# How do you solve the system x-5y=36 and 2x+y=-16 using substitution?

May 31, 2017

See a solution process below:

#### Explanation:

Step 1) Solve the first equation for $x$:

$x - 5 y = 36$

$x - 5 y + \textcolor{red}{5 y} = 36 + \textcolor{red}{5 y}$

$x - 0 = 36 + 5 y$

$x = 36 + 5 y$

Step 2) Substitute $\left(36 + 5 y\right)$ for $x$ in the second equation and solve for $y$:

$2 x + y = - 16$ becomes:

$2 \left(36 + 5 y\right) + y = - 16$

$\left(2 \times 36\right) + \left(2 \times 5 y\right) + y = - 16$

$72 + 10 y + y = - 16$

$72 + 10 y + 1 y = - 16$

$72 + \left(10 + 1\right) y = - 16$

$72 + 11 y = - 16$

$- \textcolor{red}{72} + 72 + 11 y = - \textcolor{red}{72} - 16$

$0 + 11 y = - 88$

$11 y = - 88$

$\frac{11 y}{\textcolor{red}{11}} = - \frac{88}{\textcolor{red}{11}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{11}}} y}{\cancel{\textcolor{red}{11}}} = - 8$

$y = - 8$

Step 3) Substitute $- 8$ for $y$ in the solution to the first equation at the end of Step 1 and calculate $x$:

$x = 36 + 5 y$ becomes:

$x = 36 + \left(5 \times - 8\right)$

$x = 36 + \left(- 40\right)$

$x = - 4$

The solution is: $x = - 4$ and $y = - 8$ or $\left(- 4 , - 8\right)$