 x − 3     (x≠3)

We have said x≠3 to avoid a division by zero.

Let's multiply through by (x − 3) :

2x + 3(x−3) = 6

Bring the 6 to the left:

2x + 3(x−3) − 6 = 0

Expand and solve:

2x + 3x − 9 − 6 = 0

5x − 15 = 0

5(x − 3) = 0

Which can be solved by having x=3

Let us check x=3 using the original question:

2 × 3 3 − 3 + 3  =   6 3 − 3

Hang On: 3 − 3 = 0 That means dividing by Zero!

And anyway, we said at the top that x≠3 , so ...

x = 3 does not actually work, and so:

There is No Solution!

That was interesting ... we thought we had found a solution, but when we looked back at the question we found it wasn't allowed!

This gives us a moral lesson:

"Solving" only gives us possible solutions, they need to be checked!

  • Note down where an expression is not defined (due to a division by zero, the square root of a negative number, or some other reason)
  • Show all the steps , so it can be checked later (by you or someone else)

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Equations with variables

  • Equations with variables I
  • Equations with variables II
  • Equations with variables III

In this section, you will learn how to solve equations that contain unknown variables. You will learn how to solve equations mentally by using the multiplication table and you will also learn how to identify a solution to an equation with given numbers as well as by using inverse operations.

You can solve an easy equation in your head by using the multiplication table.

$$\begin{array}{lcl} 8x=64 \end{array}$$

$$\begin{array}{lcl} 8\cdot x=64 \end{array}$$

Which number should you multiply 8 by to get a product of 64? By using the multiplication table, we know that the number is 8.

$$8\cdot 8=64$$

When we solve an equation, we figure out what value of x (or any other variable) makes the statement true (satisfies the equation).

Which of the following numbers is a solution to the equation? x = 2, 7,  or 8?

Here you are given the numbers 2, 7 and 8. One of these numbers will satisfy the equation. If you don't know the solution right away, you can investigate which of the given numbers gives results in the correct answer by plugging in the different values of x.

$$\begin{matrix} x=2\Rightarrow & 14-2=12& {\color{red} {Wrong}}\: \: \\ x=7\Rightarrow & 14-7=7\: &{\color{green} {Correct}} \\ x=8\Rightarrow & 14-8=6\: & {\color{red} {Wrong}}\: \: \end{matrix}$$

Answer: x=7

You have already solved equations where the solutions are quite easy to see, by using mental math or patterns. Most equations are harder to solve and you have to simplify the equation before you can see the solution. One way to do this is to use inverse operations.

An operation is, for example, addition, multiplication, division and subtraction. An inverse operation is an operation that reverses the effect of another operation. Addition and subtraction are inverses of each other, just like division and multiplication are inverses.

With numbers

$$18+4=22$$

$$18+4{\color{blue} \, -\, 4}=22{\color{blue} \, -\, 4}$$

With variables and numbers

$$x+4{\color{blue} \, -\, 4}=22{\color{blue} \, -\, 4}$$

We subtract 4 from both sides.

$$ x\cdot 2=10$$

$$\frac{x\cdot 2}{{\color{blue} 2}}=\frac{10}{{\color{blue} 2}}$$

We divide both sides by 2

Video lessons

Solve the following equation

$$8\cdot x-x=21$$

Solve the following equation using inverse operations

$$6x+4=28$$

  • Pre-Algebra
  • Absolute value
  • Adding and subtracting integers
  • Multiplying and dividing with integers
  • Different ways to solve equations
  • Calculating the area and the perimeter
  • Solving inequalities
  • Understanding inequalities and equations
  • Monomials and adding or subtracting polynomials
  • Powers and exponents
  • Multiplying polynomials and binomials
  • Factorization and prime numbers
  • Finding the greatest common factor
  • Finding the least common multiple
  • Integers and rational numbers
  • Learn how to estimate calculations
  • Calculating with decimals and fractions
  • Geometric sequences of numbers
  • Scientific notation
  • Fundamentals in solving Equations in one or more steps
  • Calculating the circumference of a circle
  • Linear equations in the coordinate plane
  • The slope of a linear function
  • Graphing linear inequalities
  • Solve systems of equations by graphing
  • Rates and ratios
  • Proportions and percent
  • Solving problems with percentages
  • The mean, the median and the mode
  • Stem-and-Leaf Plots and Box-and-Whiskers Plot
  • Calculating the outcome
  • Combinations and permutations
  • Finding the odds
  • Probability of events
  • Geometry – fundamental statements
  • Circle graphs
  • Angles and parallel lines
  • Quadrilaterals, polygons and transformations
  • Measure areas
  • Pyramids, prisms, cylinders and cones
  • Square roots and real numbers
  • The Pythagorean Theorem
  • Trigonometry
  • Algebra 1 Overview
  • Algebra 2 Overview
  • Geometry Overview
  • SAT Overview
  • ACT Overview

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)

Elimination Method of solving linear equations

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.

Graphical Method of Solving Linear Equations

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'.

Cross Multiplication Method of solving linear equations

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).

Solving Linear Equations Example

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.

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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

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Sat / act prep online guides and tips, single variable equations in algebra: act math strategies.

Feature_variable.png

So make sure you are prepared to tackle the ins and outs of single variable equations (no matter how they are presented on the ACT), before you take on some of the more complicated elements of ACT math.

This guide will be your complete walk-through of single variable equations for the ACT --what they are, how you’ll see them on the test, and how to set up and solve them.

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And the mystery unfolds.

What Are Single Variable Equations?

To understand a single variable equation, let us break it into its two components: the variable and the equation.

A variable is a symbolic placeholder for a number we do not yet know. It’s very common to see $x$ or $y$ used as a variable in math problems, but variables can be represented by any symbol or letter.

$x + 4 = 14$

In this case, $x$ is our variable. It represents a number that is currently unknown.

An equation sets two mathematical expressions equal to one another. This equality is represented with an equals sign (=) and each side of the expression can be as simple as a single integer or as complex as an expression with multiple variables, exponents, or anything else.

$({x +y^2})/14 - 65(x - 3) = 2$

The above is an example of an equation. Each side of the expression equals the other.

So if we put together our definitions, we know that:

A single variable equation is an equation in which there is only one variable used. (Note: the variable can be used multiple times and/or used on either side of the equation; all that matters is that the variable remains the same.)

${(x + 4)}/2 = 12$

$6x + 3 - 2x = 19$

$4y - 2 = y + 7$

These are all examples of single variable equations. You can see how some expressions used the variable multiple times or used the variable in both expressions (on either side of the equals sign).

No matter how many times the variable is used, these still count as single variable problems because the variable remains constant and there are no other variables.

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Typical Single Variable Equations on the ACT

Single variable equations will fall into two broad categories on the ACT--given equations and word problems. Let’s look at each type.

Given Equations

A given equation will provide you with the equation you need to use to solve the problem. We will go through the exact processes needed to solve this kind of problem in the next section, but for now just understand that your goal is to isolate your variable.

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(We will go through how to solve this question later in the guide)

As you can see from this problem, the isolated variable may not be your final answer . Sometimes the question will ask you to solve for $x$, sometimes the question will ask you to solve for $x$ to a different term (as in this case, where they ask you to find $2x$).

Always pay close attention to exactly what the question is asking you to find! You need to first isolate your $x$ to solve the problem, but if you stop there then you will get the final answer wrong.

Word Problems

A word problem describes a scene in which you must set up your own single variable equation to solve it. Again, your final answer may be the value of your variable ($x$ or $y$, etc.) or your variable taken to a different term ($2x$, $y/2$, etc.). 

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How to Manipulate a Single Variable Equation

In order to solve a single variable equation, we must isolate our variable on one side of the equation. And the way we do this is by shifting the rest of our terms to the other side of the equation.

In order to shift our terms (numbers), we must therefore cancel them out on their original side by performing the opposite function of the term.

Opposite function pairs are:

Addition and subtraction

Multiplication and division

So if we have a term on one side that has a plus sign (addition), we must subtract that same amount from both sides.

$x + 2 = 6$

$x + 2 - 2 = 6 - 2$

If we have a term that is multiplied, we must divide that same amount from both sides.

