June 2023 Paper 2 Q19
19 In these simultaneous equations, \(k\) is a positive constant.
\[\begin{aligned} 3x + 4y &= k \\ y &= 2kx \end{aligned}\]Solve the simultaneous equations.
Give the answers in their simplest form in terms of \(k\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Eliminates \(y\) then uses substitution | ||
| Eliminates \(y\) | M1 | eg \(3x + 4(2kx) = k\) or \(3x = k - 8kx\) or \(\dfrac{k - 3x}{4} = 2kx\) or \(x = \dfrac{k}{3 + 8k}\) |
| Substitutes their \(x\) into one of the equations | M1dep | eg \(y = 2k \times \text{their } \dfrac{k}{3 + 8k}\) or \(3 \times \text{their } \dfrac{k}{3 + 8k} + 4y = k\) |
| \(x = \dfrac{k}{3 + 8k} \qquad y = \dfrac{2k^2}{3 + 8k}\) | A1 | |
| Alternative method 2 Eliminates \(x\) then uses substitution | ||
| Eliminates \(x\) | M1 | eg \(y = 2k \times \dfrac{k - 4y}{3}\) or \(y = \dfrac{2k^2 - 8ky}{3}\) or \(3 \times \dfrac{y}{2k} + 4y = k\) or \(y = \dfrac{2k^2}{3 + 8k}\) |
| Substitutes their \(y\) into one of the equations | M1dep | eg their \(\dfrac{2k^2}{3 + 8k} = 2kx\) or \(3x + 4 \times \text{their } \dfrac{2k^2}{3 + 8k} = k\) |
| \(x = \dfrac{k}{3 + 8k} \qquad y = \dfrac{2k^2}{3 + 8k}\) | A1 | |
| Alternative method 3 Eliminates \(y\) and eliminates \(x\) | ||
| Eliminates \(y\) | M1 | eg \(3x + 4(2kx) = k\) or \(3x = k - 8kx\) or \(\dfrac{k - 3x}{4} = 2kx\) or \(x = \dfrac{k}{3 + 8k}\) |
| Eliminates \(x\) | M1 | eg \(y = 2k \times \dfrac{k - 4y}{3}\) or \(y = \dfrac{2k^2 - 8ky}{3}\) or \(3 \times \dfrac{y}{2k} + 4y = k\) or \(y = \dfrac{2k^2}{3 + 8k}\) |
| \(x = \dfrac{k}{3 + 8k} \qquad y = \dfrac{2k^2}{3 + 8k}\) | A1 | |
Additional guidance
| Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Correct answers followed by further incorrect work | M2A0 |
| Missing brackets must be recovered | |
| Alt 1 M1 scored then rearrangement errors before substituting can score the second M mark if their \(x\) is in terms of \(k\) (their expression for \(x\) must be seen) eg \(3x + 8kx = k \qquad x = \dfrac{k}{3 - 8k}\) \(y = \dfrac{2k^2}{3 - 8k}\) | M1 M1depA0 |
| Alt 2 M1 scored then rearrangement errors before substituting can score the second M mark if their \(y\) is in terms of \(k\) (their expression for \(y\) must be seen) eg \(y = 2k \times \dfrac{k - 4y}{3} \qquad y = \dfrac{2k}{3 + 8k}\) \(x = \dfrac{2k}{3 + 8k} \div 2k\) | M1 M1depA0 |
| A correct equation is needed for elimination marks to be awarded eg Alt 1 \(\begin{array}{r} 3x + 4y = k \\ \underline{4y = 8kx} \end{array}\; - \qquad\) is M0 unless \(3x = k - 8kx\) also seen |