How do you solve $4(3 y - 2) + 5( y + 8) = 0$ ?

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Answer 1 · 2016-03-04T16:59:51

$y=-32/17$

$4(3y-2) + 5(y+8)=0$

$4 xx (3y) + 4 xx (-2) + 5xx(y) + 5 xx (8)=0$

$12y -8 + 5y + 40=0$

$17y -8 + 40=0$

$17y +32=0$

$17y =-32$

$y=-32/17$

How do you solve 4(3 y - 2) + 5( y + 8) = 04(3 y - 2) + 5( y + 8) = 0?