How do you solve #4/(w+4)=-7#?

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Answer 1 · 2018-03-25T22:17:55

#w=-32/7#

Our equation is essentially #(4)/(w+4)=-7/1#, thus we can cross-multiply to solve for #w#. We get:

#-7(w+4)=4#

Distributing the #-7#, we get:

#-7w-28=4#

Adding #28# to both sides, we get:

#-7w=32#

Dividing both sides by #-7#, we get:

#w=-32/7#

Answer 2 · 2018-03-25T22:42:03

#w=-32/7#

Start by multiplying each side of the equation by #(w+4)#

#4/cancel(w+4)*cancel((w+4))=-7(w+4)#

Expand the parentheses:

#4=-7w-28#

Add #7w# and subtract #4# from each side:

#cancel4+7w-cancel4=-cancel(7w)-28+cancel(7w)-4#

#7w=-32#

Divide both sides by 7:

#(7w)/7=(-32)/7#

#w=-32/7#