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

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Solve the system:

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

`x + y = -2`



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Posted (Answer #1)

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`(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`



Then, substitute this to EQ1 and simplify.

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










Now that the resulting equation is simplified, factor it.


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`




And when y=2, x is:




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

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