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

1 Answer
Oct 9, 2017

x=-1 and y=1

Explanation:

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.