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.

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.

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.

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.