# How do you solve the system of linear equations by elimination 12x-7y=-2, 8x+11y=30?

Jun 8, 2018

$x = 1$
$y = 2$

#### Explanation:

Solving By Elimination

$12 x - 7 y = - 2$
$8 x + 11 y = 30$

First, let's have a look at our system and recall what the elimination method is. To eliminate a variable (either $x$ or $y$), we have to multiply the different equations in the system to get the least common multiple. One equation must be positive and the other negative to cancel each other out. The least common multiple of $12$ and $8$ is $24$, so I multiply the first equation of the system by $2$ to get to $24 x$, and the second equation by $- 3$ to get $- 24 x$ to cancel out the positive $24 x$.

$2 \left(12 x - 7 y = - 2\right)$
$- 3 \left(8 x + 11 y = 30\right)$

Distribute. All you have to do is multiply each term in the parentheses by the number on the outside, so:

First equation
$2 \cdot 12 x = 24 x$
$2 \cdot - 7 y = - 14 y$
$2 \cdot - 2 = - 4$

Second equation
$- 3 \cdot 8 x = - 24 x$
$- 3 \cdot 11 y = - 33 y$
$- 3 \cdot 30 = - 90$

Your equations should now look like this:
$24 x - 14 y = - 4$
$- 24 x - 33 y = - 90$

Because $24 x$ and $- 24 x$ cancel out, you're left with $- 14 y$, $- 33 y$, $- 4$, and $- 90$. Combine like terms, so:

$- 47 y = - 94$

Divide by $- 47$ to isolate for $y$.

$y = 2$

Now that you know that $y = 2$, plug it into an equation of the system to solve for $x$:

$12 x - 7 y = - 2$
$12 x - 7 \left(2\right) = - 2$
$12 x - 14 = - 2$
$12 x = 12$
$x = 1$

Now that you know that $x$ is $1$, you can plug both values back into the system to confirm they're correct:

$12 x - 7 y = - 2$
$12 \left(1\right) - 7 \left(2\right) = - 2$
$12 - 14 = - 2$
$- 2 = - 2$

$8 x + 11 y = 30$
$8 \left(1\right) + 11 \left(2\right) = 30$
$8 + 22 = 30$
$30 = 30$

They're correct!