${3x}/3 = 18/3$

Whatever you do on one side of the equation, you must do on the other. This cancels out like terms and essentially moves your terms from one side of the equation to the other.

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Steps to Solving a Single Variable Problem

Let us take a typical variable expression and break it into the steps needed to solve it.

$3y - 10 + 2y = 15$. Find $y$.

1) Combine like terms

If there is more than one term with a same variable, we must combine them in order to ultimately isolate that variable. We can add or subtract terms with a same variable in the same way we can any other numbers.

$3y - 10 + 2y = 15$

Here we have a $3y$ and a $2y$. They are both positive, so we add them together.

$3y + 2y = 5y$

So now our equation looks like this:

$5y - 10 = 15$

2) Isolate the term with your variable

Once we have combined our variables, we must isolate the variable term. If the term is simply the variable itself (e.g. $y$), then we can skip this step. But since our term her is $5y$, we must isolate the whole term first.

So we must add 10 to either side of our equation. Why? Because we have a negative 10 and addition is the opposite of subtraction. And we must do it to either side to cancel out the 10 on the first expression in order to isolate our variable.

$5y - 10 + 10 = 15 + 10$

3) Isolate your variable

Now that we’ve isolated our term ($5y$), we can further isolate the variable itself.

Again, we perform an opposite function of the term. In this case, we have $5y$, which uses multiplication. In order to isolate the variable, we must therefore use division (the opposite of multiplication) by dividing on both sides.

${5y}/5 = 25/5$

4) Double-check your variable by plugging it back in

Now that we’ve solved for our variable, let us check to make sure it is correct by plugging it back into the original equation.

$3(5) - 10 + 2(5) = 15$

$15 - 10 + 10 = 15$

Success! We have correctly isolated the variable and found its value.

5) And, finally, double-check to make sure you are answering the right question!

In this case, we are done, because our initial question asked us to find the value of $y$. But you must always double-check to make sure you are answering the right question. If they had asked us the value for $5y$ or $y/3$, then we would have gotten the answer wrong if we had stopped here at $y = 5$.

Always double-check that your variable is correct and that you are answering the question the test is asking you to answer.

Now let’s try it again with our problem from earlier:

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Step 1, combine like terms:

There are no like terms to combine, so we can skip step 1.

Step 2, isolate variable term:

$7 + 3x = 22$

$7 - 7 + 3x = 22 - 7$

Step 3, isolate variable:

${3x}/3 = 15/3$

Step 4, double-check answer:

$7 + 3(5) = 22$

$7 + 15 = 22$

Success. But wait! We’re not done just yet.

Step 5, look at what the final question is asking:

We must finish the question by finding $2x$

$2(5) = 10$

So our final answer is G , $2x = 10$

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Test Your Knowledge

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Answers:  C, G, B, G, E

Answer Explanations:

1) Ms. Lewis begins by driving 900 miles at 50 miles per hour and we want to find out how much faster she must go to travel the same amount of miles in three hours less time. Because she is driving the same amount, we can set these terms equal. 

We are also only working with the variable of miles per hour, so this is a single variable equation. 

Now, the two sides of the equation are dealing with miles and miles per hour. The first half of our equation will look like this:

$(900/50) - 3$

Why? Because Ms. Lewis is driving 900 miles at 50 miles per hour, so we need to divide the miles by mph in order to find out her travel time. And then we must reduce that amount by 3 because we are told that her new travel time will be 3 miles less than that. 

This means that the other half of our equation will look like this:

Why? Because we know that the number of miles she drives will be the same, but our unknown is her miles per hour. 

Now let's put them together and solve for our variable. 

$(900/50) - 3 = 900/x$

$18 - 3 = 900/x$

$15 = 900/x$

Now we must isolate our $x$ value. Because it is acting as a denominator, we must multiply both sides of the equation by $x$. 

$x * 15 = (900/x) * x$

$15x = 900$

Now, we can divide both sides by 15 in order to isolate our $x$ value. 

${15x}/15 = 900/15$

Finally, let us plug this value back into our original equation to double-check our answer.

$(900/50) - 3 = 900/60$

We have successfully found our $x$ value, which is the new mileage per hour that Ms. Lewis must travel. 

But wait, we're not done yet! The question asked us to find out how much faster she must drive, not the new miles per hour at which she must travel. This means we must take the difference of the original miles per hour and the new miles per hour. 

$60 - 50 = 10$

She must drive 10 miles per hour faster in order to drive the same amount in three hours less time.

So our final answer is C , 10.

2) Here we have two cable companies and we are told that we must solve for when their rates are equal after an equal number of months. That means we have a single variable (the number of months) and we have an equation because we are setting each side equal (since the question specifies that their prices will be equal after an unknown number of months).

Uptown Cable has a flat fee of 120 dollars and an additional fee of 25 dollars per month. The flat fee will be unchanged (it only happens once), but the 25 dollars will be affected by the number of months. Since the number of months is our unknown variable, let’s give it a value of $x$.

So our first expression will look like this:

$120 + 25x$

Now Downtown Cable has a 60 dollar flat fee (occurs only once) and a 35 dollar per month fee. We are trying the find the equal number of months for a Downtown Cable package and an Uptown Cable package, so our variable, $x$, will remain the same. So our second expression will look like this:

Now we set the two expressions equal to one another. (Why? Because we are told that the prices will be equal after a certain number of months.)

$120 + 25x = 60 + 35x$

Now we solve by shifting the terms on each side of the equation. First, let us combine our variable terms by subtracting 25x from each side.

$120 + 25x - 25x = 60 + 35x - 25x$

$120 = 60 + 10x$

Now, let us subtract 60 from each side.

$120 - 60 = 60 - 60 + 10x$

And finally, let us isolate our variable.

$60/10 = {10x}/10$

So our final answer is G , in exactly 6 months, the prices of each cable package will be equal.

3) This question relies on manipulating fractions. If this process is unfamiliar to you, definitely check out our guide to ACT fractions and ratios . If this is familiar to you, then let’s keep going.

${1/3}k + {1/4}k =1$

We must find a common denominator of the two fractions in order to combine our like terms. In this case, the least common factor of 3 and 4 is 12. (For more on this process, check out our guide to ACT fractions and ratios.)

${4/12}k + {3/12}k = 1$

${7/12}k = 1$

Now we have a number (7) being divided by another number (12). We know that division is the opposite of multiplication, so we must multiply each side by 12.

$12 * {7/12}k = 1 * 12$

And finally, we must divide each side by 7 to isolate our variable.

${7k}/7 = 12/7$

So our final answer is B , $12/7$

4) We have a consultant with a flat (one time) fee of 30 dollars and an additional fee of 45 dollars per hour. Because the 45 dollars is hourly, it changes based on our variable (the number of hours). We do not know the number of hours she works, but we do know that her final earnings were 210 dollars. So let’s set this up as an equation.

$30 + 45x = 210$

There are no like terms, so we can start isolating our variable.

$30 - 30 + 45x = 210 - 30$

$45x = 180$

${45x}/45 = 180/45$

So our final answer is G , she worked 4 hours to earn 210 dollars.

5) This is a single variable problem that can be solved in one of two ways--you can either distribute first and then solve, or you can solve without the need to distribute. We’ll go through both ways here.

Solve with distributing:

$9(x - 9) = -11$

First, distribute your 9 across the expression $(x - 9)$

$9(x) - 9(9) = -11$

$9x - 81 = -11$

Now, isolate your variable term as usual.

$9x - 81 + 81 = -11 + 81$

And finally, isolate your variable.

${9x}/9 = 70/9$

So our final answer is E , 70/9.

Alternatively, you can solve this problem without the need to distribute your 9 across the expression (x - 9)

Solve without distributing:

Divide each side by 9

${9(x - 9)}/9 = -11/9$

$x - 9 = -11/9$

Now, we must add 9 to each side.

$x - 9 + 9 = -11/9 + 9$

$x = -11/9 + 9$

In order to add $-11/9$ and 9, we must give them a common denominator. Again, check out the guide on fractions and ratios  if this process is unfamiliar to you.

