4MA0 Higher June 2013 Paper 3R Q25
25 Solve the simultaneous equations
\[\begin{aligned} x^2 + y^2 &= 26 \\ y &= 3 - 2x \end{aligned}\]Show clear algebraic working.
(6)
| Scheme | Marks |
|---|---|
| \(x^2 + (3 - 2x)^2 = 26\) | M1 |
| \(x^2 + 9 - 6x - 6x + 4x^2 = 26\) or \(5x^2 - 12x + 9 = 26\) | M1 |
| \(5x^2 - 12x - 17\ (= 0)\) | A1 |
\((5x - 17)(x + 1)\ (= 0)\) or \(\dfrac{--12 \pm \sqrt{(-12)^2 - 4 \times 5 \times (-17)}}{2 \times 5}\) (may be partially evaluated; condone lack of brackets around negative numbers) eg. \(\dfrac{12 \pm \sqrt{144 + 340}}{10}\) or \(\dfrac{12 \pm 22}{10}\) | M1 |
| \(x = 3.4\) oe , \(x = -1\) | A1 |
| \(x = 3.4\)oe \(x = -1\) \(y = 5\) \(y = -3.8\) oe | A1 |
| (6) |
Notes
M1: or \(y^2 + \left(\dfrac{3 - y}{2}\right)^2 = 26\)
M1: (indep) for correct expansion of \((3 - 2x)^2\) or \(\left(\dfrac{3 - y}{2}\right)^2\) even if unsimplified
eg \(y^2 + \left(\dfrac{9 - 6y + y^2}{4}\right) = 26\) or \(y^2 + \left(\dfrac{9 - 3y - 3y + y^2}{4}\right) = 26\)
A1: \(5y^2 - 6y - 95\ (= 0)\)
M1: \((5y + 19)(y - 5)\ (=0)\) oe
or \(\dfrac{--6 \pm \sqrt{(-6)^2 - 4 \times 5 \times (-95)}}{2 \times 5}\)
(may be partially evaluated; condone lack of brackets around negative numbers)
eg. \(\dfrac{6 \pm \sqrt{1936}}{10}\) or \(\dfrac{6 \pm 44}{10}\)
A1: \(y = 5\), \(y = -3.8\) oe
dep on all preceding marks
A1: NB. No marks for \(x = -1\) , \(y = 5\) with no working