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

[latex]\begin{array}{l}-x+y=-5\hfill \\ \text{ }y=x - 5\hfill \end{array}[/latex]

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

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

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

[latex]\begin{array}{l}-\left(8\right)+y=-5\hfill \\ \text{ }y=3\hfill \end{array}[/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]

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]\frac{\begin{array}{l}\hfill \\ x+2y=-1\hfill \\ -x+y=3\hfill \end{array}}{\text{}\text{}\text{}\text{}\text{}3y=2}[/latex]

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

[latex]\begin{array}{l}3y=2\hfill \\ \text{ }y=\frac{2}{3}\hfill \end{array}[/latex]

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

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

Analysis of the Solution

We gain an important perspective on systems of equations by looking at the graphical representation. See the graph below 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.

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{array}{llll}\text{ }x - 2y=11\hfill & \hfill & \hfill & \hfill \\ -3\left(x - 2y\right)=-3\left(11\right)\hfill & \hfill & \hfill & \text{Multiply both sides by }-3.\hfill \\ \text{ }-3x+6y=-33\hfill & \hfill & \hfill & \text{Use the distributive property}.\hfill \end{array}[/latex]
Now, let’s add them.

[latex]\begin{array}\ \hfill 3x+5y=−11 \\ \hfill −3x+6y=−33 \\ \text{_____________} \\ \hfill 11y=−44 \\ \hfill y=−4 \end{array}[/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{array}{c}3x+5y=-11\\ 3x+5\left(-4\right)=-11\\ 3x - 20=-11\\ 3x=9\\ x=3\end{array}[/latex]

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

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

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{array}{l} -5\left(2x+3y\right)=-5\left(-16\right)\hfill \\ \text{ }-10x - 15y=80\hfill \\ \text{ }2\left(5x - 10y\right)=2\left(30\right)\hfill \\ \text{ }10x - 20y=60\hfill \end{array}[/latex]

Then, we add the two equations together.

[latex]\begin{array}\ −10x−15y=80 \\ 10x−20y=60 \\ \text{______________} \\ \text{ }−35y=140 \\ y=−4 \end{array}[/latex]

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

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

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

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

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

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

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

[latex]\begin{array}{l}-1\left(2x-y\right)=-1\left(4\right)\hfill \\ \text{ }-2x+y=-4\hfill \end{array}[/latex]

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

[latex]\begin{array}\ \hfill 2x+y=18 \\ \hfill−2x+y=−4 \\ \text{_____________} \\ \hfill 2y=14 \\ \hfill y=7 \end{array}[/latex]

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

[latex]\begin{array}{l}2x+\left(7\right)=18\hfill \\ \text{ }2x=11\hfill \\ \text{ }x=\frac{11}{2}\hfill \\ \text{ }=7.5\hfill \end{array}[/latex]

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

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