How do you solve the following system: 3x + 2y = 1, 2x - 3y = 10 ?

1 Answer
Apr 16, 2018

$x = \frac{23}{13} , y = - \frac{28}{13}$

Explanation:

$3 x + 2 y = 1$ equation (1)
$2 x - 3 y = 10$ equation (2)

From (2): $2 x = 3 y + 10 \implies x = \frac{3 y + 10}{2}$

Substitute $x = \frac{3 y + 10}{2}$ into (1);

$3 \left(\frac{3 y + 10}{2}\right) + 2 y = 1$

$3 \left(3 y + 10\right) + 4 y = 2$

$9 y + 30 + 4 y = 2$

$13 y = - 28$

$y = - \frac{28}{13}$

Substitute $y = - \frac{28}{13}$ into $x = \frac{3 y + 10}{2}$:

$x = \frac{3 \left(- \frac{28}{13}\right) + 10}{2} = \frac{23}{13}$