# Thread: System of Equations HELP

1. ## System of Equations HELP

Solve the system:

x^2 + y^2 = 100
-x + y = 2

I don't know where to start because my teacher never taught us how to solve a system of equations where the variables are squared. If someone could show me how to do this then that'd be great. THANKS!

2. ## Re: System of Equations HELP

Write y in terms of x (from the second equation), substitute it into the first equation, then solve the quadratic.

3. ## Re: System of Equations HELP

Okay. I got:

x^2 + y^2 = 100
-x + y = 2 ------> y = x + 2

x^2 + (x + 2)^2 = 100
x^2 + x^2 + 4x + 4 = 100
2x^2 + 4x = 96
2x^2 + 4x - 96 = 0
(2x + 12)(x - 8)

2x + 12 = 0
2x = -12
x = -6

x - 8 = 0
x = 8

So from here would I plug both x-values in to my 'y = x + 2' equation? Thus giving me 2 solutions to the system of equations? Idk if I'm doing this right.

4. ## Re: System of Equations HELP

Originally Posted by Prove It
Write y in terms of x (from the second equation), substitute it into the first equation, then solve the quadratic.
Okay. I got:

x^2 + y^2 = 100
-x + y = 2 ------> y = x + 2

x^2 + (x + 2)^2 = 100
x^2 + x^2 + 4x + 4 = 100
2x^2 + 4x = 96
2x^2 + 4x - 96 = 0
(2x + 12)(x - 8)

2x + 12 = 0
2x = -12
x = -6

x - 8 = 0
x = 8

So from here would I plug both x-values in to my 'y = x + 2' equation? Thus giving me 2 solutions to the system of equations? Idk if I'm doing this right.

5. ## Re: System of Equations HELP

$(2x+12)(x-8) = 2x^2+12x-16x-96 = 2x^2-4x-96$, so you have the signs backwards. It should be $2(x-6)(x+8) = 0$ (note: I factored out the 2). Hence $x = 6, x = -8$. Now, plug those values in for $y = x+2$.

Finally, plug in both sets of values you found for $x$ and $y$ to the first equation to make sure it is correct.

6. ## Re: System of Equations HELP

Originally Posted by SlipEternal
$(2x+12)(x-8) = 2x^2+12x-16x-96 = 2x^2-4x-96$, so you have the signs backwards. It should be $2(x-6)(x+8) = 0$ (note: I factored out the 2). Hence $x = 6, x = -8$. Now, plug those values in for $y = x+2$.

Finally, plug in both sets of values you found for $x$ and $y$ to the first equation to make sure it is correct.
So therefore the solutions would be: (6, 8) & (-8, -6).

Thanks so much!