June 2022 Paper 1 Q16
16 Solve the simultaneous equations
\[\begin{aligned} x - y &= \frac{19}{4} \\ xy &= -3 \end{aligned}\]Do not use trial and improvement.
You must show your working. [6 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| Correct substitution \(x - \dfrac{-3}{x} = \dfrac{19}{4}\) or \(x\left(x - \dfrac{19}{4}\right) = -3\) | M1 | penalise no brackets unless recovered |
| \(4x^2 - 19x + 12 \;(= 0)\) | M1dep | oe eg \(4x^2 + 12 = 19x\) must be integer values unless going on to complete the square |
| \((4x + a)(x + b)\) or \((4x - 3)(4x - 16)\) | M1dep | where \(ab = 12\) or \(a + 4b = -19\) |
| \((4x - 3)(x - 4)\) | A1 | |
| \(x = \dfrac{3}{4}\) and 4 or \(x = \dfrac{3}{4}\) and \(y = -4\) or \(x = 4\) and \(y = -\dfrac{3}{4}\) | A1 | |
| \(y = -4\) and \(-\dfrac{3}{4}\) or \(x = 4\) and \(y = -\dfrac{3}{4}\) or \(x = \dfrac{3}{4}\) and \(y = -4\) | A1 | all 4 values must be correct to gain this mark |
| Alternative method 2 | ||
| Correct substitution \(x - \dfrac{-3}{x} = \dfrac{19}{4}\) or \(x\left(x - \dfrac{19}{4}\right) = -3\) | M1 | penalise no brackets unless recovered |
| \(4x^2 - 19x + 12 \;(= 0)\) | M1dep | oe eg \(4x^2 + 12 = 19x\) must be integer values unless going on to complete the square |
| \(\dfrac{19 \pm \sqrt{19^2 - 4 \times 4 \times 12}}{2 \times 4}\) | M1dep | |
| \(\dfrac{19 \pm \sqrt{169}}{8}\) | A1 | |
| \(x = \dfrac{3}{4}\) and 4 or \(x = \dfrac{3}{4}\) and \(y = -4\) or \(x = 4\) and \(y = -\dfrac{3}{4}\) | A1 | |
| \(y = -4\) and \(-\dfrac{3}{4}\) or \(x = 4\) and \(y = -\dfrac{3}{4}\) or \(x = \dfrac{3}{4}\) and \(y = -4\) | A1 | all 4 values must be correct to gain this mark |
| Alternative method 3 | ||
| Correct substitution \(x - \dfrac{-3}{x} = \dfrac{19}{4}\) or \(x\left(x - \dfrac{19}{4}\right) = -3\) | M1 | penalise no brackets unless recovered |
| \(4x^2 - 19x + 12 \;(= 0)\) or \(x^2 - \dfrac{19}{4}x + 3 \;(= 0)\) | M1dep | oe eg \(4x^2 + 12 = 19x\) must be integer values unless going on to complete the square |
| \(4\left[\left(x - \dfrac{19}{8}\right)^2 \ldots\ldots\right] \ldots\ldots\) or \(\left[\left(x - \dfrac{19}{8}\right)^2 \ldots\ldots\right] \ldots\ldots\) | M1 | oe |
| \(4\left(x - \dfrac{19}{8}\right)^2 - \dfrac{169}{16} = 0\) or \(\left[\left(x - \dfrac{19}{8}\right)^2\right] - \dfrac{169}{64} = 0\) | M1dep | |
| \(x = \dfrac{3}{4}\) and 4 or \(x = \dfrac{3}{4}\) and \(y = -4\) or \(x = 4\) and \(y = -\dfrac{3}{4}\) | A1 | |
| \(y = -4\) and \(-\dfrac{3}{4}\) or \(x = 4\) and \(y = -\dfrac{3}{4}\) or \(x = \dfrac{3}{4}\) and \(y = -4\) | A1 | all 4 values must be correct to gain this mark |
| Alternative method 4 | ||
| Correct substitution \(\dfrac{-3}{y} - y = \dfrac{19}{4}\) or \(y\left(y + \dfrac{19}{4}\right) = -3\) | M1 | penalise no brackets unless recovered |
| \(4y^2 + 19y + 12 \;(= 0)\) | M1dep | oe eg \(4y^2 + 12 = -19y\) must be integer values unless going on to complete the square |
| \((4y + a)(y + b)\) or \((4y + 3)(4y + 16)\) | M1dep | where \(ab = 12\) or \(a + 4b = 19\) |
| \((4y + 3)(y + 4)\) | A1 | |
| \(y = -\dfrac{3}{4}\) and \(-4\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\) | A1 | |
| \(x = 4\) and \(\dfrac{3}{4}\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\) | A1 | all 4 values must be correct to gain this mark |
| Alternative method 5 | ||
| Correct substitution \(\dfrac{-3}{y} - y = \dfrac{19}{4}\) or \(y\left(y + \dfrac{19}{4}\right) = -3\) | M1 | penalise no brackets unless recovered |
| \(4y^2 + 19y + 12 \;(= 0)\) | M1dep | oe eg \(4y^2 + 12 = -19y\) must be integer values unless going on to complete the square |
| \(\dfrac{-19 \pm \sqrt{19^2 - 4 \times 4 \times 12}}{2 \times 4}\) | M1dep | |
| \(-\dfrac{19 \pm \sqrt{169}}{8}\) | A1 | |
| \(y = -\dfrac{3}{4}\) and \(-4\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\) | A1 | |
| \(x = 4\) and \(\dfrac{3}{4}\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\) | A1 | all 4 values must be correct to gain this mark |
| Alternative method 6 | ||
| Correct substitution \(\dfrac{-3}{y} - y = \dfrac{19}{4}\) or \(y\left(y + \dfrac{19}{4}\right) = -3\) | M1 | penalise no brackets unless recovered |
| \(4y^2 + 19y + 12 \;(= 0)\) or \(y^2 + \dfrac{19}{4}y + 3 \;(= 0)\) | M1dep | oe eg \(4y^2 + 12 = -19y\) must be integer values unless going on to complete the square |
| \(4\left[\left(y + \dfrac{19}{8}\right)^2 \ldots\ldots\right] \ldots\ldots\) or \(\left[\left(y + \dfrac{19}{8}\right)^2 \ldots\ldots\right] \ldots\ldots\) | M1 | oe |
| \(4\left(y + \dfrac{19}{8}\right)^2 - \dfrac{169}{16} = 0\) or \(\left[\left(y + \dfrac{19}{8}\right)^2\right] - \dfrac{169}{64} = 0\) | M1dep | |
| \(y = -\dfrac{3}{4}\) and \(-4\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\) | A1 | |
| \(x = 4\) and \(\dfrac{3}{4}\) or \(y = -\dfrac{3}{4}\) and \(x = 4\) or \(y = -4\) and \(x = \dfrac{3}{4}\) | A1 | all 4 values must be correct to gain this mark |
Additional guidance
| Correct A marks must come from correct algebra in M marks |