Higher November 2016 Unit 2 Q13
13. Make \(x\) the subject of the following formula. [4]
\[a(x - b) = x(c - d)\]| Scheme | Marks | Comments |
|---|---|---|
| \(ax - ab = cx - dx\) | B1 | FT until 2nd error for equivalent level of difficulty. |
| \(ax - cx + dx = ab\) OR \(-ab = cx - dx - ax\) | B1 | Or equivalent. |
| \(x(a - c + d) = ab\) OR \(-ab = x(c - d - a)\) | B1 | Or equivalent. |
| \(x = ab/(a - c + d)\) OR \(-ab/(c - d - a) = x\) or \(\dfrac{ab}{a-c+d}\) or \(\dfrac{-ab}{c-d-a}\) | B1 | Or equivalent. Do not accept \(-x = ab/(c - d - a)\) |
Alternative method
| Scheme | Marks | Comments |
|---|---|---|
| \(ax - ab = x(c - d)\) | B1 | Only award the first B1 if the 2nd step is attempted. |
| \(-ab = x(c - d) - ax\) | B1 | |
| \(-ab = x(c - d - a)\) | B1 | |
| \(-ab/(c - d - a) = x\) | B1 |