4MA0 Higher June 2013 Paper 3 Q22
22 Make \(y\) the subject of \(\dfrac{y}{x} + \dfrac{2y}{x + 4} = 3\)
Show your working clearly and give your answer as simply as possible.
(5)
| Scheme | Marks |
|---|---|
\(\dfrac{y(x+4)}{x(x+4)} + \dfrac{2xy}{x(x+4)} = 3\) or \(\dfrac{y(x+4)}{x(x+4)} + \dfrac{2xy}{x(x+4)} = \dfrac{3x(x+4)}{x(x+4)}\) | M1 |
\(y(x + 4) + 2xy = 3x(x + 4)\) or \(\dfrac{xy+4y}{x(x+4)} + \dfrac{2xy}{x(x+4)} = 3\) or \(\dfrac{xy+4y}{x(x+4)} + \dfrac{2xy}{x(x+4)} = \dfrac{3x(x+4)}{x(x+4)}\) | M1 |
| \(xy + 4y + 2xy = 3x^2 + 12x\) or \(xy + 4y + 2xy = 3x(x+4)\) or \(3xy + 4y = 3x^2 + 12x\) or \(3xy + 4y = 3x(x+4)\) | M1 |
| \(y(3x + 4) = 3x(x + 4)\) or \(y(3x + 4) = 3x^2 + 12x\) | M1 |
| \(\dfrac{3x(x+4)}{3x+4}\) | A1 |
| (5) |
Notes
M1: LHS may be two separate fractions or one single fraction
(brackets may or may not be removed on RHS and denominator)
M1: LHS may be two separate fractions or one single fraction; if one fraction, numerator on LHS may or may not be simplified
(implies previous M1)
(brackets may or may not be removed on RHS and denominator)
M1: (brackets may or may not be removed on RHS)
(implies previous two M1s)
M1: LHS factorised correctly - expression in bracket on LHS may or may not be simplified
A1: \(\dfrac{3x(x+4)}{3x+4}\) or \(\dfrac{3x^2+12x}{3x+4}\)
a fully correct method must be seen in order to award full marks