# How do you solve the system of equations -2x - 3y = 7 and x + 8y = - 36?

Dec 2, 2017

$y = - 5 \setminus \quad , \setminus \quad x = 4$

#### Explanation:

We’ll use elimination for this problem by first cancelling out the $x$ terms, so we can solve for $y$.

Given the equations:

$- 2 x - 3 y = 7$

$x + 8 y = - 36$

We can multiply the bottom equation by $2$, so $2 x$ and $- 2 x$ will be eliminated:

$2 \left(x + 8 y\right) = \left(- 36\right) 2$

$\setminus \implies \textcolor{red}{2 x} + 16 y = - 72$

Adding it to the other equation:

$\cancel{\textcolor{red}{- 2 x}} - 3 y = 7 \setminus \quad + \setminus \quad \cancel{\textcolor{red}{2 x}} + 16 y = - 72$

$\setminus \implies 13 y = - 65$

$\setminus \implies y \equiv 5$

Now we can plug $y$ into one of the equations to solve for $x$:

$- 2 x - 3 y = 7$

$\setminus \implies - 2 x - 3 \left(- 5\right) = 7$

$\setminus \implies - 2 x + 15 = 7$

\implies -2x=-8

$\setminus \implies x = 4$

Finally, to check our answers, we can plug both $x$ and $y$ into an equation:

$x + 8 y = - 36$

\implies 4+8(-5)=-36

$\setminus \implies 4 - 40 = - 36$

$\setminus \implies - 36 = - 36$