Topics
3. Systems of Linear Equations and Inequalities
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United States of America
Grades 9-12

3.02 Solving systems using substitution

Lesson
Practice

Introduction

We can solve systems of equations by graphing, as seen in lesson  4.01 Writing and graphing systems of equations  . However, systems can be solved algebraically, which may provide an advantage in efficiency or accuracy depending on the context.

Solving systems using substitution

Exploration

Consider the following system of equations:

\begin{cases} x+ 3y = 6 \\ y = 9 \end{cases}

  1. How could you solve this without graphing?
  2. Would that method still work if the second equation was y=9-2x?

When at least one equation in a system of equations has an isolated variable (or a variable that can be easily isolated), the system can be solved efficiently by using a method called substitution.

Substitution method

A method of solving a system of equations by replacing a variable in one equation with an equivalent expression from another equation

The following steps can be used to solve a system of equations using substitution:

StepsExample
\text{Given system}x-y=5, 2x+3y=6
\text{1. Isolate a variable in one of the equations}x-y=5
\text{ }x=5+y
\text{2. Substitute the resulting expression into the other equation}2\left(5+y\right)+3y=6
\text{3. Solve the equation for the variable}10+2y+3y=6
\text{ }10+5y=6
\text{ }5y=-4
\text{ }y=-\dfrac{4}{5}
\text{4. Substitute the value into one of the original equations}x-\left(-\dfrac{4}{5}\right)=5
\text{5. Solve for the remaining variable}x=\dfrac{21}{5}
\text{6. Write the solution as an ordered pair}\left(\dfrac{21}{5},-\dfrac{4}{5}\right)

Recall that a solution to a system of equations is the set of ordered pairs that make all equations in the system true and that a system of linear equations can have three types of solutions:

-4
-3
-2
-1
1
2
3
4
x
-4
-3
-2
-1
1
2
3
4
y

A system of linear equations with one solution.

Solving algebraically will result in equations of the form x=a and y=b, where a and b could be the same number.

-4
-3
-2
-1
1
2
3
4
x
-4
-3
-2
-1
1
2
3
4
y

A system of linear equations with no solution.

Solving algebraically will result in an equation of the form a=b, where a and b are different numbers.

-4
-3
-2
-1
1
2
3
4
x
-4
-3
-2
-1
1
2
3
4
y

A system of linear equations with infinitely many solutions.

Solving algebraically will result in equations of the form a=a, where a is a number.

When the solution to a system of equations does not consist of integer values it is difficult to determine the exact solution by graphing, so solving algebraically using a method like substitution is necessary.

Examples

Example 1

Solve the following systems of equations using the substitution method.

a

\begin{cases}y=x+11\\y=3x+19 \end{cases}

Worked Solution
Create a strategy

We first want to number our equations to make them easier to work with.

1\displaystyle y\displaystyle =\displaystyle x + 11
2\displaystyle y\displaystyle =\displaystyle 3x + 19

Since both equations already have y isolated, we can start by substituting equation 1 into equation 2 to eliminate y from the equation. We can then solve for x, and substitute this value back into one of the equations to solve for y.

Apply the idea

First we will solve for x:

\displaystyle y\displaystyle =\displaystyle 3x + 19Equation 2
\displaystyle x + 11\displaystyle =\displaystyle 3x + 19Substitute y=x+11
\displaystyle 11\displaystyle =\displaystyle 2x + 19Subtract x from both sides
\displaystyle -8\displaystyle =\displaystyle 2xSubtract 19 from both sides
\displaystyle -4\displaystyle =\displaystyle xDivide both sides by 2

Now we will use x=-4 to solve for y:

\displaystyle y\displaystyle =\displaystyle x + 11Equation 1
\displaystyle y\displaystyle =\displaystyle -4 + 11Substitute x=-4
\displaystyle y\displaystyle =\displaystyle 7Evaluate the addition

So the solution to the system of equations is x = -4, y = 7 and written as (-4,7).

b

\begin{cases} 2x-3y=5 \\ x = 8 + 2y \end{cases}

Worked Solution
Create a strategy

We first want to number our equations to make them easier to work with.

1\displaystyle 2x-3y\displaystyle =\displaystyle 5
2\displaystyle x\displaystyle =\displaystyle 8+2y

Since equation 2 already has x isolated, we can start by substituting equation 2 into equation 1 to eliminate x from the equation. We can then solve for y, and substitute this value back into one of the equations to solve for x.

Apply the idea

First we will solve for y:

\displaystyle 2x-3y\displaystyle =\displaystyle 5Equation 1
\displaystyle 2(8+2y)-3y\displaystyle =\displaystyle 5Substitute x=8+2y
\displaystyle 16+4y-3y\displaystyle =\displaystyle 5Distributive property
\displaystyle 16+y\displaystyle =\displaystyle 5Combine like terms
\displaystyle y\displaystyle =\displaystyle -11Subtract 16 from both sides

Now we will use y=-11 to solve for x:

\displaystyle x\displaystyle =\displaystyle 8+2yEquation 2
\displaystyle x\displaystyle =\displaystyle 8+2(-11)Substitute y=-11
\displaystyle x\displaystyle =\displaystyle -14Evaluate the multiplication and addition

So the solution to the system of equations is x = -14, y = -11 and written as (-14,-11).

c

\begin{cases}y = \dfrac{1}{2}x\\ 4x-8y=20 \end{cases}

Worked Solution
Create a strategy

The first equation in the system is already isolated for y, so we can substitute the expression \dfrac{1}{2}x for the value of y in the second equation.

