How do you solve this system of equation using the elimination method: $2 x + 4 y = 16, 2 x - 4 y = 0$ $2x + 4y = 16, 2x - 4y = 0$ ?

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Answer 1 · 2017-07-04T22:37:16

See below.

We notice that there is a $+ 4 y$ $+4y$ and a $- 4 y$ $-4y$ . If we added the two equations, we would eliminate $y$ $y$ .

$2 x + 4 y = 16$ $2x+4y=16$
$+$ $+$
$2 x - 4 y = 0$ $2x-4y=0$

$4 x = 16$ $4x=16$

$x = 4$ $x=4$

Plugging this in,

$2 (4) + 4 y = 16$ $2(4)+4y=16$

$4 y = 8$ $4y=8$

$y = 2$ $y=2$

Thus, our solution set is $(4, 2)$ $(4,2)$

How do you solve this system of equation using the elimination method: 2x+4y=16,2x−4y=02x + 4y = 16, 2x - 4y = 0?