# How do you use substitution to solve the following equations: y=-6x-32, 2y=10x+46?

May 18, 2017

See a solution process below:

#### Explanation:

Step 1) Because the first equation is already solved for $y$ we can substitute $- 6 x - 32$ for $y$ in the second equation and solve for $x$:

$2 y = 10 x + 46$ becomes:

$2 \left(- 6 x - 32\right) = 10 x + 46$

$\left(2 \cdot - 6 x\right) - \left(2 \cdot 32\right) = 10 x + 46$

$- 12 x - 64 = 10 x + 46$

$\textcolor{red}{12 x} - 12 x - 64 - \textcolor{red}{46} = \textcolor{red}{12 x} + 10 x + 46 - \textcolor{red}{46}$

$0 - 110 = \left(\textcolor{red}{12} + 10\right) x + 0$

$- 110 = 22 x$

$- \frac{110}{\textcolor{red}{22}} = \frac{22 x}{\textcolor{red}{22}}$

$- 5 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{22}}} x}{\cancel{\textcolor{red}{22}}}$

$- 5 = x$

$x = - 5$

Step 2) Substitute $- 5$ for $x$ in the first equation and calculate $y$:

$y = - 6 x - 32$ becomes:

$y = \left(- 6 \times - 5\right) - 32$

$y = 30 - 32$

$y = - 2$

The solution is: $x = - 5$ and $y = - 2$ or $\left(- 5 , - 2\right)$