$x = -11/9 + 9/1(9/9)$

$x = -11/9 + 81/9$

So again, our answer is E , 70/9.

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The Take-Aways

Single variations make up the backbone of many other ACT problems. By knowing how to manipulate these kinds of expressions, you’ll be able to build on these techniques to solve much more complex problems and equations.

Just remember to always perform the same act to each side of the equation and save isolating your variable for last. Now take your single variable knowledge and conquer the rest of our math guides. You’ve got this.

What’s Next?

You’ve build up your mathematical foundation and now you’re raring to take on more. Before you start in on another guide to an ACT math topic, make sure you have a good idea of all the topics covered on the ACT math .

Think you might need a tutor? Check out the best ways to shop around for a tutor who suits your needs , whether online or in person .

Taken a practice test and don’t know how you match up for schools? Make sure you have a good idea of what your ideal score truly is .

And if you feel like you’ve got a handle on the math itself, but struggle with the timing , then be sure to check out on our article on how to stop running out of time on the ACT.

Want to improve your ACT score by 4 points?

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How to Solve Multivariable Linear Equations in Algebra

Last Updated: February 19, 2024 Fact Checked

This article was co-authored by Taylor Klein . Taylor Klein is an Advanced Math Teacher based in Philadelphia, Pennsylvania. She has worked in the education field for over 10 years, including eight years as a middle school Advanced Math Teacher. She has a master’s degree in Instructional Technology and Design and a master’s degree in Educational Leadership and Administration. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 94,475 times.

Multivariable linear equations are equations that have two or more unknowns (generally represented by 'x' and 'y'). There are multiple ways that you can solve these equations, including elimination and substitution.

Understanding the Components of Linear Equations

Step 1 Understand what multi-variable equations are.

  • 8x - 3y = -3
  • 5x - 2y = -1
  • These are two linear equations that you must solve at the same time, meaning you must use both equations to solve both equations.

Step 2 Know that you are trying to figure out the values of the variables, or unknowns.

  • In the case of our example, you are trying to find out what numbers ‘x’ and ‘y’ represent that will make both of the equations true. In the case of this example, x = -3 and y = -7. Plug them in. 8(-3) - 3(-7) = -3. This is TRUE. 5(-3) -2(-7) = -1. This is also TRUE.

Step 3 Know what a numerical coefficient is.

  • 8 and 3 for the first equation; 5 and 2 for the second equation.

Step 4 Understand the difference between solving with elimination and solving with substitution.

  • Substitution, on the other hand, is where you begin working with only one equation so that you can again solve for one variable. Once you solve one equation, you can plug in your findings to the other equation, effectively making one large equation out of your two smaller ones. Again, don’t worry—this will be covered in detail in Method 3.

Step 5 Understand that there can be linear equations that have three or more variables.

Solving a Linear Equation with Elimination

Step 1 Look at your equation.

  • In 8x - 3y = -3 (equation A) and 5x - 2y = -1 (equation B), you can multiply equation A with 2 and equation B with 3 so that you get 6y in equation A and 6y in equation B.
  • This would look like: equation A: 2(8x - 3y =-3) = 16x -6y = -6.
  • Equation B: 3(5x - 2y = -1) = 15x -6y =-3

Step 3 Add or subtract the two equations to remove one of the variables and solve the other variable.

  • (16x - 6y = -6) - (15x - 6y = -3) = 1x = -3. Therefore x = -3.
  • For other cases, if the numerical coefficient of x is not 1 after we add or subtract, we must divide both sides by the numerical coefficient to simplify the equation.

Step 4 Plug in your solution to solve for the remaining variable.

  • Equation B: 5(-3) - 2y = -1 so -15 -2y = -1. Add 15 to both sides so -2y = 14. Divide both sides by -2 so that y = -7.
  • Therefore x = -3 and y = -7.

Step 5 Plug your findings into both equations to make sure that they are correct.

  • 8(-3) - 3(-7) = -3 so -24 +21 = -3 TRUE.
  • 5(-3) -2(-7) = -1 so -15 + 14 = -1 TRUE.
  • Therefore, the variables we have found are correct.

Solving a Linear Equation with Substitution

Step 1 Begin by solving one equation for either variable.

  • x - 2y = 10 (equation A) and -3x -4y = 10 (equation B). You would choose to work with x - 2y = 10 because the coefficient of x in this equation is 1.
  • Solving for x in equation A would meaning adding 2y to both sides. Therefore, x = 10 + 2y.

Step 2 Substitute your findings in Step 1 into the other equation.

  • Insert the ‘x’ of equation B into equation A: -3(10 + 2y) -4y = 10. You can see that we have taken ‘x’ out of the equation and inserted what ‘x’ equals.

Step 3 Solve for the other variable.

  • -3(10 + 2y) -4y = 10 so -30 -6y -4y = 10.
  • Combine the y’s: -30 - 10y = 10.
  • Move the -30 over to the other side: -10y = 40.
  • Solve for y: y = -4.

Step 4 Solve the second variable.

  • Solve for ‘x’ in equation A by plugging in y = -4: x - 2(-4) = 10.
  • Simply the equation: x + 8 = 10.
  • Solve for x: x = 2.

Step 5 Double check that the variables you have found work for both equations.

  • Equation A: 2 - 2(-4) = 10 is TRUE.
  • Equation B: -3(2) -4(-4) = 10 is TRUE.

Community Q&A

Donagan

  • Be careful of your signs, since we use a lot of basic operations, sign changes can affect every step of your computation. Thanks Helpful 0 Not Helpful 0
  • Check your finals answers. You can do this by replacing the values obtained in the final answer to their corresponding variables in any of the original equations, if the left hand side and the right hand side matches, your final answer is correct. Thanks Helpful 0 Not Helpful 0

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Thanks for reading our article! If you’d like to learn more about mathematics, check out our in-depth interview with Taylor Klein .

  • ↑ Steward, J., Lothar, R., Watson, S., Algebra and Trigonometry. Second Edition. Singapore: Thomson Learning Asia,
  • ↑ https://www.khanacademy.org/test-prep/praxis-math/praxis-math-lessons/praxis-math-algebra/a/gtp--praxis-math--article--linear-equations--lesson
  • ↑ http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut49_systwo.htm
  • ↑ https://people.richland.edu/james/lecture/m116/systems/gaussian.html
  • ↑ https://www.mathplanet.com/education/algebra-1/systems-of-linear-equations-and-inequalities/the-substitution-method-for-solving-linear-systems
  • ↑ https://www.purplemath.com/modules/systlin4.htm

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How to solve systems of 3 variable equations

Using Elimination

Types of solutions for systems of planes (3 variable equations)

What is a solution of system of equations with 3 variables.

Solution icon

Solution for system of lines

Just as the solution system of lines is where those lines meet, a solution for a system of 3 variable equations (planes), is again, just where these planes meet.

types of solutions

read more here

Why 3 planes?

If you want to solve a linear equation with 2 variables, you need 2 equations.

You can's solve $$ x + y = 1$$ , right? That's because you need equations to solve for 2 variables.

Similarly, if you have an equation with 3 variables, ( graphically represented by 3 planes), you're going to need 3 equations to solve it.

Two important terms

Means that there is at least 1 intersection (solutions).

Means that there are no intersections (solutions).

Ok, so how do we find that point of intersection?

Before attempting to solve systems of three variable equations using elimination, you should probably be comfortable solving 2 variable systems of linear equations using elimination .

Video Tutorial on using Elimination

Example of how to solve a system of three variable equations using elimination.

Steps to solve system of 3 equations

Practice Problems

Use elimination to solve the following system of three variable equations.

  • A) 4x + 2y – 2z = 10
  • B) 2x + 8y + 4z = 32
  • C) 30x + 12y – 4z = 24

steps to solve 3 variable system by elimination

  • A) x - y + z = -1
  • B) x + y + z = 3
  • C) 4x + 2y + z = 8
  • A) 2x + 2y + 2z = -4

Although you can indeed solve 3 variable systems using elimination and substitution as shown on this page, you may have noticed that this method is quite tedious. The most efficient method is to use matrices or, of course, you can use this online system of equations solver . ( all of our pictures on this topic )

Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!

