Question

What is the solution set to the following system? {x+y=3 {x^2+y^2=9

Answer 1

There are two solutions

The first solution is (x,y) = (3,0)

The second solution is (x,y) = (0,3)

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Work Shown:

Isolate x in the first equation (subtract y from both sides)

x+y = 3

x = -y+3

Now plug this into the second equation. Isolate y.

x^2 + y^2 = 9

(-y+3)^2 + y^2 = 9 ... replace x with -y+3

y^2 - 6y + 9 + y^2 = 9 ... FOIL rule

2y^2 - 6y + 9 = 9

2y^2 - 6y = 0

2y(y-3) = 0

2y = 0 or y-3 = 0 .... zero product property

y = 0 or y = 3

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If y = 0, then x is

x = -y+3

x = -0+3

x = 3

Therefore, x = 3 and y = 0 pair up to get the ordered pair (x,y) = (3,0) as one solution of this system

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If y = 3, then x is

x = -y+3

x = -3+3

x = 0

Therefore, x = 0 and y = 3 pair up to get the ordered pair (x,y) = (0,3) as the other solution of this system

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If you were to graph the two equations, you should find that the curves cross each other at (0,3) and (3,0). I've marked these two points as A and B respectively. The order of the solutions does not matter.