4MA0 Higher January 2017 Paper 3 Q23
23 Solve the simultaneous equations
\[\begin{aligned} x^2 + y^2 &= 52 \\ 2x + y &= 8 \end{aligned}\]Show clear algebraic working.
(6)
| Scheme | Marks |
|---|---|
| \(x^2 + (8 - 2x)^2 = 52\) | M1 |
| \(x^2 + 4x^2 - 16x - 16x + 64 = 52\) | M1 |
| \(5x^2 - 32x + 12\) (= 0) | A1 |
\((5x - 2)(x - 6)\) (=0) or \(\dfrac{- -32 \pm \sqrt{(-32)^2 - 4 \times 5 \times 12}}{2 \times 5}\) (may be partially evaluated, condone lack of brackets around negative numbers) | M1 |
| \(x = \dfrac{2}{5}\) or \(x = 6\) | A1 |
\(x = \dfrac{2}{5}\) oe, \(y = \dfrac{36}{5}\) oe \(x = 6\), \(y = -4\) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for elimination of one variable
\(\left(\dfrac{8 - y}{2}\right)^2 + y^2 = 52\)
M1: (indep) for a correct expansion of
\((8 - 2x)^2\) or \(\left(\dfrac{8 - y}{2}\right)^2\)
\(\left(\dfrac{64 - 8y - 8y + y^2}{4}\right) + y^2 = 52\)
A1: for correct simplified 3 term quadratic equation in any form (may not be equated to zero)
\(5y^2 - 16y - 144\) (= 0)
M1: \((5y - 36)(y + 4)\) (=0)
or \(\dfrac{- -16 \pm \sqrt{(-16)^2 - 4 \times 5 \times -144}}{2 \times 5}\)
(may be partially evaluated, condone lack of brackets around negative numbers)
NB: can ft for this mark only provided M1 awarded and a 3 term quadratic
A1: for both \(x\) values (or both \(y\) values)
\(y = \dfrac{36}{5}\) or \(y = -4\)
A1: for both solutions with \(x\) and \(y\) values correctly paired