November 2021 Paper 1 Q14
14 Solve the simultaneous equations.
\[\begin{aligned} 4a - b + 3c &= 27 \\ 3a + 2b - c &= 5 \\ 2a - 5c &= -7 \end{aligned}\]Do not use trial and improvement.
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Eliminates \(b\) from first two equations before eliminating a second variable | ||
| Correct attempt to eliminate \(b\) from LHS of first two equations | M1 | eg \(2(4a - b + 3c) + 3a + 2b - c\) or \(11a + 5c\) adding or subtracting the two equations can be implied from two terms correct |
| Correct attempt to eliminate \(a\) or \(c\) from LHS of third equation and their equation in \(a\) and \(c\) | M1dep | eg \(11a + 5c + 2a - 5c\) or \(2(11a + 5c) - 11(2a - 5c)\) |
| Correct equation in \(a\) or \(c\) | M1dep | eg \(13a = 52\) or \(65c = 195\) implied by \(a = 4\) or \(c = 3\) with M2 |
| Two correct values with M3 | A1 | eg \(a = 4\) and \(c = 3\) with M3 |
| \(a = 4\) and \(b = -2\) and \(c = 3\) with M3 | A1 | |
| Alternative method 2 Eliminates \(a\) or \(c\) before eliminating a second variable | ||
| Two correct attempts to eliminate the same variable (\(a\) or \(c\)) from LHS | M1 | eg (eliminating \(a\)) \(4a - b + 3c - 2(2a - 5c)\) and \(2(3a + 2b - c) - 3(2a - 5c)\) or \(-b + 13c\) and \(4b + 13c\) |
| Correct attempt to eliminate a second variable from LHS of their two equations | M1dep | eg \(-b + 13c - (4b + 13c)\) |
| Correct equation in one variable | M1dep | eg \(-5b = 10\) implied by \(b = -2\) with M2 |
| Two correct values with M3 | A1 | eg \(b = -2\) and \(a = 4\) with M3 or \(b = -2\) and \(c = 3\) with M3 |
| \(a = 4\) and \(b = -2\) and \(c = 3\) with M3 | A1 | |
Additional guidance
| For the first two marks ignore the RHS of the equations | |
| First two method marks may be seen in one attempt eg Alt1 \(2(4a - b + 3c) + 3a + 2b - c + 2a - 5c\) | M1M1 |
| Elimination may be seen from other approaches eg1 Alt 1 (equates expressions for \(2b\) from first two equations) \(2(4a + 3c - 27) = 5 - 3a + c\) eg2 Alt 2 (rearranges third equation to \(a = 2.5c - 3.5\) and substitutes into first two equations) \(4(2.5c - 3.5) - b + 3c\) and \(3(2.5c - 3.5) + 2b - c\) | M1 M1 |
| Correct values with no working | M0A0 |