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

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Answer 1 · 2016-02-10T10:55:00

$(-4,-24)$

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

$2x- y/2=4$ ------------(2)

multiply $eq^n$ (2) by 2 we get,

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

subtract equation (1) from equation (2)

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

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

$x=8-12$

$x=-4$

Substitute in equation (1)

$3x-y=12$

$3(-4) -y =12$

$-12-y = 12$

$-12-12 = y$

$y=-24$

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