Example
Solve the following system of equations by substitution.
−x+y=−5 2x−5y=1
Answer:
First, we will solve the first equation for y.
−x+y=−5 y=x−5
Now, we can substitute the expression
x−5 for
y in the second equation.
2x−5y=12x−5(x−5)=12x−5x+25=1−3x=−24 x=8
Now, we substitute
x=8 into the first equation and solve for
y.
−(8)+y=−5 y=3
Our solution is
(8,3).
Check the solution by substituting
(8,3) into both equations.
−x+y=−5−(8)+(3)=−52x−5y=12(8)−5(3)=1TrueTrue
The substitution method can be used to solve any linear system in two variables, but the method works best if one of the equations contains a coefficient of 1 or
–1 so that we do not have to deal with fractions.
In the following video, you will be given an example of solving a system of two equations using the substitution method.
https://youtu.be/MIXL35YRzRw
If you had chosen the other equation to start with in the previous example, you would still be able to find the same solution. It is really a matter of preference, because sometimes solving for a variable will result in having to work with fractions. As you become more experienced with algebra, you will be able to anticipate what choices will lead to more desirable outcomes.
Recall that an
Example
Solve the following system of equations.
x=9−2yx+2y=13
Answer:
We can approach this problem in two ways. Because one equation is already solved for x, the most obvious step is to use substitution.
We can substitute the expression 9−2y for x in the second equation.
x+2y=13(9−2y)+2y=139+0y=139=13
Clearly, this statement is a contradiction because
9=13. Therefore, the system has no solution.
The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.
x=9−2y2y=−x+9 y=−21x+29
We then convert the second equation to slope-intercept form.
x+2y=13 2y=−x+13 y=−21x+213
Comparing the equations, we see that they have the same slope but different
y-intercepts. Therefore, the lines are parallel and do not intersect.
y=−21x+29y=−21x+213
Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown below.
In the next video, we show another example of using substitution to solve a system that has no solution.
https://youtu.be/kTtKfh5gFUc
In our next video, we show that a system can have an infinite number of solutions.
https://youtu.be/Pcqb109yK5Q
Consider a skateboard manufacturer’s
Example
Given the cost function
C(x)=0.85x+35,000 and the revenue function
R(x)=1.55x, find the break-even point.
Answer:
Write the system of equations using y to replace function notation.
y=0.85x+35,000y=1.55x
Substitute the expression
0.85x+35,000 from the first equation into the second equation and solve for
x.
0.85x+35,000=1.55x35,000=0.7x50,000=x
Then, substitute
x=50,000 into either the cost function or the revenue function.
1.55(50,000)=77,500
The break-even point is
(50,000,77,500).
The cost to produce
50,000 units is
$77,500, and the revenue from the sales of
50,000 units is also
$77,500. To make a profit, the business must produce and sell more than
50,000 units.

The company will make a profit after
50,000 units are produced.
In the next example, we will show how to write a system of linear equations given attendance and ticket cost data. We will then find the number of tickets purchased based on our system.
Example
The cost of a ticket to the circus is
$25.00 for children and
$50.00 for adults. On a certain day, attendance at the circus is
2,000 and the total gate revenue is
$70,000. How many children and how many adults bought tickets?
Answer:
Let c= the number of children and a= the number of adults in attendance.
The total number of people is 2,000. We can use this to write an equation for the number of people at the circus that day.
c+a=2,000
The revenue from all children can be found by multiplying
$25.00 by the number of children,
25c. The revenue from all adults can be found by multiplying
$50.00 by the number of adults,
50a. The total revenue is
$70,000. We can use this to write an equation for the revenue.
25c+50a=70,000
We now have a system of linear equations in two variables.
c+a=2,00025c+50a=70,000
In the first equation, the coefficient of both variables is
1. We can quickly solve the first equation for either
c or
a. We will solve for
a.
c+a=2,000a=2,000−c
Substitute the expression
2,000−c in the second equation for
a and solve for
c.
25c+50(2,000−c)=70,00025c+100,000−50c=70,000−25c=−30,000c=1,200
Substitute
c=1,200 into the first equation to solve for
a.
1,200+a=2,000a=800
We find that
1,200 children and
800 adults bought tickets to the circus that day.
In our last video example, we show how to set up a system of linear equations that represents the total cost for admission to a museum.
https://youtu.be/euh9ksWrq0A
In the next section, we will introduce more methods for solving systems of equations that cannot be easily solved by substitution.