# How do you solve the system x + y = 12 and y = 2x?

$x = 4 , \setminus \setminus y = 8$

#### Explanation:

Given equations:

$x + y = 12 \setminus \ldots . . \left(1\right)$

$y = 2 x$

$2 x - y = 0 \setminus \ldots \ldots \ldots \left(2\right)$

Adding (1) & (2), we get

$x + y + 2 x - y = 12 + 0$

$3 x = 12$

$x = \frac{12}{3}$

$x = 4$

setting $x = 4$ in (1), we get

$4 + y = 12$

$y = 8$

hence the solution is

$x = 4 , \setminus \setminus y = 8$

Jul 28, 2018

$\left(4 , 8\right)$

#### Explanation:

We already have $y$ defined in terms of $x$, so we can plug this value into the first equation to get

$x + 2 x = 12$

Combining like terms, this simplifies to

$3 x = 12$

Dividing both sides by $3$m we get

$x = 4$

We can plug this value into the second equation to get

$y = 2 \left(4\right) \implies y = 8$

Therefore, the solution of these two equations is at the point

$\left(4 , 8\right)$

Hope this helps!