# How do you solve the system of equations-8x+5y=-8 and -16x-y=-16 by elimination?

Jul 31, 2018

$x = 1$ and $y = 0$

#### Explanation:

$2 \cdot \left(- 8 x + 5 y\right) - \left(- 16 x - y\right) = 2 \cdot \left(- 8\right) - \left(- 16\right)$

$- 16 x + 10 y + 16 y + y = - 16 + 16$

$11 y = 0$, so $y = \frac{0}{11} = 0$

Hence,

$- 8 x + 5 \cdot 0 = - 8$

$- 8 x = - 8$, thus $x = \frac{- 8}{- 8} = 1$

Jul 31, 2018

$x = 1 \mathmr{and} y = 0$

#### Explanation:

The perfect scenario for the elimination method is if one of the variables can be additive inverses.

Multiply the first equation by $- 2$ to achieve this:

$\textcolor{w h i t e}{\times \times \times \times \times x} - 8 x + 5 y \text{ "=-8" } \ldots . A$
$\textcolor{w h i t e}{\times \times \times \times . \times x} \textcolor{b l u e}{- 16 x} - y \text{ "=-16" } \ldots . B$
$A \times - 2 : \textcolor{w h i t e}{\times \times x . x} \textcolor{b l u e}{16 x} - 10 y = + 16 \text{ } \ldots . C$

$B + C : \textcolor{w h i t e}{\times \times \times \times . x} - 11 y = 0$

$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times x . x} y = 0$

If $y = 0$ then $- 8 x = - 8$

Which gives $x = 1$