Apply the idea
\displaystyle 4x-8y\displaystyle =\displaystyle 20Equation 2
\displaystyle 4x-8\left(\dfrac{1}{2}x\right)\displaystyle =\displaystyle 20Substitute y=\dfrac{1}{2}x
\displaystyle 4x-4x\displaystyle =\displaystyle 20Substitute y=\dfrac{1}{2}x
\displaystyle 0\displaystyle =\displaystyle 20Combine like terms

In this situation, the x-term is eliminated when solving, and we have a false statement. This means there is no solution to the system of equations.

Reflect and check

Converting the second equation in the system to slope-intercept form would show that the equations have the same slope and are actually parallel when graphed.

\displaystyle 4x-8y\displaystyle =\displaystyle 20Equation 2
\displaystyle -8y\displaystyle =\displaystyle -4x +20Subtract 4x from both sides
\displaystyle y\displaystyle =\displaystyle \dfrac{1}{2}x - \dfrac{5}{2}Divide both sides by -8

Both equations have the slope \dfrac{1}{2} and different y-intercepts.

Example 2

The length of a rectangle is 3 inches less than twice its width. If the perimeter of the rectangle is 48 inches, find the length.

Worked Solution
Create a strategy

For this real-world situation, we should start by defining the variables, then build a system of equations to model the situation.

The length of the rectangle in inches is unknown, so we can give it the variable l. Since the length and width make up the perimeter of the rectangle and the width in inches is also unknown, we can give it the variable w.

Apply the idea

Based on the first sentence in the problem, we know that the length is equal to 3 inches less than twice its width. The equation that models this relationship is l=2w-3.

We know that the perimeter of a rectangle is equal to the sum of twice the length and twice the width, so the equation that models this relationship is 2l + 2w=48.

We can write our system of equations as: \begin{cases} l=2w-3 \\ 2l+2w=48 \end{cases}

Solve for w:

\displaystyle 2l+2w\displaystyle =\displaystyle 48Second equation
\displaystyle 2(2w-3)+2w\displaystyle =\displaystyle 48Substitute l=2w-3
\displaystyle 4w-6+2w\displaystyle =\displaystyle 48Distributive property
\displaystyle 6w-6\displaystyle =\displaystyle 48Combine like terms
\displaystyle 6w\displaystyle =\displaystyle 54Add 6 to both sides
\displaystyle w\displaystyle =\displaystyle 9Divide both sides by 6

Solve for l:

\displaystyle l\displaystyle =\displaystyle 2w-3First equation
\displaystyle l\displaystyle =\displaystyle 2(9)-3Substitute w=9
\displaystyle l\displaystyle =\displaystyle 15Evaluate the multiplication and subtraction

The length of the rectangle is 15 inches.

Example 3

A theater club at a high school charges a student rate and an adult rate to attend the spring musical. The cost for a student ticket is \$3 and the cost for an adult ticket is \$7. If 200 people attended the show and the theater club at the school raised \$700, determine how many students and how many adults attended.

Worked Solution
Create a strategy

A system of equations will help us solve for the unknown variables, which are the number of students, x, and the number of adults, y.

Based on the total amounts provided in the problem, we will write two equations in standard form to model the situation and solve the system to find the solution.

Apply the idea

A system of equations that models the situation follows:

\begin{cases} x+ y = 200 \\ 3x + 7y = 700 \end{cases}

To solve the system using the substitution method, at least one equation must have an isolated variable. We can easily isolate either x or y in the first equation because they do not have coefficients.

\displaystyle x+y\displaystyle =\displaystyle 200First equation
\displaystyle x\displaystyle =\displaystyle 200-ySubtract y from both sides

We will substitute the expression 200-y for x in the second equation to solve for y.

\displaystyle 3x+7y\displaystyle =\displaystyle 700Second equation
\displaystyle 3(200-y)+7y\displaystyle =\displaystyle 700Substitute x=200-y
\displaystyle 600-3y+7y\displaystyle =\displaystyle 700Distributive property
\displaystyle 4y\displaystyle =\displaystyle 100Combine like terms
\displaystyle y\displaystyle =\displaystyle 25Divide both sides by 4

Now we will solve for x using y=25.

\displaystyle x+y\displaystyle =\displaystyle 200First equation
\displaystyle x+25\displaystyle =\displaystyle 200Substitute y=25
\displaystyle x\displaystyle =\displaystyle 175Subtract 25 from both sides

So the solution to the system of equations is x = 175, y = 25 and written as (175,25). Based on the context, this means that 175 students attended the spring musical and 25 adults attended.

Idea summary

Solving systems of equations using the substitution method leads to one of three solutions:

  • One solution: When solving algebraically an equation of the form x=a or y=b is reached
  • No solutions: When solving algebraically an equation of the form a=b is reached
  • Infinitely many solutions: When solving algebraically an equation of the form a=a is reached

Outcomes

A.CED.A.3

Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context.

A.REI.C.6

Solve systems of linear equations exactly and approximately (e.g. With graphs), focusing on pairs of linear equations in two variables.

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