# How do you solve the following system: x-4y=2, 3x+4y=-10 ?

Mar 18, 2018

See a solution process below:

#### Explanation:

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

$x - 4 y + \textcolor{red}{4 y} = 2 + \textcolor{red}{4 y}$

$x - 0 = 2 + 4 y$

$x = 2 + 4 y$

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

$3 x + 4 y = - 10$ becomes:

$3 \left(2 + 4 y\right) + 4 y = - 10$

$\left(3 \times 2\right) + \left(3 \times 4 y\right) + 4 y = - 10$

$6 + 12 y + 4 y = - 10$

$6 + \left(12 + 4\right) y = - 10$

$6 + 16 y = - 10$

$6 - \textcolor{red}{6} + 16 y = - 10 - \textcolor{red}{6}$

$0 + 16 y = - 16$

$16 y = - 16$

$\frac{16 y}{\textcolor{red}{16}} = - \frac{16}{\textcolor{red}{16}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{16}}} y}{\cancel{\textcolor{red}{16}}} = - 1$

$y = - 1$

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

$x = 2 + 4 y$ becomes:

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

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

$x = 2 - 4$

$x = - 2$

The Solution Is:

$x = - 2$ and $y = - 1$

Or

$\left(- 2 , - 1\right)$