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4.2: Solve Systems of Linear Equations with Two Variables

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Learning Objectives

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

  • Determine whether an ordered pair is a solution of a system of equations
  • Solve a system of linear equations by graphing
  • Solve a system of equations by substitution
  • Solve a system of equations by elimination
  • Choose the most convenient method to solve a system of linear equations

Be Prepared 4.1

Before you get started, take this readiness quiz.

For the equation y = 2 3 x − 4 , y = 2 3 x − 4 , ⓐ Is ( 6 , 0 ) ( 6 , 0 ) a solution? ⓑ Is ( −3 , −2 ) ( −3 , −2 ) a solution? If you missed this problem, review Example 3.2.

Be Prepared 4.2

Find the slope and y -intercept of the line 3 x − y = 12 . 3 x − y = 12 . If you missed this problem, review Example 3.16.

Be Prepared 4.3

Find the x- and y -intercepts of the line 2 x − 3 y = 12 . 2 x − 3 y = 12 . If you missed this problem, review Example 3.8.

Determine Whether an Ordered Pair is a Solution of a System of Equations

In Solving Linear Equations, we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations .

System of Linear Equations

When two or more linear equations are grouped together, they form a system of linear equations .

In this section, we will focus our work on systems of two linear equations in two unknowns. We will solve larger systems of equations later in this chapter.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

{ 2 x + y = 7 x − 2 y = 6 { 2 x + y = 7 x − 2 y = 6

A linear equation in two variables, such as 2 x + y = 7 , 2 x + y = 7 , has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs ( x , y ) ( x , y ) that make both equations true. These are called the solutions of a system of equations .

Solutions of a System of Equations

The solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair ( x , y ) . ( x , y ) .

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Example 4.1

Determine whether the ordered pair is a solution to the system { x − y = −1 2 x − y = −5 . { x − y = −1 2 x − y = −5 .

ⓐ ( −2 , −1 ) ( −2 , −1 ) ⓑ ( −4 , −3 ) ( −4 , −3 )

The equations are x minus y equals minus 1 and 2 x minus y equals minus 5. We substitute x equal to minus 2 and y equal to minus 1 into both equations. So, x minus y equals minus 1 becomes minus 2 minus open parentheses minus 1 close parentheses equal to or not equal to minus 1. Simplifying, we get minus 1 equals minus 1 which is correct. The equation 2 x minus y equals minus 5 becomes 2 times minus 2 minus open parentheses minus 1 close parentheses equal to or not equal to minus 5. Simplifying, we get minus 3 not equal to minus 5. Hence, the ordered pair minus 2, minus 1 does not make both equations true. So, it is not a solution.

Determine whether the ordered pair is a solution to the system { 3 x + y = 0 x + 2 y = −5 . { 3 x + y = 0 x + 2 y = −5 .

ⓐ ( 1 , −3 ) ( 1 , −3 ) ⓑ ( 0 , 0 ) ( 0 , 0 )

Determine whether the ordered pair is a solution to the system { x − 3 y = −8 − 3 x − y = 4 . { x − 3 y = −8 − 3 x − y = 4 .

ⓐ ( 2 , −2 ) ( 2 , −2 ) ⓑ ( −2 , 2 ) ( −2 , 2 )

Solve a System of Linear Equations by Graphing

In this section, we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.

The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.

Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.

Figure shows three graphs. In the first, the lines intersect at point 3, minus 1. The intersecting lines have one point in common. There is one solution to the system. In the second graph, the lines are parallel. Parallel lines have no points in common. There is no solution to the system. The third graph has only one line. Here, both equations give the same line. Because we have only one line, there are infinite many solutions.

Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end of the section you’ll decide which method was the most convenient way to solve this system.

Example 4.2

How to solve a system of equations by graphing.

Solve the system by graphing { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Step 1 is to graph the first equation. To graph the first line, write the equation in slope intercept form. So, 2 x plus y equals 7 becomes y equal to minus 2 x plus 7. Here, m is minus 2 and b is 7. So the graph will be a line with slope equal to minus 2 and y intercept equal to 7.

Solve the system by graphing: { x − 3 y = −3 x + y = 5 . { x − 3 y = −3 x + y = 5 .

Solve the system by graphing: { − x + y = 1 3 x + 2 y = 12 . { − x + y = 1 3 x + 2 y = 12 .

The steps to use to solve a system of linear equations by graphing are shown here.

Solve a system of linear equations by graphing.

  • Step 1. Graph the first equation.
  • Step 2. Graph the second equation on the same rectangular coordinate system.
  • Step 3. Determine whether the lines intersect, are parallel, or are the same line.
  • If the lines intersect, identify the point of intersection. This is the solution to the system.
  • If the lines are parallel, the system has no solution.
  • If the lines are the same, the system has an infinite number of solutions.
  • Step 5. Check the solution in both equations.

In the next example, we’ll first re-write the equations into slope–intercept form as this will make it easy for us to quickly graph the lines.

Example 4.3

Solve the system by graphing: { 3 x + y = − 1 2 x + y = 0 . { 3 x + y = − 1 2 x + y = 0 .

We’ll solve both of these equations for y y so that we can easily graph them using their slopes and y -intercepts.

Solve the system by graphing: { − x + y = 1 2 x + y = 10 . { − x + y = 1 2 x + y = 10 .

Solve the system by graphing: { 2 x + y = 6 x + y = 1 . { 2 x + y = 6 x + y = 1 .

In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.

Example 4.4

Solve the system by graphing: { y = 1 2 x − 3 x − 2 y = 4 . { y = 1 2 x − 3 x − 2 y = 4 .

Solve the system by graphing: { y = − 1 4 x + 2 x + 4 y = − 8 . { y = − 1 4 x + 2 x + 4 y = − 8 .

Solve the system by graphing: { y = 3 x − 1 6 x − 2 y = 6 . { y = 3 x − 1 6 x − 2 y = 6 .

Sometimes the equations in a system represent the same line. Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to the system.

Example 4.5

Solve the system by graphing: { y = 2 x − 3 − 6 x + 3 y = − 9 . { y = 2 x − 3 − 6 x + 3 y = − 9 .

If you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y -intercept.

Solve the system by graphing: { y = − 3 x − 6 6 x + 2 y = − 12 . { y = − 3 x − 6 6 x + 2 y = − 12 .

Try It 4.10

Solve the system by graphing: { y = 1 2 x − 4 2 x − 4 y = 16 . { y = 1 2 x − 4 2 x − 4 y = 16 .

When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident . Coincident lines have the same slope and same y- intercept.

Coincident Lines

Coincident lines have the same slope and same y- intercept.

The systems of equations in Example 4.2 and Example 4.3 each had two intersecting lines. Each system had one solution.

In Example 4.5, the equations gave coincident lines, and so the system had infinitely many solutions.

The systems in those three examples had at least one solution. A system of equations that has at least one solution is called a consistent system.

A system with parallel lines, like Example 4.4, has no solution. We call a system of equations like this inconsistent. It has no solution.

Consistent and Inconsistent Systems

A consistent system of equations is a system of equations with at least one solution.

An inconsistent system of equations is a system of equations with no solution.

We also categorize the equations in a system of equations by calling the equations independent or dependent . If two equations are independent, they each have their own set of solutions. Intersecting lines and parallel lines are independent.

If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines.

Let’s sum this up by looking at the graphs of the three types of systems. See below and Table 4.1.

The figure shows three graphs. The first one has two intersecting line. The second one has two parallel lines. The third one has only one line. This is labeled coincident.

Example 4.6

Without graphing, determine the number of solutions and then classify the system of equations.

ⓐ { y = 3 x − 1 6 x − 2 y = 12 { y = 3 x − 1 6 x − 2 y = 12 ⓑ { 2 x + y = − 3 x − 5 y = 5 { 2 x + y = − 3 x − 5 y = 5

ⓐ We will compare the slopes and intercepts of the two lines.

A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.

ⓑ We will compare the slope and intercepts of the two lines.

