Higher June 2024 Paper 1 Q17
17 Rearrange \(\quad y = \dfrac{3x + 7}{x} \quad\) to make \(x\) the subject. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: multiplies by \(x\) first | ||
| \(xy = 3x + 7\) | M1 | allow \(yx\) for \(xy\) throughout |
| \(xy - 3x = 7\) or \(3x - xy = -7\) | M1dep | oe collection of terms |
| \(x(y - 3) = 7\) or \(x(3 - y) = -7\) or \(\dfrac{7}{y - 3}\) or \(\dfrac{-7}{3 - y}\) | M1dep | |
| \(x = \dfrac{7}{y - 3}\) or \(x = \dfrac{-7}{3 - y}\) | A1 | oe in the form \(x =\) may have brackets on the denominator |
| Alternative method 2: splits up the fraction first | ||
| \(y = 3 + \dfrac{7}{x}\) or \(y - \dfrac{7}{x} = 3\) | M1 | allow \(\dfrac{3x}{x}\) for 3 |
| \(y - 3 = \dfrac{7}{x}\) or \(3 - y = -\dfrac{7}{x}\) | M1dep | |
| \(\dfrac{1}{y - 3} = \dfrac{x}{7}\) or \(x(y - 3) = 7\) or \(x(3 - y) = -7\) or \(\dfrac{7}{y - 3}\) or \(\dfrac{-7}{3 - y}\) | M1dep | |
| \(x = \dfrac{7}{y - 3}\) or \(x = \dfrac{-7}{3 - y}\) | A1 | oe in the form \(x =\) may have brackets on the denominator |
Additional guidance
| Up to M2 may be awarded for correct work with no answer or incorrect answer if this is seen amongst multiple attempts | |
| \(\dfrac{7}{y - 3}\) on answer line with \(x = \dfrac{7}{y - 3}\) in working | M3A1 |
| Allow the equation with \(x\) on the right, eg \(\dfrac{7}{y - 3} = x\) | M3A1 |
| Condone \(x = 7/y - 3\) if not from incorrect working | M3A1 |
| Allow appropriate \(\times\) or \(\div\) signs throughout for up to M3 |