Solve the simultaneous equations $x=2y$ and $x^2−5xy=−24$
Solve for the first variable in one of the equations, then substitute the result into the other equation.
Since $x=2y$ in the first equation, substitute for $x$ in the second equation
\[(2y)^2-5(2y)y=-24\]
removing brackets yields
\[4y^2-10y^2=-24\]
which simplifies to
\[-6y^2=-24\]
dividing both sides by $-6$ gives
\[y^2=4\]
square-root both sides
\[\sqrt{y^2}=\pm \sqrt{4}\]
therefore $y=2$ or $y=-2$
Remember that $x=2y$
so $x=4$ when $y=2$ and $x=-4$ when $y=-2$