How do you solve the following system: $3 x - y = 12, 2 x - \frac{y}{2} = 4$ $3x-y=12, 2x-y/2=4$ ?

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Answer 1 · 2016-02-10T10:55:00

$(- 4, - 24)$ $(-4,-24)$

$3 x - y = 12$ $3x-y=12$ -----(1)

$2 x - \frac{y}{2} = 4$ $2x- y/2=4$ ------------(2)

multiply $e q^{n}$ $eq^n$ (2) by 2 we get,

$4 x - y = 8$ $4x-y=8$ ----------(3)

subtract equation (1) from equation (2)

( $4 x - y = 8) - (3 x - y = 12)$ $4x-y=8) - (3x-y=12)$

$4 x - 3 x - y + y = 8 - 12$ $4x-3x-y+y=8-12$

$x = 8 - 12$ $x=8-12$

$x = - 4$ $x=-4$

Substitute in equation (1)

$3 x - y = 12$ $3x-y=12$

$3 (- 4) - y = 12$ $3(-4) -y =12$

$- 12 - y = 12$ $-12-y = 12$

$- 12 - 12 = y$ $-12-12 = y$

$y = - 24$ $y=-24$

How do you solve the following system: 3x−y=12,2x−y2=43x-y=12, 2x-y/2=4 ?