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#