Higher June 2022 Paper 2R Q21
21 Solve the simultaneous equations
\[x - 2y = 3\] \[x^2 - y^2 + 2x = 10\]Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
\((3 + 2y)^2 - y^2 + 2(3 + 2y) = 10\) or \(x^2 - \left(\dfrac{x - 3}{2}\right)^2 + 2x = 10\) | M1 |
| eg \(3y^2 + 16y + 5\;(= 0)\) or eg \(3x^2 + 14x - 49\;(= 0)\) \(3x^2 + 14x = 49\) | A1 |
eg \((3y + 1)(y + 5)\;(= 0)\) or \(\dfrac{-16 \pm \sqrt{16^2 - 4 \times 3 \times 5}}{2 \times 3}\) or \(3\left[\left(y + \dfrac{8}{3}\right)^2 - \left(\dfrac{8}{3}\right)^2\right] + 5 = 0\) (should give \((y =)\;-\dfrac{1}{3}, -5\)) or eg \((3x - 7)(x + 7)\;(= 0)\) or \(\dfrac{-14 \pm \sqrt{14^2 - 4 \times 3 \times (-49)}}{2 \times 3}\) or \(3\left[\left(x + \dfrac{7}{3}\right)^2 - \left(\dfrac{7}{3}\right)^2\right] - 49 = 0\) (should give \((x =)\;\dfrac{7}{3}, -7\)) | M1 |
eg \(x = 3 + 2 \times -5\) and \(x = 3 + 2 \times -\dfrac{1}{3}\) or eg \(\dfrac{7}{3} - 2 \times y = 3\) \(-7 - 2 \times y = 3\) | M1ft |
Working required Answer: \(x = \dfrac{7}{3}\), \(y = -\dfrac{1}{3}\) \(x = -7\), \(y = -5\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for using correct substitution of a linear equation into the quadratic – all terms shown correctly
A1: for a correct 3 term quadratic
M1: dep on M1 method to solve their 3 term quadratic using any correct method (allow one sign error and some simplification – allow as far as eg \(\dfrac{-16 \pm \sqrt{256 - 60}}{6}\) or \(\dfrac{-14 \pm \sqrt{196 + 588}}{6}\) or if factorising allow brackets which expanded give 2 out of 3 terms correct)
or correct values for \(x\) or correct values for \(y\)
M1ft: dep on previous M1 for substituting their 2 found values of \(x\) or \(y\) in a suitable equation (use 2dp or better for substitution)
or fully correct values for the other variable (correct labels for \(x\) / \(y\))
A1: dep on M1 (allow coordinates)
must be paired correctly
allow \(x = -7\), \(y = -5\)
\(x = 2.33(3\ldots)\), \(y = -0.33(3\ldots)\)