Substitute the ordered pair [latex]\left(5,1\right)[/latex] into both equations.

[latex]\begin{align}&\left(5\right)+3\left(1\right)= 5+3=8 && \text{True} \\ &2\left(5\right)-9=10-9=1&& \text{True} \end{align}[/latex]

The ordered pair [latex]\left(5,1\right)[/latex] satisfies both equations, so it is the solution to the system.

Analysis of the Solution

We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines.

Figure 3

Solve the first equation for [latex]y[/latex].

[latex]\begin{gathered}2x+y=-8\\ y=-2x - 8\end{gathered}[/latex]

Solve the second equation for [latex]y[/latex].

[latex]\begin{gathered}x-y=-1\\ y=x+1\end{gathered}[/latex]

Graph both equations on the same set of axes as in Figure 4.

Figure 4

The lines appear to intersect at the point [latex]\left(-3,-2\right)[/latex]. We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.

[latex]\begin{align}2\left(-3\right)+\left(-2\right)&=-6-2=-8&& \text{True} \\ \left(-3\right)-\left(-2\right)&=-3+2=-1&& \text{True} \end{align}[/latex]

The solution to the system is the ordered pair [latex]\left(-3,-2\right)[/latex], so the system is independent.

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

[latex]\begin{gathered}-x+y=-5 \\ y=x - 5 \end{gathered}[/latex]

Now we can substitute the expression [latex]x - 5[/latex] for [latex]y[/latex] in the second equation.

[latex]\begin{gathered}2x - 5y=1\\ 2x - 5\left(x - 5\right)=1\\ 2x - 5x+25=1\\ -3x=-24\\ x=8\end{gathered}[/latex]

Now, we substitute [latex]x=8[/latex] into the first equation and solve for [latex]y[/latex].

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

Both equations are already set equal to a constant. Notice that the coefficient of [latex]x[/latex] in the second equation, –1, is the opposite of the coefficient of [latex]x[/latex] in the first equation, 1. We can add the two equations to eliminate [latex]x[/latex] without needing to multiply by a constant.

[latex]\begin{align} x+2y&=-1 \\ -x+y&=3 \\ \hline 3y&=2\end{align}[/latex]

Now that we have eliminated [latex]x[/latex], we can solve the resulting equation for [latex]y[/latex].

[latex]\begin{gathered}3y=2 \\ y=\frac{2}{3} \end{gathered}[/latex]

Then, we substitute this value for [latex]y[/latex] into one of the original equations and solve for [latex]x[/latex].

[latex]\begin{gathered}-x+y=3 \\ -x+\frac{2}{3}=3\\ -x=3-\frac{2}{3}\\ -x=\frac{7}{3}\\ x=-\frac{7}{3} \end{gathered}[/latex]

The solution to this system is [latex]\left(-\frac{7}{3},\frac{2}{3}\right)[/latex].

Check the solution in the first equation.

[latex]\begin{align}x+2y&=\left(-\frac{7}{3}\right)+2\left(\frac{2}{3}\right) \\&=-\frac{7}{3}+\frac{4}{3} \\ &=-\frac{3}{3}=-1&& \text{True} \end{align}[/latex]

Analysis of the Solution

We gain an important perspective on systems of equations by looking at the graphical representation. See Figure 5 to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.

Figure 5

Adding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[/latex] in it and the second equation has [latex]x[/latex]. So if we multiply the second equation by [latex]-3,\text{}[/latex] the x-terms will add to zero.

[latex]\begin{align}x - 2y&=11 \\ -3\left(x - 2y\right)&=-3\left(11\right)&& \text{Multiply both sides by }-3. \\ -3x+6y&=-33&& \text{Use the distributive property}.\end{align}[/latex]

Now, let’s add them.

[latex]\begin{align} 3x+5y&=−11 \\ −3x+6y&=−33 \\ \hline 11y&=−44 \\ y&=−4 \end{align}[/latex]

For the last step, we substitute [latex]y=-4[/latex] into one of the original equations and solve for [latex]x[/latex].

[latex]\begin{align}3x+5y&=-11\\ 3x+5\left(-4\right)&=-11\\ 3x - 20&=-11\\ 3x&=9\\ x&=3\end{align}[/latex]

Our solution is the ordered pair [latex]\left(3,-4\right)[/latex]. Check the solution in the original second equation.

[latex]\begin{align}x - 2y&=\left(3\right)-2\left(-4\right)\\&=3+8\\ &=11&& \text{True} \end{align}[/latex]

Figure 6

One equation has [latex]2x[/latex] and the other has [latex]5x[/latex]. The least common multiple is [latex]10x[/latex] so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate [latex]x[/latex] by multiplying the first equation by [latex]-5[/latex] and the second equation by [latex]2[/latex].

