$\displaystyle y-2x=3$ $\displaystyle x^2+y^2=18$ Following Mr Fantastic's previous advice I did $\displaystyle -9=4x^2-y^2$ Sub into the second equation $\displaystyle 9=4x^2+x^2$ $\displaystyle \pm\sqrt{\frac{9}{5}}=x$
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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 $\displaystyle \pm\sqrt{3}=x$
Originally Posted by Mukilab $\displaystyle y-2x=3$ $\displaystyle x^2+y^2=18$ Following Mr Fantastic's previous advice I did $\displaystyle -9=4x^2-y^2$ No $\displaystyle (y-2x)^2 \neq y^2-4x^2$ $\displaystyle y=2x+3$ $\displaystyle y^2=(2x+3)^2$ Expand and substitute into the 2nd equation to get a quadratic with x as unknown
y=4x^2+12x+9 Substituted 0=5x^2+12x-9
Yes Now you need to solve 0=5x^2+12x-9
yes but I get a horrendous equation along the lines of $\displaystyle \frac{-12\pm\sqrt{324}}{10}$ This is a no calculator paper. Maybe I should just use that?
Originally Posted by Mukilab yes but I get a horrendous equation along the lines of $\displaystyle \frac{-12\pm\sqrt{324}}{10}$ This is a no calculator paper. Maybe I should just use that? Are you kidding us? SQRT(324) = 18
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