# How do you solve the system 2x-y= 7 and 3x-4y= -7?

May 19, 2018

$x = 7 \mathmr{and} y = 7$

#### Explanation:

$2 x - y = 7 - - - e q n 1$

$3 x - 4 y = - 7 - - - e q n 2$

Multiply $e q n 1$ by $3$ and $e q n 2$ by $2$

$3 \left(2 x - y = 7\right)$

$2 \left(3 x - 4 y = - 7\right)$

$6 x - 3 y = 21 - - - e q n 3$

$6 x - 8 y = - 14 - - - e q n 4$

Subtracting $e q n 4$ from $e q n 3$

(6x - 6x) + (-3y - (-8y)) = 21 - (-14))

$0 - 3 y + 8 y = 21 + 14$

$5 y = 35$

$y = \frac{35}{5}$

$y = 7$

Substiting the value of $y$ into $e q n 1$

$2 x - y = 7 - - - e q n 1$

$2 x - 7 = 7$

$2 x = 7 + 7$

$2 x = 14$

$x = \frac{14}{2}$

$x = 7$

Therefore;

$x = 7 \mathmr{and} y = 7$