4MA0 Higher January 2016 Paper 4R Q24
24 Solve the simultaneous equations
\[\begin{aligned} y &= 3x + 2 \\ x^2 + y^2 &= 20 \end{aligned}\]Show clear algebraic working.
(6)
| Scheme | Marks |
|---|---|
| \(x^2 + (3x + 2)^2 = 20\) | M1 |
| \(x^2 + 9x^2 + 6x + 6x + 4 = 20\) | M1 |
| \(10x^2 + 12x - 16 = 0\) \((5x^2 + 6x - 8 = 0)\) | A1 |
eg. \((10x - 8)(x + 2)\) or \((5x - 4)(2x + 4)\) or \((5x - 4)(x + 2)\) or \(\dfrac{-12 \pm \sqrt{12^2 - 4 \times 10 \times -16}}{2 \times 10}\) oe | M1 |
| A1 | |
\(x = \dfrac{4}{5}\), \(y = \dfrac{22}{5}\) or \(x = -2\), \(y = -4\) | A1 |
| (6) |
Notes
M1: for elimination of one variable
\(\left(\dfrac{y - 2}{3}\right)^2 + y^2 = 20\)
M1: indep for correct expansion
ie. \(9x^2 + 6x + 6x + 4\) or \(\dfrac{y^2 - 2y - 2y + 4}{9}\)
A1: Correct simplified 3 part quadratic equation (may not be equated to zero)
\(10y^2 - 4y - 176 = 0\) or
\(5y^2 - 2y - 88 = 0\)
M1: for correct factorisation or substitution into quadratic formula
ft (dep on at least 1 previous M1 scored) a 3 term quadratic for this mark only
eg. \((5y - 22)(y + 4) = 0\)
A1: for both \(x\) values (or both \(y\) values)
A1: for both solutions with \(x\) and \(y\) values correctly paired