# How do solve the following linear system?:  -6x+6y=6 , -6x+3y=-12 ?

Apr 4, 2018

$\left(x , y\right) = \left(5 , 6\right)$

#### Explanation:

$6 y = 6 x + 6$
$y = x + 1$ solve for $y$ in terms of $x$ in the first equation
$- 6 x + 3 \left(x + 1\right) = - 12$ plug into the other equation
$- 6 x + 3 x + 3 = - 12$ distribute
$3 - 3 x = - 12$ combine like terms
$- 3 x = - 15$ subtract 3 from both sides
$x = 5$ divide both sides by -3

Now, going back to the other equation
$- 6 \left(5\right) + 6 y = 6$ plug in the value 5 for x
$- 30 + 6 y = 6$ multiply
$6 y = 36$ subtract 30 from both sides
$y = 6$ divide both sides by 6

Apr 4, 2018

Subtract equation 2 from equation 1 to solve for $y$, and then back-solve for $x$ to find that $x = 5$ and $y = 6$

#### Explanation:

Let's use linear algebra to solve the system of equations. This will involve subtracting whole equations from each other:

$- 6 x + 6 y = 6$
$\underline{- \left(- 6 x + 3 y = - 12\right)}$

$- 6 x + 6 y = 6$
$\underline{+ 6 x - 3 y = 12}$
$\cancel{0 x} + 3 y = 18$

Now that $x$ is eliminated, we can solve for $y$:

$3 y = 18$

$\Rightarrow \frac{\cancel{3} y}{\cancel{\textcolor{red}{3}}} = \frac{18}{\textcolor{red}{3}}$

$\Rightarrow \textcolor{b l u e}{y = 6}$

Finally, we can plug our solution for $y$ back into either equation to solve for $x$:

$- 6 x + 3 y = - 12$

$\Rightarrow - 6 x + 3 \left(\textcolor{b l u e}{6}\right) = - 12$

$\Rightarrow - 6 x + 18 = - 12$

$\Rightarrow - 6 x \cancel{+ 18 \textcolor{red}{- 18}} = - 12 \textcolor{red}{- 18}$

$\Rightarrow - 6 x = - 30$

$\Rightarrow \frac{\cancel{- 6} x}{\cancel{\textcolor{red}{- 6}}} = \frac{- 30}{\textcolor{red}{- 6}}$

$\Rightarrow \textcolor{g r e e n}{x = 5}$