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

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Answer 1 · 2017-10-09T15:02:29

$x = - 1 and y = 1$ $x=-1 and y=1$

Lets take the equation $3 y + x^{2} = {(3 + x)}^{2}$ $3y+x^2=(3+x)^2$

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

We get $3 y + x^{2} = 9 + x^{2} + 6 x$ $3y+x^2=9+x^2+6x$

$3 y - 6 x = 9$ $3y-6x=9$

Divide both sides of the equation by 3.

$y - 2 x = 3$ $y-2x=3$

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

We have two same linear equations.

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

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

How do you solve this system of equations: −2x+y=3and3y+x2=(3+x)2-2x + y = 3 and 3y + x ^ { 2} = ( 3+ x ) ^ { 2}?