1. ## Solve the equation

$y-2x=3$
$x^2+y^2=18$

Following Mr Fantastic's previous advice I did

$-9=4x^2-y^2$

Sub into the second equation
$
9=4x^2+x^2$

$\pm\sqrt{\frac{9}{5}}=x$

2. Did it again and got a different answer

y^2=9+2x^2
-y^2=x^2-18

Add them together to eliminate y

0=3x^2-9
9=3x^2
$\pm\sqrt{3}=x$

3. Originally Posted by Mukilab
$y-2x=3$
$x^2+y^2=18$

Following Mr Fantastic's previous advice I did

$-9=4x^2-y^2$
No $(y-2x)^2 \neq y^2-4x^2$

$y=2x+3$

$y^2=(2x+3)^2$

Expand and substitute into the 2nd equation to get a quadratic with x as unknown

4. y=4x^2+12x+9

Substituted
0=5x^2+12x-9

5. Yes
Now you need to solve 0=5x^2+12x-9

6. yes but I get a horrendous equation along the lines of

$\frac{-12\pm\sqrt{324}}{10}$

This is a no calculator paper.

Maybe I should just use that?

7. Originally Posted by Mukilab
yes but I get a horrendous equation along the lines of

$\frac{-12\pm\sqrt{324}}{10}$

This is a no calculator paper.

Maybe I should just use that?
Are you kidding us? SQRT(324) = 18