# How do you solve the following system?: 3x+3y=-21 , 6x+5y=18

$x = 53 , \setminus \setminus y = - 60$

#### Explanation:

Given equations

$3 x + 3 y = - 21 \setminus \ldots \ldots . . \left(1\right)$

$6 x + 5 y = 18 \setminus \ldots \ldots . . \left(2\right)$

Multiplying (1) by $2$ & subtracting from (2) as follows

$6 x + 5 y - 2 \left(3 x + 3 y\right) = 18 - 2 \left(- 21\right)$

$6 x + 5 y - 6 x - 6 y = 18 + 42$

$- y = 60$

$y = - 60$

Setting the value of $y$ in (1), we get

$3 x + 3 \left(- 60\right) = - 21$

$3 x - 180 = - 21$

$3 x = 180 - 21$

$x = \setminus \frac{159}{3}$

$x = 53$

Jul 22, 2018

$x = 53$ and $y = - 60$

#### Explanation:

$2 \cdot \left(3 x + 3 y\right) - \left(6 x + 5 y\right) = 2 \cdot \left(- 21\right) - 18$

$\left(6 x + 6 y\right) - \left(6 x + 5 y\right) = \left(- 42\right) - 18$

$y = - 60$

So,

$3 x + 3 \cdot \left(- 60\right) = - 21$

$3 x - 180 = - 21$

$3 x = 159$ or $x = 53$