Example

Solve the following system of equations by substitution.

Answer: First, we will solve the first equation for [latex]y[/latex].

Now, we can substitute the expression [latex]x - 5[/latex] for [latex]y[/latex] in the second equation. Now, we substitute [latex]x=8[/latex] into the first equation and solve for [latex]y[/latex]. Our solution is [latex]\left(8,3\right)[/latex]. Check the solution by substituting [latex]\left(8,3\right)[/latex] into both equations.

[latex]\begin{array}{llll}-x+y=-5\hfill & \hfill & \hfill & \hfill \\ -\left(8\right)+\left(3\right)=-5\hfill & \hfill & \hfill & \text{True}\hfill \\ 2x - 5y=1\hfill & \hfill & \hfill & \hfill \\ 2\left(8\right)-5\left(3\right)=1\hfill & \hfill & \hfill & \text{True}\hfill \end{array}[/latex]

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 [latex]–1[/latex] so that we do not have to deal with fractions.

Example

Solve the following system of equations.

[latex]\begin{array}{l}x=9 - 2y\hfill \\ x+2y=13\hfill \end{array}[/latex]

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 [latex]9-2y[/latex] for [latex]x[/latex] in the second equation.

[latex]\begin{array}{r}x+2y=13\hfill \\ \left(9 - 2y\right)+2y=13\hfill \\ 9+0y=13\hfill \\ 9=13\hfill \end{array}[/latex]

Clearly, this statement is a contradiction because [latex]9\ne 13[/latex]. 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.

[latex]\begin{array}{l}\text{ }x=9 - 2y\hfill \\ 2y=-x+9\hfill \\ \text{ }y=-\dfrac{1}{2}x+\dfrac{9}{2}\hfill \end{array}[/latex]

We then convert the second equation to slope-intercept form.

[latex]\begin{array}{l}x+2y=13\hfill \\ \text{ }2y=-x+13\hfill \\ \text{ }y=-\dfrac{1}{2}x+\dfrac{13}{2}\hfill \end{array}[/latex]

Comparing the equations, we see that they have the same slope but different y-intercepts. Therefore, the lines are parallel and do not intersect.

[latex]\begin{array}{l}\begin{array}{l}\\ y=-\dfrac{1}{2}x+\dfrac{9}{2}\end{array}\hfill \\ y=-\dfrac{1}{2}x+\dfrac{13}{2}\hfill \end{array}[/latex]

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.

Example

Given the cost function [latex]C\left(x\right)=0.85x+35,000[/latex] and the revenue function [latex]R\left(x\right)=1.55x[/latex], find the break-even point.

Answer: Write the system of equations using [latex]y[/latex] to replace function notation.

[latex]\begin{array}{l}\begin{array}{l}\\ y=0.85x+35,000\end{array}\hfill \\ y=1.55x\hfill \end{array}[/latex]

Substitute the expression [latex]0.85x+35,000[/latex] from the first equation into the second equation and solve for [latex]x[/latex].

[latex]\begin{array}{c}0.85x+35,000=1.55x\\ 35,000=0.7x\\ 50,000=x\end{array}[/latex]

Then, substitute [latex]x=50,000[/latex] into either the cost function or the revenue function.

[latex]1.55\left(50,000\right)=77,500[/latex]

The break-even point is [latex]\left(50,000,77,500\right)[/latex]. The cost to produce [latex]50,000[/latex] units is [latex]$77,500[/latex], and the revenue from the sales of [latex]50,000[/latex] units is also [latex]$77,500[/latex]. To make a profit, the business must produce and sell more than [latex]50,000[/latex] units. The company will make a profit after [latex]50,000[/latex] units are produced.  

Example

The cost of a ticket to the circus is [latex]$25.00[/latex] for children and [latex]$50.00[/latex] for adults. On a certain day, attendance at the circus is [latex]2,000[/latex] and the total gate revenue is [latex]$70,000[/latex]. How many children and how many adults bought tickets?

Answer: Let [latex]c=[/latex] the number of children and [latex]a=[/latex] the number of adults in attendance. The total number of people is [latex]2,000[/latex]. We can use this to write an equation for the number of people at the circus that day.

[latex]c+a=2,000[/latex]

The revenue from all children can be found by multiplying [latex]$25.00[/latex] by the number of children, [latex]25c[/latex]. The revenue from all adults can be found by multiplying [latex]$50.00[/latex] by the number of adults, [latex]50a[/latex]. The total revenue is [latex]$70,000[/latex]. We can use this to write an equation for the revenue.

[latex]25c+50a=70,000[/latex]

We now have a system of linear equations in two variables.

[latex]\begin{array}{c}c+a=2,000\\ 25c+50a=70,000\end{array}[/latex]

In the first equation, the coefficient of both variables is [latex]1[/latex]. We can quickly solve the first equation for either [latex]c[/latex] or [latex]a[/latex]. We will solve for [latex]a[/latex].

[latex]\begin{array}{c}c+a=2,000\\ a=2,000-c\end{array}[/latex]

Substitute the expression [latex]2,000-c[/latex] in the second equation for [latex]a[/latex] and solve for [latex]c[/latex].

[latex]\begin{array}{l} 25c+50\left(2,000-c\right)=70,000\hfill \\ 25c+100,000 - 50c=70,000\hfill \\ -25c=-30,000\hfill \\ c=1,200\hfill \end{array}[/latex]

Substitute [latex]c=1,200[/latex] into the first equation to solve for [latex]a[/latex].

[latex]\begin{array}{l}1,200+a=2,000\hfill \\ a=800\hfill \end{array}[/latex]

We find that [latex]1,200[/latex] children and [latex]800[/latex] adults bought tickets to the circus that day.