Question 846641
To solve this system of equations, first, rewrite the second equation in terms of either x or y. I will rewrite it in terms of y.  So, I will rewrite x + y = 2, as y = 2 - x.  Next, replace y in equation 1 with 2 - x:


{{{x^2+(2-x)^2=9}}} ----->


{{{x^2+(2-x)*(2-x)=9}}} ----->


{{{x^2+4-4x+x^2=9}}}


Next, combine like terms:


{{{2x^2-4x+4=9}}}


Now, subtract 9 from both sides, giving us this equation in quadratic form:


{{{2x^2-4x+4-9=0}}} ----->


{{{2x^2-4x-5=0}}}


You will notice this equation cannot be factored with integers, so we can use the quadratic equation to find the values of x:


*[invoke quadratic "x", 2, -4, -5 ]


Rounding to two decimal points gives us x: 2.87 or -0.87


To find y, replace each of these values for x in our second equation x + y = 2:


2.87 + y = 2 ----->


y = 2 - 2.87 ----->


y = -.87


AND


-0.87 + y = 2 ----->


y = 2 + 0.87 ----->


y = 2.87


SO, x = 2.87 or -0.87 and y = 2.87 or -0.87