Higher June 2025 Paper 1 Q22
22 Solve the simultaneous equations
\[\begin{aligned} x^2 + 3y + y^2 &= 7 \\ y &= x + 2 \end{aligned}\]Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
| eg \(x^2 + 3(x + 2) + (x + 2)^2 = 7\) or eg \((y - 2)^2 + 3y + y^2 = 7\) | M1 |
| eg \(2x^2 + 7x + 3 (= 0)\) \(2x^2 + 7x = -3\) or eg \(2y^2 - y - 3 (= 0)\) \(2y^2 - y = 3\) | M1ft |
eg \((2x + 1)(x + 3)(= 0)\) or \(x = \dfrac{-7 \pm \sqrt{(7)^2 - 4 \times 2 \times 3}}{2 \times 2}\) or \(\left(x + \dfrac{7}{4}\right)^2 - \left(\dfrac{7}{4}\right)^2 = -\dfrac{3}{2}\) \(\left(x = -\dfrac{1}{2} \text{ and } x = -3\right)\) or eg \((y + 1)(2y - 3)(= 0)\) or \(y = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 2 \times -3}}{2 \times 2}\) or \(\left(y - \dfrac{1}{4}\right)^2 - \left(\dfrac{1}{4}\right)^2 = \dfrac{3}{2}\) \(\left(y = -1 \text{ and } y = \dfrac{3}{2}\right)\) | M1ft |
\(y = \text{``}{-\dfrac{1}{2}}\text{''} + 2\left(= \dfrac{3}{2}\right)\) and \(y = \text{``}{-3}\text{''} + 2\;(= -1)\) or \(x = \text{``}{-1}\text{''} - 2\;(= -3)\) and \(x = \text{``}{\dfrac{3}{2}}\text{''} - 2\left(= -\dfrac{1}{2}\right)\) | M1ft |
Working required Answer: \(x = -\dfrac{1}{2}\) and \(y = \dfrac{3}{2}\) \(x = -3\) and \(y = -1\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for substitution of \(y = x + 2\) (or \(x = \pm y \pm 2\)) into \(x^2 + 3y + y^2 = 7\) to obtain an equation in \(x\) only (or \(y\) only)
M1ft: dep on previous M1 for multiplying out and collecting terms, forming a three term quadratic in any form of \(ax^2 + bx + c\;(= 0)\) where at least 2 coefficients (\(a\) or \(b\) or \(c\)) are correct
M1ft: dep on previous M1 for substituting their 2 found values of \(x\) or \(y\) into one of the two given equations
or fully correct values for the other variable (correct labels for \(x\) / \(y\))
or for one correct pair of values
A1: oe dep on M2 (allow coordinates)
If they find the values of \(y\) but think they are the values of \(x\) then the maximum mark is 3