Question

X^2+2y=9
x-y+3=0
could someone show me the steps to solving this?

Answer 1

If you are finding x and y then :

1) call the first equation(a) and second equation(b)
2) solve for y in equation(b) and you'll get:

\[y = x + 3\]

then substitute y in equation(a) and you'll get :

\[x^2 + 2(x + 3) - 9 = 0 \]
\[x^2 + 2x -3 = 0\]\[(x-1)(x+3) = 0\]
then you have : x = 1 and x = -3

Substitute both x's in equation (b) to find y, so for the first x = 1, you'll get :
y = 1 + 3
y = 4

and for the second x = -3, you'll get :
y = -3 + 3
y = 0

In summary, your 2 points are :

(1,4) and (-3,0)

^_^

Answer 2

do you understand how to solve such problems now? :)

Answer 3

absolutly! thanks!