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

Try It 4.11

ⓐ { y = −2 x − 4 4 x + 2 y = 9 { y = −2 x − 4 4 x + 2 y = 9 ⓑ { 3 x + 2 y = 2 2 x + y = 1 { 3 x + 2 y = 2 2 x + y = 1

Try It 4.12

ⓐ { y = 1 3 x − 5 x − 3 y = 6 { y = 1 3 x − 5 x − 3 y = 6 ⓑ { x + 4 y = 12 − x + y = 3 { x + 4 y = 12 − x + y = 3

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

Solve a System of Equations by Substitution

We will now solve systems of linear equations by the substitution method.

We will use the same system we used first for graphing.

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

Example 4.7

How to solve a system of equations by substitution.

Solve the system by substitution: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

The equations are 2 x plus y equals 7 and x minus 2y equals 6. Step 1 is to solve one of the equations for either variable. We’ll solve the first equation for y. We get y equals 7 minus 2 x.

Try It 4.13

Solve the system by substitution: { − 2 x + y = −11 x + 3 y = 9 . { − 2 x + y = −11 x + 3 y = 9 .

Try It 4.14

Solve the system by substitution: { 2 x + y = −1 4 x + 3 y = 3 . { 2 x + y = −1 4 x + 3 y = 3 .

Solve a system of equations by substitution.

  • Step 1. Solve one of the equations for either variable.
  • Step 2. Substitute the expression from Step 1 into the other equation.
  • Step 3. Solve the resulting equation.
  • Step 4. Substitute the solution in Step 3 into either of the original equations to find the other variable.
  • Step 5. Write the solution as an ordered pair.
  • Step 6. Check that the ordered pair is a solution to both original equations.

Be very careful with the signs in the next example.

Example 4.8

Solve the system by substitution: { 4 x + 2 y = 4 6 x − y = 8 . { 4 x + 2 y = 4 6 x − y = 8 .

We need to solve one equation for one variable. We will solve the first equation for y .

Try It 4.15

Solve the system by substitution: { x − 4 y = −4 − 3 x + 4 y = 0 . { x − 4 y = −4 − 3 x + 4 y = 0 .

Try It 4.16

Solve the system by substitution: { 4 x − y = 0 2 x − 3 y = 5 . { 4 x − y = 0 2 x − 3 y = 5 .

Solve a System of Equations by Elimination

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d .

if a = b and c = d then a + c = b + d . if a = b and c = d then a + c = b + d .

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

{ 3 x + y = 5 2 x − y = 0 ————— 5 x = 5 { 3 x + y = 5 2 x − y = 0 ————— 5 x = 5

The y ’s add to zero and we have one equation with one variable.

Let’s try another one:

{ x + 4 y = 2 2 x + 5 y = −2 { x + 4 y = 2 2 x + 5 y = −2

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2 , −2 , we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2 . −2 .

Minus 2 open parentheses x plus 4y close parentheses is minus 2 times 2. And, 2 x plus 5y is minus 2.

Then rewrite the system of equations.

Minus 2 x minus 8y is minus 4 and 2 x plus 5y is minus 2.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Minus 2 x minus 8y is minus 4 and 2 x plus 5y is minus 2. Adding these, we get minus 3y equals minus 6.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

Example 4.9

How to solve a system of equations by elimination.

Solve the system by elimination: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

The equations are 2 x plus y equals 7 and x minus 2y equals 6. Step 1 is to write both equations in standard form. Both equations are in standard form, Ax plus By equals C. If any coefficients are fractions, clear them. There are no fractions.

Try It 4.17

Solve the system by elimination: { 3 x + y = 5 2 x − 3 y = 7 . { 3 x + y = 5 2 x − 3 y = 7 .

Try It 4.18

Solve the system by elimination: { 4 x + y = − 5 − 2 x − 2 y = − 2 . { 4 x + y = − 5 − 2 x − 2 y = − 2 .

The steps are listed here for easy reference.

Solve a system of equations by elimination.

  • Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
  • Decide which variable you will eliminate.
  • Multiply one or both equations so that the coefficients of that variable are opposites.
  • Step 3. Add the equations resulting from Step 2 to eliminate one variable.
  • Step 4. Solve for the remaining variable.
  • Step 5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
  • Step 6. Write the solution as an ordered pair.
  • Step 7. Check that the ordered pair is a solution to both original equations.

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Example 4.10

Solve the system by elimination: { 4 x − 3 y = 9 7 x + 2 y = −6 . { 4 x − 3 y = 9 7 x + 2 y = −6 .

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants to get the opposites.

Try It 4.19

Solve the system by elimination: { 3 x − 4 y = − 9 5 x + 3 y = 14 . { 3 x − 4 y = − 9 5 x + 3 y = 14 .

Try It 4.20

Solve each system by elimination: { 7 x + 8 y = 4 3 x − 5 y = 27 . { 7 x + 8 y = 4 3 x − 5 y = 27 .

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the LCD of all the fractions in the equation.

Example 4.11

Solve the system by elimination: { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 . { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 .

In this example, both equations have fractions. Our first step will be to multiply each equation by the LCD of all the fractions in the equation to clear the fractions.

Try It 4.21

Solve each system by elimination: { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 . { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 .

Try It 4.22

Solve each system by elimination: { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 . { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 .

When we solved the system by graphing, we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

The same is true using substitution or elimination. If the equation at the end of substitution or elimination is a true statement, we have a consistent but dependent system and the system of equations has infinitely many solutions. If the equation at the end of substitution or elimination is a false statement, we have an inconsistent system and the system of equations has no solution.

Example 4.12

Solve the system by elimination: { 3 x + 4 y = 12 y = 3 − 3 4 x . { 3 x + 4 y = 12 y = 3 − 3 4 x .

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Try It 4.23

Solve the system by elimination: { 5 x − 3 y = 15 y = − 5 + 5 3 x . { 5 x − 3 y = 15 y = − 5 + 5 3 x .

Try It 4.24

Solve the system by elimination: { x + 2 y = 6 y = − 1 2 x + 3 . { x + 2 y = 6 y = − 1 2 x + 3 .

Choose the Most Convenient Method to Solve a System of Linear Equations

When you solve a system of linear equations in in an application, you will not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

Choose the Most Convenient Method to Solve a System of Linear Equations Graphing ———— Substitution ————— Elimination ————— Use when you need a Use when one equation is Use when the equations are picture of the situation. already solved or can be in standard form. easily solved for one variable. Choose the Most Convenient Method to Solve a System of Linear Equations Graphing ———— Substitution ————— Elimination ————— Use when you need a Use when one equation is Use when the equations are picture of the situation. already solved or can be in standard form. easily solved for one variable.

Example 4.13

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 3 x + 8 y = 40 7 x − 4 y = −32 { 3 x + 8 y = 40 7 x − 4 y = −32 ⓑ { 5 x + 6 y = 12 y = 2 3 x − 1 { 5 x + 6 y = 12 y = 2 3 x − 1

{ 3 x + 8 y = 40 7 x − 4 y = −32 { 3 x + 8 y = 40 7 x − 4 y = −32

Since both equations are in standard form, using elimination will be most convenient.

{ 5 x + 6 y = 12 y = 2 3 x − 1 { 5 x + 6 y = 12 y = 2 3 x − 1

Since one equation is already solved for y , using substitution will be most convenient.

Try It 4.25

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 4 x − 5 y = −32 3 x + 2 y = −1 { 4 x − 5 y = −32 3 x + 2 y = −1 ⓑ { x = 2 y − 1 3 x − 5 y = −7 { x = 2 y − 1 3 x − 5 y = −7

Try It 4.26

ⓐ { y = 2 x − 1 3 x − 4 y = − 6 { y = 2 x − 1 3 x − 4 y = − 6 ⓑ { 6 x − 2 y = 12 3 x + 7 y = −13 { 6 x − 2 y = 12 3 x + 7 y = −13

Section 4.1 Exercises

Practice makes perfect.

In the following exercises, determine if the following points are solutions to the given system of equations.

