# Solve the system: `(x+3)^2 + (y+2)^2 = 17` `x + y = -2`

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lemjay | Certified Educator

`(x+3)^2 + (y+2)^2 = 17`    (Let this be EQ1.)

`x + y = -2`                          (Let this be EQ2.)

To solve this system of equation, substitution method can be applied.

To do so, solve for x in EQ2.

`x+ y =-2`

`x+y-y=-2-y`

`x=-y-2`

Then, substitute this to EQ1 and simplify.

`(x+3)^2 + (y+2)^2 = 17`

`(-y-2+3)^2+(y+2)^2=17`

`(-y+1)^2+(y+2)^2=17`

`y^2-2y+1+y^2+4y+4=17`

`2y^2+2y+5=17`

`2y^2+2y+5-17=17-17`

`2y^2+2y-12=0`

`2(y^2+y-6)=0`

`(2(y^2+y-6)/2)=0/2`

`y^2+y-6=0`

Now that the resulting equation is simplified, factor it.

`(y+3)(y-2)=0`

Then, set each factor equal to zero and solve for y.

`y+3=0`     and     `y-2=0`

`y=-3`                         `y=2`

Plug-in these values of y to EQ2 get the corresponding values of x.

So when y=-3 , x is:

`x + y = -2`

`x-3=-2`

`x-3+3=-2+3`

`x=1`

And when y=2, x is:

`x+2=-2`

`x+2-2=-2-2`

`x=-4`

Hence, the solution are `(1,-3)` and `(-4,2` ).

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