4MA0 Higher January 2015 Paper 3R Q24
24 Solve the simultaneous equations
\[\begin{aligned} y &= 2x - 3 \\ x^2 + y^2 &= 41 \end{aligned}\]Show clear algebraic working.
(6)
| Scheme | Marks |
|---|---|
| \(x^2 + (2x - 3)^2 = 41\) | M1 |
| \(x^2 + 4x^2 - 6x - 6x + 9 = 41\) or \(x^2 + 4x^2 - 12x + 9 = 41\) | M1 |
| \(5x^2 - 12x - 32\ (= 0)\) | A1 |
\((5x + 8)(x - 4)\ (= 0)\) or \(5x(x - 4) + 8(x - 4)\ (= 0)\) or \(x(5x + 8) - 4(5x + 8)\) or \(\dfrac{--12 \pm \sqrt{(-12)^2 - 4 \times 5 \times -32}}{2 \times 5}\) (may be partially evaluated condone lack of brackets around negative numbers) eg \(\dfrac{12 \pm \sqrt{144 + 640}}{10}\) or \(\dfrac{12 \pm 28}{10}\) | M1 |
| \(x = -\dfrac{8}{5}\) or \(x = 4\) | A1 |
\(x = -\dfrac{8}{5}\), \(y = -\dfrac{31}{5}\) \(x = 4\), \(y = 5\) | A1 |
| (6 marks) |
Notes
M1: \(\left(\dfrac{y + 3}{2}\right)^2 + y^2 = 41\)
M1: \(\left(\dfrac{y^2 + 6y + 9}{4}\right) + y^2 = 41\)
or \(\left(\dfrac{y^2 + 3y + 3y + 9}{4}\right) + y^2 = 41\)
M1 (indep) for correct expansion of \((2x - 3)^2\) or \(\left(\dfrac{y + 3}{2}\right)^2\) even if unsimplified
A1: \(5y^2 + 6y - 155\ (=0)\)
M1: \((5y + 31)(y - 5)\ (=0)\) oe
or \(5y(y - 5) + 31(y - 5)\ (=0)\)
or \(y(5y + 31) - 5(5y + 31)\ (=0)\)
or \(\dfrac{-6 \pm \sqrt{6^2 - 4 \times 5 \times -155}}{2 \times 5}\) or better
Condone omission of ‘=0’
A1: \(y = -\dfrac{31}{5}\) or \(y = 5\)
dep on all preceding method marks