{ 2 x − 6 y = 0 3 x − 4 y = 5 { 2 x − 6 y = 0 3 x − 4 y = 5

ⓐ ( 3 , 1 ) ( 3 , 1 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

{ − 3 x + y = 8 − x + 2 y = −9 { − 3 x + y = 8 − x + 2 y = −9

ⓐ ( −5 , −7 ) ( −5 , −7 ) ⓑ ( −5 , 7 ) ( −5 , 7 )

{ x + y = 2 y = 3 4 x { x + y = 2 y = 3 4 x

ⓐ ( 8 7 , 6 7 ) ( 8 7 , 6 7 ) ⓑ ( 1 , 3 4 ) ( 1 , 3 4 )

{ 2 x + 3 y = 6 y = 2 3 x + 2 { 2 x + 3 y = 6 y = 2 3 x + 2 ⓐ ( −6 , 2 ) ( −6 , 2 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

In the following exercises, solve the following systems of equations by graphing.

{ 3 x + y = −3 2 x + 3 y = 5 { 3 x + y = −3 2 x + 3 y = 5

{ − x + y = 2 2 x + y = −4 { − x + y = 2 2 x + y = −4

{ y = x + 2 y = −2 x + 2 { y = x + 2 y = −2 x + 2

{ y = x − 2 y = −3 x + 2 { y = x − 2 y = −3 x + 2

{ y = 3 2 x + 1 y = − 1 2 x + 5 { y = 3 2 x + 1 y = − 1 2 x + 5

{ y = 2 3 x − 2 y = − 1 3 x − 5 { y = 2 3 x − 2 y = − 1 3 x − 5

{ x + y = −4 − x + 2 y = −2 { x + y = −4 − x + 2 y = −2

{ − x + 3 y = 3 x + 3 y = 3 { − x + 3 y = 3 x + 3 y = 3

{ − 2 x + 3 y = 3 x + 3 y = 12 { − 2 x + 3 y = 3 x + 3 y = 12

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

{ x + 3 y = −6 y = − 4 3 x + 4 { x + 3 y = −6 y = − 4 3 x + 4

{ − x + 2 y = −6 y = − 1 2 x − 1 { − x + 2 y = −6 y = − 1 2 x − 1

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

{ 3 x + 5 y = 10 y = − 3 5 x + 1 { 3 x + 5 y = 10 y = − 3 5 x + 1

{ 4 x − 3 y = 8 8 x − 6 y = 14 { 4 x − 3 y = 8 8 x − 6 y = 14

{ x + 3 y = 4 − 2 x − 6 y = 3 { x + 3 y = 4 − 2 x − 6 y = 3

{ x = −3 y + 4 2 x + 6 y = 8 { x = −3 y + 4 2 x + 6 y = 8

{ 4 x = 3 y + 7 8 x − 6 y = 14 { 4 x = 3 y + 7 8 x − 6 y = 14

{ 2 x + y = 6 − 8 x − 4 y = −24 { 2 x + y = 6 − 8 x − 4 y = −24

{ 5 x + 2 y = 7 − 10 x − 4 y = −14 { 5 x + 2 y = 7 − 10 x − 4 y = −14

{ y = 2 3 x + 1 − 2 x + 3 y = 5 { y = 2 3 x + 1 − 2 x + 3 y = 5

{ y = 3 2 x + 1 2 x − 3 y = 7 { y = 3 2 x + 1 2 x − 3 y = 7

{ 5 x + 3 y = 4 2 x − 3 y = 5 { 5 x + 3 y = 4 2 x − 3 y = 5

{ y = − 1 2 x + 5 x + 2 y = 10 { y = − 1 2 x + 5 x + 2 y = 10

{ 5 x − 2 y = 10 y = 5 2 x − 5 { 5 x − 2 y = 10 y = 5 2 x − 5

In the following exercises, solve the systems of equations by substitution.

{ 2 x + y = −4 3 x − 2 y = −6 { 2 x + y = −4 3 x − 2 y = −6

{ 2 x + y = −2 3 x − y = 7 { 2 x + y = −2 3 x − y = 7

{ x − 2 y = −5 2 x − 3 y = −4 { x − 2 y = −5 2 x − 3 y = −4

{ x − 3 y = −9 2 x + 5 y = 4 { x − 3 y = −9 2 x + 5 y = 4

{ 5 x − 2 y = −6 y = 3 x + 3 { 5 x − 2 y = −6 y = 3 x + 3

{ − 2 x + 2 y = 6 y = −3 x + 1 { − 2 x + 2 y = 6 y = −3 x + 1

{ 2 x + 5 y = 1 y = 1 3 x − 2 { 2 x + 5 y = 1 y = 1 3 x − 2

{ 3 x + 4 y = 1 y = − 2 5 x + 2 { 3 x + 4 y = 1 y = − 2 5 x + 2

{ 2 x + y = 5 x − 2 y = −15 { 2 x + y = 5 x − 2 y = −15

{ 4 x + y = 10 x − 2 y = −20 { 4 x + y = 10 x − 2 y = −20

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

{ y = x − 6 y = − 3 2 x + 4 { y = x − 6 y = − 3 2 x + 4

{ x = 2 y 4 x − 8 y = 0 { x = 2 y 4 x − 8 y = 0

{ 2 x − 16 y = 8 − x − 8 y = −4 { 2 x − 16 y = 8 − x − 8 y = −4

{ y = 7 8 x + 4 − 7 x + 8 y = 6 { y = 7 8 x + 4 − 7 x + 8 y = 6

{ y = − 2 3 x + 5 2 x + 3 y = 11 { y = − 2 3 x + 5 2 x + 3 y = 11

In the following exercises, solve the systems of equations by elimination.

{ 5 x + 2 y = 2 − 3 x − y = 0 { 5 x + 2 y = 2 − 3 x − y = 0

{ 6 x − 5 y = −1 2 x + y = 13 { 6 x − 5 y = −1 2 x + y = 13

{ 2 x − 5 y = 7 3 x − y = 17 { 2 x − 5 y = 7 3 x − y = 17

{ 5 x − 3 y = −1 2 x − y = 2 { 5 x − 3 y = −1 2 x − y = 2

{ 3 x − 5 y = −9 5 x + 2 y = 16 { 3 x − 5 y = −9 5 x + 2 y = 16

{ 4 x − 3 y = 3 2 x + 5 y = −31 { 4 x − 3 y = 3 2 x + 5 y = −31

{ 3 x + 8 y = −3 2 x + 5 y = −3 { 3 x + 8 y = −3 2 x + 5 y = −3

{ 11 x + 9 y = −5 7 x + 5 y = −1 { 11 x + 9 y = −5 7 x + 5 y = −1

{ 3 x + 8 y = 67 5 x + 3 y = 60 { 3 x + 8 y = 67 5 x + 3 y = 60

{ 2 x + 9 y = −4 3 x + 13 y = −7 { 2 x + 9 y = −4 3 x + 13 y = −7

{ 1 3 x − y = −3 x + 5 2 y = 2 { 1 3 x − y = −3 x + 5 2 y = 2

{ x + 1 2 y = 3 2 1 5 x − 1 5 y = 3 { x + 1 2 y = 3 2 1 5 x − 1 5 y = 3

{ x + 1 3 y = −1 1 3 x + 1 2 y = 1 { x + 1 3 y = −1 1 3 x + 1 2 y = 1

{ 1 3 x − y = −3 2 3 x + 5 2 y = 3 { 1 3 x − y = −3 2 3 x + 5 2 y = 3

{ 2 x + y = 3 6 x + 3 y = 9 { 2 x + y = 3 6 x + 3 y = 9

{ x − 4 y = −1 − 3 x + 12 y = 3 { x − 4 y = −1 − 3 x + 12 y = 3

{ − 3 x − y = 8 6 x + 2 y = −16 { − 3 x − y = 8 6 x + 2 y = −16

{ 4 x + 3 y = 2 20 x + 15 y = 10 { 4 x + 3 y = 2 20 x + 15 y = 10

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

ⓐ { 8 x − 15 y = −32 6 x + 3 y = −5 { 8 x − 15 y = −32 6 x + 3 y = −5 ⓑ { x = 4 y − 3 4 x − 2 y = −6 { x = 4 y − 3 4 x − 2 y = −6

ⓐ { y = 7 x − 5 3 x − 2 y = 16 { y = 7 x − 5 3 x − 2 y = 16 ⓑ { 12 x − 5 y = −42 3 x + 7 y = −15 { 12 x − 5 y = −42 3 x + 7 y = −15

ⓐ { y = 4 x + 9 5 x − 2 y = −21 { y = 4 x + 9 5 x − 2 y = −21 ⓑ { 9 x − 4 y = 24 3 x + 5 y = −14 { 9 x − 4 y = 24 3 x + 5 y = −14

ⓐ { 14 x − 15 y = −30 7 x + 2 y = 10 { 14 x − 15 y = −30 7 x + 2 y = 10 ⓑ { x = 9 y − 11 2 x − 7 y = −27 { x = 9 y − 11 2 x − 7 y = −27

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S = solve( eqn , var ) solves the equation eqn for the variable var . If you do not specify var , the symvar function determines the variable to solve for. For example, solve(x + 1 == 2, x) solves the equation x  + 1 = 2 for x .

