# How do you solve the system of equations 8x+4y=13 and -2x=y+4?

May 3, 2017

See the entire solution process below:

#### Explanation:

First, sole the second equation for $y$:

$- 2 x = y + 4$

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

$- 2 x - 4 = y + 0$

$- 2 x - 4 = y$

$y = - 2 x - 4$

Step 2) Substitute $- 2 x - 4$ for $y$ in the first equation and solve for $x$:

$8 x + 4 y = 13$ becomes:

$8 x + 4 \left(- 2 x - 4\right) = 13$

$8 x + \left(4 \cdot - 2 x\right) - \left(4 \cdot 4\right) = 13$

$8 x - 8 x - 16 = 13$

$0 - 16 = 13$

$- 16 \ne 13$

Therefore, there are no solutions to this problem. Or, the solution is the empty or null set: $\left\{\emptyset\right\}$.

This means the two lines defined by these equations are parallel and are not the same line.