How do you solve the system of equations $3 x + 4 y = - 8$ $3x + 4y = - 8$ and $3 x + 16 y = - 32$ $3x + 16y = - 32$ ?

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Answer 1 · 2018-05-30T17:38:44

$x = 0$ $x=0$
$y = - 2$ $y=-2$

simply subtract each term in the second equation from the first equation:

$3 x + 4 y = - 8$ $3x + 4y = - 8$
$3 x + 16 y = - 32$ $3x + 16y = - 32$

$(3 x - 3 x) + (4 y - 16 y) = (- 8 - (- 32))$ $(3x-3x) + (4y-16y) = (-8 - (-32))$

$(0) + (- 12 y) = (- 8 + 32)$ $(0) + (-12y) = (-8 +32)$

$- 12 y = 24$ $-12y = 24$

$y = - 2$ $y=-2$

now use either original equation to solve for $x$ $x$ :

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

$3 x + 4 (- 2) = - 8$ $3x + 4(-2) = - 8$

$3 x - 8 = - 8$ $3x -8 = - 8$

$3 x = 0$ $3x = 0$

$x = 0$ $x=0$

How do you solve the system of equations 3x+4y=−83x + 4y = - 8 and 3x+16y=−323x + 16y = - 32?