S = solve( eqn , var , Name,Value ) uses additional options specified by one or more Name,Value pair arguments.

Y = solve( eqns , vars ) solves the system of equations eqns for the variables vars and returns a structure that contains the solutions. If you do not specify vars , solve uses symvar to find the variables to solve for. In this case, the number of variables that symvar finds is equal to the number of equations eqns .

Y = solve( eqns , vars , Name,Value ) uses additional options specified by one or more Name,Value pair arguments.

[ y1,...,yN ] = solve( eqns , vars ) solves the system of equations eqns for the variables vars . The solutions are assigned to the variables y1,...,yN . If you do not specify the variables, solve uses symvar to find the variables to solve for. In this case, the number of variables that symvar finds is equal to the number of output arguments N .

[ y1,...,yN ] = solve( eqns , vars , Name,Value ) uses additional options specified by one or more Name,Value pair arguments.

[ y1,...,yN , parameters , conditions ] = solve( eqns , vars ,' ReturnConditions ',true) returns the additional arguments parameters and conditions that specify the parameters in the solution and the conditions on the solution.

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Solve Quadratic Equation

Solve the quadratic equation without specifying a variable to solve for. solve chooses x to return the solution.

( - b + b 2 - 4   a   c 2   a - b - b 2 - 4   a   c 2   a )

Specify the variable to solve for and solve the quadratic equation for a .

- c + b   x x 2

Solve Polynomial and Return Real Solutions

Solve a fifth-degree polynomial. It has five solutions.

( 5 - σ 1 - 5 4 - 5   2   5 - 5   i 4 - σ 1 - 5 4 + 5   2   5 - 5   i 4 σ 1 - 5 4 - 5   2   5 + 5   i 4 σ 1 - 5 4 + 5   2   5 + 5   i 4 ) where    σ 1 = 5   5 4

Return only real solutions by setting 'Real' option to true . The only real solutions of this equation is 5 .

Numerically Solve Equations

When solve cannot symbolically solve an equation, it tries to find a numeric solution using vpasolve . The vpasolve function returns the first solution found.

Try solving the following equation. solve returns a numeric solution because it cannot find a symbolic solution.

Plot the left and the right sides of the equation. Observe that the equation also has a positive solution.

Figure contains an axes object. The axes object contains 2 objects of type functionline.

Find the other solution by directly calling the numeric solver vpasolve and specifying the interval.

Solve Multivariate Equations and Assign Outputs to Structure

When solving for multiple variables, it can be more convenient to store the outputs in a structure array than in separate variables. The solve function returns a structure when you specify a single output argument and multiple outputs exist.

Solve a system of equations to return the solutions in a structure array.

Access the solutions by addressing the elements of the structure.

Using a structure array allows you to conveniently substitute solutions into other expressions.

Use the subs function to substitute the solutions S into other expressions.

If solve returns an empty object, then no solutions exist.

Solve Inequalities

The solve function can solve inequalities and return solutions that satisfy the inequalities. Solve the following inequalities.

x 2 + y 2 + x y < 1

Set 'ReturnConditions' to true to return any parameters in the solution and conditions on the solution.

u - 3   v 2 2 - v 2

The parameters u and v do not exist in MATLAB® workspace and must be accessed using S.parameters .

Check if the values u = 7/2 and v = 1/2 satisfy the condition using subs and isAlways .

isAlways returns logical 1 ( true ) indicating that these values satisfy the condition. Substitute these parameter values into S.x and S.y to find a solution for x and y .

Solve Multivariate Equations and Assign Outputs to Variables

Solve the system of equations.

2 u 2 + v 2 = 0

When solving for more than one variable, the order in which you specify the variables defines the order in which the solver returns the solutions. Assign the solutions to variables solv and solu by specifying the variables explicitly. The solver returns an array of solutions for each variable.

( - 2 3 - 2   i 3 - 2 3 + 2   i 3 )

( 1 3 - 2   i 3 1 3 + 2   i 3 )

Entries with the same index form the pair of solutions.

( - 2 3 - 2   i 3 1 3 - 2   i 3 - 2 3 + 2   i 3 1 3 + 2   i 3 )

Use Parameters and Conditions to Refine Solution

Return the complete solution of an equation with parameters and conditions of the solution by specifying 'ReturnConditions' as true .

Solve the equation sin ( x ) = 0 . Provide two additional output variables for output arguments parameters and conditions .

The solution π k contains the parameter k , where k must be an integer. The variable k does not exist in the MATLAB® workspace and must be accessed using parameters .

Restrict the solution to 0 < x < 2 π . Find a valid value of k for this restriction. Assume the condition, conditions , and use solve to find k . Substitute the value of k found into the solution for x .

Alternatively, determine the solution for x by choosing a value of k . Check if the value chosen satisfies the condition on k using isAlways .

Check if k = 4 satisfies the condition on k .

isAlways returns logical 1( true ), meaning that 4 is a valid value for k . Substitute k with 4 to obtain a solution for x . Use vpa to obtain a numeric approximation.

Shorten Result with Simplification Rules

Solve the equation exp ( log ( x ) log ( 3 x ) ) = 4 .

By default, solve does not apply simplifications that are not valid for all values of x . In this case, the solver does not assume that x is a positive real number, so it does not apply the logarithmic identity log ( 3 x ) = log ( 3 ) + log ( x ) . As a result, solve cannot solve the equation symbolically.

Set 'IgnoreAnalyticConstraints' to true to apply simplification rules that might allow solve to find a solution. For details, see Algorithms .

( 3   e - log ( 256 ) + log ( 3 ) 2 2 3 3   e log ( 256 ) + log ( 3 ) 2 2 3 )

solve applies simplifications that allow the solver to find a solution. The mathematical rules applied when performing simplifications are not always valid in general. In this example, the solver applies logarithmic identities with the assumption that x is a positive real number. Therefore, the solutions found in this mode should be verified.

Ignore Assumptions on Variables

The sym and syms functions let you set assumptions for symbolic variables.

Assume that the variable x is positive.

When you solve an equation for a variable under assumptions, the solver only returns solutions consistent with the assumptions. Solve this equation for x .

Allow solutions that do not satisfy the assumptions by setting 'IgnoreProperties' to true .

For further computations, clear the assumption that you set on the variable x by recreating it using syms .

Solve Polynomial Equations of High Degree

When you solve a polynomial equation, the solver might use root to return the solutions. Solve a third-degree polynomial.

( root ( z 3 + z 2 + a , z , 1 ) root ( z 3 + z 2 + a , z , 2 ) root ( z 3 + z 2 + a , z , 3 ) )

Try to get an explicit solution for such equations by calling the solver with 'MaxDegree' . The option specifies the maximum degree of polynomials for which the solver tries to return explicit solutions. The default value is 2 . Increasing this value, you can get explicit solutions for higher order polynomials.

Solve the same equations for explicit solutions by increasing the value of 'MaxDegree' to 3 .

( 1 9   σ 1 + σ 1 - 1 3 - 1 18   σ 1 - σ 1 2 - 1 3 - 3   1 9   σ 1 - σ 1   i 2 - 1 18   σ 1 - σ 1 2 - 1 3 + 3   1 9   σ 1 - σ 1   i 2 ) where    σ 1 = a 2 + 1 27 2 - 1 729 - a 2 - 1 27 1 / 3

Return One Solution

Solve the equation sin ( x ) + cos ( 2 x ) = 1 .

