+2446 When I solve for m in this equation \(\sqrt{\frac{2m}{5-m}}=t\), I agree with Mark Scheme's answer of \(m=\frac{5t^2}{t^2+2}\).
| \(\sqrt{\frac{2m}{5-m}}=t\) | Squaring both sides is definitely the first step. This will make it easier to make m the subject of this equation. |
| \(\frac{2m}{5-m}=t^2\) | Let's multiply both sides by the denominator 5-m. |
| \(2m=t^2(5-m)\) | Let's distribute. |
| \(2m=5t^2-t^2m\) | It is generally easiest to place all terms with an "m" in it on the same side of the equation. Let's do that by adding \(t^2m\) |
| \(t^2m+2m=5t^2\) | Factor out the greatest common factor of the left hand side of the equation. This is why you should place all terms with an "m" in it on the same side. |
| \(m(t^2+2)=5t^2\) | Divide by \(t^2+2\) |
| \(m=\frac{5t^2}{t^2+2},t\geq 0\) | I place this restriction on t because the original equation had a square root equation in it, which can only result in a nonnegative answer. |
You did not provide any work on how you arrived at your answer, so I do not know which step went awry. Maybe this work will help guide you, and it may help you spot your error. Good luck!
