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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`
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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