Solve the equation

**Mukilab** · June 6th 2010, 01:15 AM

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

Following Mr Fantastic's previous advice I did

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

Sub into the second equation
$<br /> 9=4x^2+x^2$

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

**Mukilab** · June 6th 2010, 01:18 AM

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
$\pm\sqrt{3}=x$

**running-gag** · June 6th 2010, 01:19 AM

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

**Mukilab** · June 6th 2010, 01:32 AM

y=4x^2+12x+9

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

**running-gag** · June 6th 2010, 01:34 AM

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

**Mukilab** · June 6th 2010, 02:20 AM

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?

**Wilmer** · June 6th 2010, 03:14 AM

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

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