[latex]\begin{align} -5\left(2x+3y\right)&=-5\left(-16\right) \\ -10x - 15y&=80 \\ 2\left(5x - 10y\right)&=2\left(30\right) \\ 10x - 20y&=60 \end{align}[/latex]

Then, we add the two equations together.

[latex]\begin{align}\ −10x−15y&=80 \\ 10x−20y&=60 \\ \hline −35y&=140 \\ y&=−4 \end{align}[/latex]

Substitute [latex]y=-4[/latex] into the original first equation.

[latex]\begin{gathered}2x+3\left(-4\right)=-16\\ 2x - 12=-16\\ 2x=-4\\ x=-2\end{gathered}[/latex]

The solution is [latex]\left(-2,-4\right)[/latex]. Check it in the other equation.

[latex]\begin{align}5x - 10y&= 5\left(-2\right)-10\left(-4\right) \\ &=-10+40 \\ &=30\end{align}[/latex]

Figure 7

First clear each equation of fractions by multiplying both sides of the equation by the least common denominator.

[latex]\begin{align}6\left(\frac{x}{3}+\frac{y}{6}\right)&=6\left(3\right) \\ 2x+y&=18 \\ 4\left(\frac{x}{2}-\frac{y}{4}\right)&=4\left(1\right) \\ 2x-y&=4 \end{align}[/latex]

Now multiply the second equation by [latex]-1[/latex] so that we can eliminate the x-variable.

[latex]\begin{align}-1\left(2x-y\right)&=-1\left(4\right) \\ -2x+y&=-4 \end{align}[/latex]

Add the two equations to eliminate the x-variable and solve the resulting equation.

[latex]\begin{align} 2x+y&=18 \\ −2x+y&=−4 \\ \hline 2y&=14 \\ y&=7 \end{align}[/latex]

Substitute [latex]y=7[/latex] into the first equation.

[latex]\begin{gathered}2x+\left(7\right)=18 \\ 2x=11\\ x=\frac{11}{2} \end{gathered}[/latex]

The solution is [latex]\left(\frac{11}{2},7\right)[/latex]. Check it in the other equation.

[latex]\begin{align}\frac{x}{2}-\frac{y}{4}&=\frac{\frac{11}{2}}{2}-\frac{7}{4}\\ &=\frac{11}{4}-\frac{7}{4} \\&=\frac{4}{4}\\&=1\end{align}[/latex]

We can approach this problem in two ways. Because one equation is already solved for [latex]x[/latex], the most obvious step is to use substitution.

[latex]\begin{gathered}x+2y=13 \\ \left(9 - 2y\right)+2y=13 \\ 9+0y=13 \\ 9=13 \end{gathered}[/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{align}x&=9 - 2y \\ 2y&=-x+9 \\ y&=-\frac{1}{2}x+\frac{9}{2} \end{align}[/latex]

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

[latex]\begin{gathered}x+2y=13 \\ 2y=-x+13 \\ y=-\frac{1}{2}x+\frac{13}{2} \end{gathered}[/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{gathered} y=-\frac{1}{2}x+\frac{9}{2} \\ y=-\frac{1}{2}x+\frac{13}{2}\end{gathered}[/latex]

Analysis of the Solution

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 in Figure 8.

Figure 8

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

[latex]\begin{align} y&=0.85x+35,000 \\ y&=1.55x \end{align}[/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{gathered}0.85x+35,000=1.55x\\ 35,000=0.7x\\ 50,000=x\end{gathered}[/latex]

Then, we 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 profit function is found using the formula [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex].

[latex]\begin{align}P\left(x\right)&=1.55x-\left(0.85x+35,000\right) \\ &=0.7x - 35,000 \end{align}[/latex]

The profit function is [latex]P\left(x\right)=0.7x - 35,000[/latex].

Analysis of the Solution

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.

Figure 11

We see from the graph in Figure 12 that the profit function has a negative value until [latex]x=50,000[/latex], when the graph crosses the x-axis. Then, the graph emerges into positive y-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.

Figure 12

Let c = the number of children and a = 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 $25.00 by the number of children, [latex]25c[/latex]. The revenue from all adults can be found by multiplying $50.00 by the number of adults, [latex]50a[/latex]. The total revenue is $70,000. 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{gathered}c+a=2,000\\ 25c+50a=70,000\end{gathered}[/latex]

In the first equation, the coefficient of both variables is 1. 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{gathered}c+a=2,000\\ a=2,000-c\end{gathered}[/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{gathered} 25c+50\left(2,000-c\right)=70,000 \\ 25c+100,000 - 50c=70,000\\-25c=-30,000 \\ c=1,200 \end{gathered}[/latex]

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

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

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