Higher June 2023 Paper 1 Q17
17 Make \(x\) the subject of the formula \(y = \dfrac{4(2x - 7)}{5x + 3}\) (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(x = \dfrac{3y + 28}{8 - 5y}\) | M1 | for clearing the fraction eg \(y(5x + 3) = 4(2x - 7)\) or \(5xy + 3y = 8x - 28\) |
| M1 | (dep M1) for isolating \(x\) terms in a correct equation eg \(3y + 28 = 8x - 5xy\) | |
| M1 | (dep on two terms in \(x\)) for factorising eg eg \(x(8 - 5y) = 3y + 28\) | |
| A1 | for \(x = \dfrac{3y + 28}{8 - 5y}\) oe eg \(x = \dfrac{-3y - 28}{5y - 8}\) |
Additional guidance
Condone error in expansion of RHS for this mark