How do you solve this system of equations: #-2x + y = 3 and 3y + x ^ { 2} = ( 3+ x ) ^ { 2}#?

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Answer 1 · 2017-10-09T15:02:29

#x=-1 and y=1#

Lets take the equation #3y+x^2=(3+x)^2#

we'll apply #(a+b)^2=a^2+b^2+2ab# on the RHS of the equation.

We get #3y+x^2=9+x^2+6x#

#3y-6x=9#

Divide both sides of the equation by 3.

#y-2x=3#

As you can see its the same as the first equation that is #-2x+y=3#

We have two same linear equations.

Therefore we'll have to try hit and trial method on this linear equation.

If we keep #x=-1 and y=1# it satisfies our equation.