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!
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!
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.
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.
$\displaystyle (2x+12)(x-8) = 2x^2+12x-16x-96 = 2x^2-4x-96$, so you have the signs backwards. It should be $\displaystyle 2(x-6)(x+8) = 0$ (note: I factored out the 2). Hence $\displaystyle x = 6, x = -8$. Now, plug those values in for $\displaystyle y = x+2$.
Finally, plug in both sets of values you found for $\displaystyle x$ and $\displaystyle y$ to the first equation to make sure it is correct.