Given #(x^2+2xy+y^2)(x+6) = 64#, how do you find the value of #x+y#?

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Answer 1 · 2017-05-05T18:15:14

See below.

Suppose that you are interested in integer solutions. So you can do

#(x^2+2xy+y^2)(x+6)=(x+y)^2(x+6)=8^2=4^3=2^6#

so you have many choices.

Now making a choice like that

#{((x+y)^2= 16),(x+6=4):}-> {(x+y = pm 4),(x+6=4):}#

we will have two solutions

#(x = -2, y = -2)# and #(x = -2, y = 6)#

So there is a lot of solutions.