Higher June 2025 Paper 1R Q22
22 Solve the simultaneous equations
\[\begin{aligned} x^2 + y^2 &= 41 \\ 2x + y &= 3 \end{aligned}\]Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
\(x^2 + (-2x + 3)^2 = 41\) or \(\left(\dfrac{-y + 3}{2}\right)^2 + y^2 = 41\) | M1 |
| eg \(5x^2 - 12x - 32 (= 0)\) oe or \(5x^2 - 12x = 32\) or eg \(5y^2 - 6y - 155 (= 0)\) or \(5y^2 - 6y = 155\) | M1ft |
\((5x + 8)(x - 4) (= 0)\) or \((x =)\dfrac{12 \pm \sqrt{(-12)^2 - 4 \times 5 \times (-32)}}{2 \times 5}\) or \(5\left[\left(x - \dfrac{6}{5}\right)^2 - \left(\dfrac{6}{5}\right)^2\right] - 32 (= 0)\) (should give \((x =) -\dfrac{8}{5}, 4\)) or eg \((5y - 31)(y + 5) (= 0)\) or \(\dfrac{6 \pm \sqrt{(-6)^2 - 4 \times 5 \times (-155)}}{2 \times 5}\) or \(5\left[\left(y - \dfrac{3}{5}\right)^2 - \left(\dfrac{3}{5}\right)^2\right] - 155 (= 0)\) (should give \((y =) \dfrac{31}{5}, -5\)) | M1ft |
eg \(2 \times 4 + y = 3\) and \(2 \times -\dfrac{8}{5} + y = 3\) or eg \(2x + \dfrac{31}{5} = 3\) and \(2x - 5 = 3\) | M1ft |
Working required if the correct answers come from incorrectly using \(y = 2x - 3\) oe award M4A0. Answer: \(x = -\dfrac{8}{5}\), \(y = \dfrac{31}{5}\), \(x = 4\), \(y = -5\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: substitution of \(y = \pm 2x \pm 3\) (or \(x = \dfrac{\pm y \pm 3}{2}\)) into \(x^2 + y^2 = 41\) 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 M1 method to solve their 3 term quadratic using any correct method (allow one sign error and some simplification – allow as far as eg \(\dfrac{12 \pm \sqrt{144 + 640}}{10}\) or \(\dfrac{6 \pm \sqrt{36 + 3100}}{10}\) or if factorising allow brackets which expanded give 2 out of 3 terms correct) or correct values for \(x\) or correct values for \(y\)
M1ft: dep on previous M1 for substituting their 2 found values of \(x\) or \(y\) into one of the two given equations or their rearranged equation used in the substitution
or for one correct pair of values
A1: oe dep on M2 for all 4 values (allow coordinates)
If they find the values of \(y\) but think they are the values of \(x\) then the maximum mark is 3