4MA0 Higher January 2014 Paper 4 Q22
22 Solve the simultaneous equations \[\begin{aligned} 2x - y &= 7 \\ x^2 + y^2 &= 34 \end{aligned}\]
Show clear algebraic working.
(7)
| Scheme | Marks |
|---|---|
| \(y = 2x - 7\) | M1 |
| \(x^2 + 4x^2 - 14x - 14x + 49 = 34\) | M2 |
| \(5x^2 - 28x + 15\;(= 0)\) | A1 |
| \((5x - 3)(x - 5)\;(=0)\) | M1 |
| \(x = 0.6\;\; x = 5\) Answer: \(x = 0.6\) & \(y = -5.8\) \(x = 5\) & \(y = 3\) | A1 A1 |
| (7) | |
| (7 marks) |
Notes
M2: M1 for \(4x^2 - 14x - 14x + 49\) or better
A1: correct 3 part quadratic
M1: or \(\dfrac{--28 \pm \sqrt{(-28)^2 - 4 \times 5 \times 15}}{2 \times 5}\)
or better or \(5x(x - 5) - 3(x - 5)\)
condone no brackets around negative numbers.
A1 A1: A1 Dependent on previous M1 (both \(x\) values correct).
A1 Dependent on previous M1 (both complete solutions correct).
Alternative
| Scheme | Marks |
|---|---|
| Alt: \(x = (y + 7)/2\) | M1 |
| \(0.25 \times (y^2 + 14y + 49) + y^2 = 34\) | M2 |
| \(5y^2 + 14y - 87 = 0\) | A1 |
| \((y - 3)(5y + 29)\;(=0)\) | M1 |
| \(y = 3\;\; y = -5.8\) Answer: \(x = 0.6\) & \(y = -5.8\) \(x = 5\) & \(y = 3\) | A1 A1 |
Notes
M2: M1 for \(0.25 \times (y^2 + 14y + 49)\)
A1: correct 3 part quadratic
M1: or \(\dfrac{-14 \pm \sqrt{14^2 - 4 \times 5 \times -87}}{2 \times 5}\)
or better or \(y(5y + 29) - 3(5y + 29)\)
A1 A1: A1 Dependent on previous M1 (both \(y\) values correct).
A1 Dependent on previous M1 (both complete solutions correct)