How do you solve the following system: $3x+2y =0, y + 4x = 16$ ?

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Answer 1 · 2018-06-22T16:05:36

$x=32/5,y=-48/5$

By the first equation we get

$y=-3/2x$
plugging this in the second one:

$-3/2x+4x=16$
$x=32/5$

$y=-48/5$

Answer 2 · 2018-06-22T16:09:42

$x = 6(2/5), y = -9(3/5)$

$3x + 2y = 0$ (1)

$2y = -3x$

$y + 4x = 16$ (2)

Substituting y in terms of x in (2),

$-(3/2)x + 4x = 16$

$-3x + 8x = 32$

$5x = 32$ or $x = 6(2/5)$

$y = 16 - 4x = 16 - (128/5) = -48/5 = -9(3/5)$

How do you solve the following system: 3x+2y =0, y + 4x = 16 3x+2y =0, y + 4x = 16 ?