Higher June 2021 Paper 2 Q19
19 Solve the simultaneous equations
\(y = 3 - 2x\)
\(x^2 + y^2 = 18\)
Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
\(x^2 + (3 - 2x)^2 = 18\) or \(\left(\dfrac{3 - y}{2}\right)^2 + y^2 = 18\) | M1 |
| \(5x^2 - 12x - 9\ [= 0]\) oe or \(5y^2 - 6y - 63\ [= 0]\) oe | M1 |
\((5x + 3)(x - 3)\ [= 0]\) \(\dfrac{-(-12) \pm \sqrt{(-12)^2 - 4 \times 5 \times (-9)}}{2 \times 5}\) \(5\left[\left(x - \dfrac{12}{10}\right)^2 - \dfrac{144}{100}\right] - 9 = 0\) oe or \((5y - 21)(y + 3)\ [= 0]\) \(\dfrac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 5 \times (-63)}}{2 \times 5}\) \(5\left[\left(y - \dfrac{6}{10}\right)^2 - \dfrac{36}{100}\right] - 63 = 0\) oe | M1ft |
| \(x = -0.6\) and \(x = 3\) OR \(y = 4.2\) and \(y = -3\) | A1 |
| Working must be shown Answer: \(x = -0.6,\ y = 4.2\) \(x = 3,\ y = -3\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: substitution of linear equation into quadratic
M1: simplified to a correct 3 term quadratic
M1ft: dep on M1 for solving their 3 term quadratic equation using any correct method (if factorising, allow brackets which expanded give 2 out of 3 terms correct) (if using formula allow one sign error and some simplification – allow as far as \(\dfrac{12 \pm \sqrt{144 + 180}}{10}\) or \(\dfrac{6 \pm \sqrt{36 + 1260}}{10}\)) (if completing the square allow as far as shown)
A1: oe dep on M2 for both \(x\)-values OR both \(y\)-values
A1: oe dep on M2 (must be clearly shown as correct pairs), accept answers given as coordinates