How do you solve for x in #(a-3y)/ b +4=a+y#?

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Answer 1 · 2016-10-01T10:57:09

I assume the question is to solve the equation for #y# (not #x#), and I show it in the explanation section.

First step is to multiply by #b# to change both sides to polynomials.

#a-3y+4b=ab+yb#

Now we move the unknown #y# to left and other terms to right side:

#-3y-by=ab-a-4b#

Now we have to factorize left side to get #y# in one term only:

#-y(b+3)=ab-a-4b#

Next step is to multiply by #(-1)#:

#y(b+3)=a+4b-ab#

Finally we divide both sides by #(b+3)# to leave #y# alone

Answer: #y=(a+4b-ab)/(b+3)#