# How do you solve 6x + 2y = -2 and 4x + y = 1 using substitution?

Jun 20, 2017

See a solution process below:

#### Explanation:

Step 1) Solve the second equation for $y$:

$4 x + y = 1$

$- \textcolor{red}{4 x} + 4 x + y = - \textcolor{red}{4 x} + 1$

$0 + y = - 4 x + 1$

$y = - 4 x + 1$

Step 2) Substitute $\left(- 4 x + 1\right)$ for $y$ in the first equation and solve for $x$:

$6 x + 2 y = - 2$ becomes:

$6 x + 2 \left(- 4 x + 1\right) = - 2$

$6 x + \left(2 \cdot - 4 x\right) + \left(2 \cdot 1\right) = - 2$

$6 x + \left(- 8 x\right) + 2 = - 2$

$6 x - 8 x + 2 = - 2$

$\left(6 - 8\right) x + 2 = - 2$

$- 2 x + 2 = - 2$

$- 2 x + 2 - \textcolor{red}{2} = - 2 - \textcolor{red}{2}$

$- 2 x + 0 = - 4$

$- 2 x = - 4$

$\frac{- 2 x}{\textcolor{red}{- 2}} = \frac{- 4}{\textcolor{red}{- 2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 2}}} x}{\cancel{\textcolor{red}{- 2}}} = 2$

$x = 2$

Step 3) Substitute $2$ for $x$ in the solution to the second equation at the end of Step 1 and calculate $y$:

$y = - 4 x + 1$ becomes:

$y = \left(- 4 \cdot 2\right) + 1$

$y = - 8 + 1$

$y = - 7$

The solution is: $x = 2$ and $y = - 7$ or $\left(2 , - 7\right)$