# How do you solve [(2x), (2x+3y)]=[(y), (12)]?

Aug 5, 2017

$x = \frac{3}{2} \text{ and } y = 3$

#### Explanation:

Here we have the system

$\left[\begin{matrix}2 x \\ 2 x + 3 y\end{matrix}\right] = \left[\begin{matrix}y \\ 12\end{matrix}\right] {\to}_{R 2 - R 1}$

We can subtract the first equation from the second equation to cancel out the $2 x$ factor

$\left[\begin{matrix}0 \\ 0 + 3 y\end{matrix}\right] = \left[\begin{matrix}0 \\ 12 - y\end{matrix}\right]$

Which then leaves us with the equation

$3 y = 12 - y$

$\iff$ Add $y$ to both sides

$4 y = 12$

$\iff$ Divide both sides by $4$

$y = 3$

Then, since we know that $y = 2 x$ from the first equation, we can plug in

$2 x = 3$

$\iff$ Divide both sides by $2$

$x = \frac{3}{2}$

We can check out answer by plugging in $\left(\frac{3}{2} , 3\right)$ into the second equation

$2 \left(\frac{3}{2}\right) + 3 \left(3\right) = 12$

$\iff$

$3 + 9 = 12$

$\iff$

$12 = 12$