June 2025 Paper 2 Q17
17 Solve
\[\begin{aligned} 2a + 4b + c &= 12 \\ 3a - b + 2c &= -2 \\ a - 5b - 3c &= 6 \end{aligned}\]Do not use trial and improvement.
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| One correct equation in two variables with terms collected and simplified | M1 | eg \(14b + 7c = 0\) or \(14a + 13c = -16\) or \(a + 9b = 26\) |
| Two correct equations each in two variables with terms collected and simplified | M1dep | eg \(14b + 7c = 0\) and \(14b + 11c = -20\) or \(14a + 13c = -16\) and \(a + 9b = 26\) |
| Correct equation in one variable with terms collected and simplified | M1dep | eg \(4c = -20\) |
| Correct equation in one variable with terms collected and simplified and any two of \(a = 3.5 \quad b = 2.5 \quad c = -5\) | A1 | implied by final A1 oe values |
| Correct equation in one variable with terms collected and simplified and \(a = 3.5 \quad b = 2.5 \quad c = -5\) | A1 | oe values |
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 | |
| 3rd M1 Do not award if the only correct equation is one of the values eg only correct equation in one variable is \(c = -5\) | 3rd M0 |
| Some correct equations in two variables are \(14b - c = 40 \qquad 14a + 9c = 4 \qquad 6a - 2b = 16\) \(14b + 3c = 20 \qquad 14a - 7c = 84 \qquad 7a + 7b = 42\) \(14b + 7c = 0 \qquad 14a + 13c = -16 \qquad a + 9b = 26\) \(14b + 11c = -20 \qquad 14a + 5c = 24 \qquad 11a - 13b = 6\) All equations have equivalents | |
| M3 can be implied if two variables are eliminated in one step eg \(4c = -20\) (from equation 2 – equation 1 – equation 3) | M3 |
| Correct values only | Zero |
| Escalate any responses using a matrix method |