Higher November 2020 Paper 2 Q16
16 Solve the simultaneous equations
\(3xy - y^2 = 8\)
\(x - 2y = 1\)
Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
\(3y(2y + 1) - y^2 = 8\) or \(x = \dfrac{8 + y^2}{3y} \rightarrow \dfrac{8 + y^2}{3y} - 2y = 1\) or \(3xy - y^2 = 8\) \(3xy - 3y \times 2y = 3y \times 1\) oe (subtract) or \(3x\left(\dfrac{x - 1}{2}\right) - \left(\dfrac{x - 1}{2}\right)^2 = 8\) oe | M1 |
| eg \(5y^2 + 3y - 8\;(= 0)\) or eg \(5x^2 - 4x - 33\;(= 0)\) | A1 |
\((5y + 8)(y - 1)\;(= 0)\) or \(\dfrac{-3 \pm \sqrt{3^2 - 4 \times 5 \times (-8)}}{2 \times 5}\) or \((5x + 11)(x - 3)\;(= 0)\) or \(\dfrac{4 \pm \sqrt{(-4)^2 - 4 \times 5 \times (-33)}}{2 \times 5}\) | M1ft |
\(y = -\dfrac{8}{5}\) and \(y = 1\) (both) or \(x = -\dfrac{11}{5}\) and \(x = 3\) (both) | A1 |
Working required Answer: \(x = -\dfrac{11}{5}\), \(y = -\dfrac{8}{5}\) \(x = 3\), \(y = 1\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: correct first step eg substitution by eg \(x = 1 + 2y\) or \(y = \dfrac{x - 1}{2}\) to get an equation in a single variable
or
writing 2nd equation with \(x\) the subject and substituting into 1st
or
multiplying 2nd equation by \(3y\) and subtracting from 1st oe
A1: for a correct simplified quadratic
M1ft: dep on M1 for solving their 3 term quadratic equation using any correct method (allow one sign error and some simplification – allow as far as \(\dfrac{-3 \pm \sqrt{9 + 160}}{10}\)) or if factorising, allow brackets which expanded give 2 out of 3 terms correct)
A1: dep on first M1
A1: oe dep on first M1
Must be paired correctly