Question

solve system of equations by substitution method
-x+2y=3
x^2+y=3

Answer 1

Re-arrange the 2nd equation so it reads y=......

Answer 2

Or you can start from the first equation,
\[y=(3+x)/2\] And then substitute in the second equation.

Answer 3

then in the first equation where there is y substitute with the expression you derived above.

That will give you an equation in x only - so you can solve for x

Answer 4

i got stuck at -2x^2-x+9=0

Answer 5

did i make a mistake or does it have 2 x values?

Answer 6

The equation should be,
\[x^2+\frac{3+x}{2}=3\Rightarrow 2x^2+x-3\] And then solve the second grade equation.

Answer 7

I mean,
\[2x^2+x−3=0\]

Answer 8

how would i deal with the 2x^2?

Answer 9

With the formula,
\[ax^2+bx+c=0\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 10

Ohhhhhhhh i see. thanks for all the help!! i think i got it :)))

Answer 11

Ok, ;).