4MA0 Higher June 2017 Paper 3R Q18
18 Solve the simultaneous equations
\[\begin{aligned} y^2 + 4x &= 12 \\ 2x + 3y &= 10 \end{aligned}\]Show clear algebraic working.
(6)
| Scheme | Marks |
|---|---|
\(y^2 + 4x = 12\) \(4x + 6y = 20\) with subtraction or \(\left(\dfrac{10 - 2x}{3}\right)^2 + 4x = 12\) or \(2 \times \left(\dfrac{12 - y^2}{4}\right) + 3y = 10\) oe | M1 |
| E.g. \(y^2 - 6y = -8\) or \(4x^2 - 4x - 8 = 0\) | A1 |
| E.g. \((y - 2)(y - 4)\) (= 0) or \(4(x - 2)(x + 1)\) (=0) | M1 |
| A1 | |
(\(x =\)) \(\dfrac{10 - 3 \times 2}{2}\) or \(\dfrac{10 - 3 \times 4}{2}\) (\(y =\)) \(\dfrac{10 - 2 \times 2}{3}\) or \(\dfrac{10 - 2 \times (-1)}{3}\) | M1 |
| \(x = 2\), \(y = 2\) or \(x = -1\), \(y = 4\) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for eliminating one variable
multiplication of equation(s) to get same multiple of \(y\) with subtraction (condone one arithmetic error) or
either rearrangement of one equation and then correct substitution into second equation (condone algebraic error in rearrangement)
A1: reduction to a correct 3 term quadratic;
terms may not all be ‘on the same side’
M1: ft if first M1 awarded and equation is quadratic
for correct factorisation or
correct substitution into formula
A1: for \(y = 2\), \(y = 4\) or \(x = 2\), \(x = -1\)
correct \(x\) or \(y\) values implies previous M1
M1: (dep on the previous M1)
for correct substitution to find both values
A1: values for \(x\) and \(y\) must be correctly paired
dep on M1 awarded