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

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Answer 1 · 2018-06-22T16:05:36

$x = \frac{32}{5}, y = - \frac{48}{5}$ $x=32/5,y=-48/5$

By the first equation we get

$y = - \frac{3}{2} x$ $y=-3/2x$
plugging this in the second one:

$- \frac{3}{2} x + 4 x = 16$ $-3/2x+4x=16$
$x = \frac{32}{5}$ $x=32/5$

$y = - \frac{48}{5}$ $y=-48/5$

Answer 2 · 2018-06-22T16:09:42

$x = 6 (\frac{2}{5}), y = - 9 (\frac{3}{5})$ $x = 6(2/5), y = -9(3/5)$

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

$2 y = - 3 x$ $2y = -3x$

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

Substituting y in terms of x in (2),

$- (\frac{3}{2}) x + 4 x = 16$ $-(3/2)x + 4x = 16$

$- 3 x + 8 x = 32$ $-3x + 8x = 32$

$5 x = 32$ $5x = 32$ or $x = 6 (\frac{2}{5})$ $x = 6(2/5)$

$y = 16 - 4 x = 16 - (\frac{128}{5}) = - \frac{48}{5} = - 9 (\frac{3}{5})$ $y = 16 - 4x = 16 - (128/5) = -48/5 = -9(3/5)$

How do you solve the following system: 3x+2y=0,y+4x=163x+2y =0, y + 4x = 16 ?