# How do you solve the system of equations algebraically 3x+4y=-1, 6x-2y=3?

Feb 14, 2017

$x = \frac{1}{3}$
$y = \frac{- 1}{2}$

#### Explanation:

Given -

$3 x + 4 y = - 1$------------------------(1)
$6 x - 2 y = 3$------------------------(2)

$6 x - 2 y = 3$------------------------(2)$\left(\times 2\right)$
$12 x - 4 y = 6$------------------------(3)

$3 x + 4 y = - 1$-------------------------(1)
$12 x - 4 y = 6$------------------------(3) $\left(1 + 3\right)$
$15 x = 5$
$x = \frac{5}{15} = \frac{1}{3}$
$x = \frac{1}{3}$

Substitute $x = \frac{1}{3}$ in equation (1)

$3 \left(\frac{1}{3}\right) + 4 y = - 1$
$1 + 4 y = - 1$
$4 y = - 1 - 1 = - 2$
$y = \frac{- 2}{4}$
$y = \frac{- 1}{2}$