Re: Simultaneous Equation

Well, what have you tried? If you replace "y" in the second equation by "3x", you get a very simple quadratic equation in x alone. What quadratic equation and how do you solve it?

Re: Simultaneous Equation

I tried subbing 3x in as Y.

Ended up something like:

$\displaystyle 2(3x)(3x)-3x^2=15$

$\displaystyle 6x^2-3x^2=15$

$\displaystyle 6x^2-3x^2-15=0$

But I then dont know where to go from there, or if the answers are ever right.

Re: Simultaneous Equation

$\displaystyle y = 3x$

$\displaystyle 2y^2 - xy = 15$

substitute $\displaystyle 3x$ for $\displaystyle y$ in the second equation ...

$\displaystyle 2(3x)^2 - x(3x) = 15$

mind your order of operations ...

$\displaystyle 18x^2 - 3x^2 = 15$

$\displaystyle 15x^2 - 15 = 0$

$\displaystyle 15(x^2-1) = 0$

$\displaystyle 15(x-1)(x+1) = 0$

finish it