4MA0 Higher June 2017 Paper 3 Q14
14
Show clear algebraic working. (4)
| Scheme | Marks |
|---|---|
| \(x^4\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(6 + 4y = 3(5 - 2y)\) | M1 |
| \(6 + 4y = 15 - 6y\) | M1 |
| \(4y + 6y = 15 - 6\) or \(10y = 9\) | M1 |
| \(\dfrac{9}{10}\) oe | A1 |
| (4) |
Notes
M1: for removing fraction
M1: for correct expansion of bracket in a correct equation
M1: for a correct equation with \(y\) terms isolated on one side
ft their equation if first M1 awarded
A1: dep on at least M2
SC: B2 for an answer of
\(y = 1.5\) oe with working shown or
\(y = -0.1\)oe with working shown
Alternative scheme
| Scheme | Marks |
|---|---|
| \(\dfrac{6}{3} + \dfrac{4y}{3} = 5 - 2y\) | M1 |
| \(\dfrac{4y}{3} + 2y = 5 - \dfrac{6}{3}\) | M1 |
| \(\dfrac{10y}{3} = 3\) | M1 |
| \(\dfrac{9}{10}\) oe | A1 |
Notes
M1: for dividing both terms on LHS by 3
allow 1.3(3…)
M1: for a correct equation with \(y\) terms isolated on one side
allow 1.3(3…)
M1: for \(y\) terms collated
allow 3.3(3…)
A1: dep on at least M2
| Scheme | Marks |
|---|---|
| \(g - gh = 3h + 1\) or \(-1 - 3h = gh - g\) | M1 |
| \(g(1 - h) = 3h + 1\) or \(-1 - 3h = g(h - 1)\) | M1 |
| \(g = \dfrac{3h + 1}{(1 - h)}\) oe | A1 |
| (3) | |
| (8 marks) |
Notes
M1: for a correct equation with terms in \(g\) isolated on one side of the equation
M1: for taking \(g\) out as a common factor (must be two terms in \(g\) but terms may not be correct (terms in \(g\) may not be isolated))
A1: for \(g = \dfrac{3h + 1}{(1 - h)}\) oe e.g. \(g = \dfrac{-1 - 3h}{(h - 1)}\)