Solving Quadratic Simultaneous Equations

When you've got quadratic simultaneous equations like these, you're dealing with a circle and a straight line that intersect at two points. The key is using substitution to turn this into a problem you can actually solve.

Start by taking the simpler equation (usually the linear one) and substituting it into the more complex one. Here, we substitute x = 2y + 6 into the circle equation x² + y² = 36. This gives us $2y + 6$² + y² = 36.

Expand the brackets carefully: $2y + 6$² becomes 4y² + 24y + 36. Add the y² term to get 5y² + 24y + 36 = 36. Subtract 36 from both sides and you've got 5y² + 24y = 0.

Top Tip: Always check your algebra by substituting your answers back into the original equations - it's the best way to catch silly mistakes!

Factorise this quadratic: y$5y + 24$ = 0. This gives you y = 0 or y = -24/5 = -4⅘. Use x = 2y + 6 to find the corresponding x-values: when y = 0, x = 6; when y = -4⅘, x = -3⅗. So your two solutions are (6, 0) and $-3⅗, -4⅘$.