Instead of returning an infinite set of periodic solutions, the solver picks three solutions that it considers to be the most practical.

( 0 π 6 5   π 6 )

Choose only one solution by setting 'PrincipalValue' to true .

Input Arguments

Eqn — equation to solve symbolic expression | symbolic equation.

Equation to solve, specified as a symbolic expression or symbolic equation. The relation operator == defines symbolic equations. If eqn is a symbolic expression (without the right side), the solver assumes that the right side is 0, and solves the equation eqn == 0 .

var — Variable for which you solve equation symbolic variable

Variable for which you solve an equation, specified as a symbolic variable. By default, solve uses the variable determined by symvar .

eqns — System of equations symbolic expressions | symbolic equations

System of equations, specified as symbolic expressions or symbolic equations. If any elements of eqns are symbolic expressions (without the right side), solve equates the element to 0 .

vars — Variables for which you solve an equation or system of equations symbolic vector | symbolic matrix

Variables for which you solve an equation or system of equations, specified as a symbolic vector or symbolic matrix. By default, solve uses the variables determined by symvar .

The order in which you specify these variables defines the order in which the solver returns the solutions.

Name-Value Arguments

Example: 'Real',true specifies that the solver returns real solutions.

Real — Flag for returning only real solutions false (default) | true

Flag for returning only real solutions, specified as the comma-separated pair consisting of 'Real' and one of these values.

See Solve Polynomial and Return Real Solutions .

ReturnConditions — Flag for returning parameters and conditions false (default) | true

Flag for returning parameters in solution and conditions under which the solution is true, specified as the comma-separated pair consisting of 'ReturnConditions' and one of these values.

See Solve Inequalities .

Example: [v1, v2, params, conditions] = solve(sin(x) +y == 0,y^2 == 3,'ReturnConditions',true) returns the parameters in params and conditions in conditions .

IgnoreAnalyticConstraints — Simplification rules applied to expressions and equations false (default) | true

Simplification rules applied to expressions and equations, specified as the comma-separated pair consisting of 'IgnoreAnalyticConstraints' and one of these values.

See Shorten Result with Simplification Rules .

IgnoreProperties — Flag for returning solutions inconsistent with properties of variables false (default) | true

Flag for returning solutions inconsistent with the properties of variables, specified as the comma-separated pair consisting of 'IgnoreProperties' and one of these values.

See Ignore Assumptions on Variables .

MaxDegree — Maximum degree of polynomial equations for which solver uses explicit formulas 2 (default) | positive integer smaller than 5

Maximum degree of polynomial equations for which solver uses explicit formulas, specified as a positive integer smaller than 5. The solver does not use explicit formulas that involve radicals when solving polynomial equations of a degree larger than the specified value.

See Solve Polynomial Equations of High Degree .

PrincipalValue — Flag for returning one solution false (default) | true

Flag for returning one solution, specified as the comma-separated pair consisting of 'PrincipalValue' and one of these values.

See Return One Solution .

Output Arguments

S — solutions of equation symbolic array.

Solutions of an equation, returned as a symbolic array. The size of a symbolic array corresponds to the number of the solutions.

Y — Solutions of system of equations structure

Solutions of a system of equations, returned as a structure. The number of fields in the structure correspond to the number of independent variables in a system. If 'ReturnConditions' is set to true , the solve function returns two additional fields that contain the parameters in the solution, and the conditions under which the solution is true.

y1,...,yN — Solutions of system of equations symbolic variables

Solutions of a system of equations, returned as symbolic variables. The number of output variables or symbolic arrays must be equal to the number of independent variables in a system. If you explicitly specify independent variables vars , then the solver uses the same order to return the solutions. If you do not specify vars , the toolbox sorts independent variables alphabetically, and then assigns the solutions for these variables to the output variables.

parameters — Parameters in solution vector of generated parameters

Parameters in a solution, returned as a vector of generated parameters. This output argument is only returned if ReturnConditions is true . If a single output argument is provided, parameters is returned as a field of a structure. If multiple output arguments are provided, parameters is returned as the second-to-last output argument. The generated parameters do not appear in the MATLAB ® workspace. They must be accessed using parameters .

Example: [solx, params, conditions] = solve(sin(x) == 0, 'ReturnConditions', true) returns the parameter k in the argument params .

conditions — Conditions under which solutions are valid vector of symbolic expressions

Conditions under which solutions are valid, returned as a vector of symbolic expressions. This output argument is only returned if ReturnConditions is true . If a single output argument is provided, conditions is returned as a field of a structure. If multiple output arguments are provided, conditions is returned as the last output argument.

Example: [solx, params, conditions] = solve(sin(x) == 0, 'ReturnConditions', true) returns the condition in(k, 'integer') in conditions . The solution in solx is valid only under this condition.

If solve cannot find a solution and ReturnConditions is false , the solve function internally calls the numeric solver vpasolve that tries to find a numeric solution. For polynomial equations and systems without symbolic parameters, the numeric solver returns all solutions. For nonpolynomial equations and systems without symbolic parameters, the numeric solver returns only one solution (if a solution exists).

If solve cannot find a solution and ReturnConditions is true , solve returns an empty solution with a warning. If no solutions exist, solve returns an empty solution without a warning.

If the solution contains parameters and ReturnConditions is true , solve returns the parameters in the solution and the conditions under which the solutions are true. If ReturnConditions is false , the solve function either chooses values of the parameters and returns the corresponding results, or returns parameterized solutions without choosing particular values. In the latter case, solve also issues a warning indicating the values of parameters in the returned solutions.

If a parameter does not appear in any condition, it means the parameter can take any complex value.

The output of solve can contain parameters from the input equations in addition to parameters introduced by solve .

Parameters introduced by solve do not appear in the MATLAB workspace. They must be accessed using the output argument that contains them. Alternatively, to use the parameters in the MATLAB workspace use syms to initialize the parameter. For example, if the parameter is k , use syms k .

The variable names parameters and conditions are not allowed as inputs to solve .

To solve differential equations, use the dsolve function.

When solving a system of equations, always assign the result to output arguments. Output arguments let you access the values of the solutions of a system.

MaxDegree only accepts positive integers smaller than 5 because, in general, there are no explicit expressions for the roots of polynomials of degrees higher than 4.

The output variables y1,...,yN do not specify the variables for which solve solves equations or systems. If y1,...,yN are the variables that appear in eqns , then there is no guarantee that solve(eqns) will assign the solutions to y1,...,yN using the correct order. Thus, when you run [b,a] = solve(eqns) , you might get the solutions for a assigned to b and vice versa.

To ensure the order of the returned solutions, specify the variables vars . For example, the call [b,a] = solve(eqns,b,a) assigns the solutions for a to a and the solutions for b to b .

When you use IgnoreAnalyticConstraints , the solver applies some of these rules to the expressions on both sides of an equation.

log( a ) + log( b ) = log( a · b ) for all values of a and b . In particular, the following equality is valid for all values of a , b , and c :

   ( a · b ) c  =  a c · b c .

log( a b ) =  b ·log( a ) for all values of a and b . In particular, the following equality is valid for all values of a , b , and c :

   ( a b ) c  =  a b · c .

If f and g are standard mathematical functions and f ( g ( x )) =  x for all small positive numbers, f ( g ( x )) =  x is assumed to be valid for all complex values x . In particular:

log( e x ) =  x

asin(sin( x )) =  x , acos(cos( x )) =  x , atan(tan( x )) =  x

asinh(sinh( x )) =  x , acosh(cosh( x )) =  x , atanh(tanh( x )) =  x

W k ( x · e x ) =  x for all branch indices k of the Lambert W function.

The solver can multiply both sides of an equation by any expression except 0 .

The solutions of polynomial equations must be complete.

Version History

R2018a: support for character vectors has been removed.

Support for character vector or string inputs has been removed. Instead, use syms to declare variables and replace inputs such as solve('2*x == 1','x') with solve(2*x == 1,x) .

  • dsolve | isolate | linsolve | root | subs | symvar | vpasolve

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IMAGES

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