June 2022 Paper 2 Q17
17 Solve the simultaneous equations
\[\begin{aligned} a + 3b - 2c &= 4 \\ 4a - 3b + 5c &= -5 \\ 2a + b + 3c &= 9 \end{aligned}\]Do not use trial and improvement.
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| Elimination of one variable making an equation with at least two terms correct | M1 | eg1 (elimination of \(b\) by adding 1st and 2nd equations) \(5a + 3c = -1\) with at least two terms correct eg2 (elimination of \(a\) by doubling 1st equation and subtracting 3rd equation) \(5b - 7c = -1\) with at least two terms correct |
| Elimination of one variable making an equation with at least two terms correct and elimination of the same variable making a different equation with at least two terms correct | M1dep | eg1 (elimination of \(b\) by adding 1st and 2nd equations and elimination of \(b\) by trebling 3rd equation and subtracting 1st equation) \(5a + 3c = -1\) with at least two terms correct and \(5a + 11c = 23\) with at least two terms correct eg2 (elimination of \(a\) by doubling 1st equation and subtracting 3rd equation and elimination of \(a\) by doubling 3rd equation and subtracting 2nd equation) \(5b - 7c = -1\) with at least two terms correct and \(5b + c = 23\) with at least two terms correct |
| Correct equation in one variable with two correct equations in the same two variables | M1dep | eg \(3c - 11c = -1 - 23\) or \(-8c = -24\) or \(c = 3\) with \(5a + 3c = -1\) and \(5a + 11c = 23\) |
| Two correct values with two correct equations in the same two variables | A1 | eg \(c = 3\) and \(a = -2\) with \(5a + 3c = -1\) and \(5a + 11c = 23\) |
| \(a = -2 \quad b = 4 \quad c = 3\) with two correct equations in the same two variables | A1 | eg \(a = -2 \quad b = 4 \quad c = 3\) with \(5a + 3c = -1\) and \(5a + 11c = 23\) |
Additional guidance
The two correct equations in the same two variables referred to in the scheme are a pair from one of these columns
eg equivalents for \(5a + 3c = -1\) include \(-5a - 3c = 1\) and \(5a = -1 - 3c\) | ||||||||||
| All equations in two variables must have terms collected eg \(a + 4a - 2c + 5c = 4 - 5\) requires simplification to \(5a + 3c = -1\) | ||||||||||
| \(0a + 15b - 13c = 21\) is equivalent to \(15b - 13c = 21\) etc | ||||||||||
| Equations with two terms correct include eg1 (For \(5b + c = 23\)) \(\quad 5b + c = 10\) and \(-5b - c = 2\) and \(5b - 3c = 23\) eg2 (For \(5a + 3c = -1\)) \(\quad 5a + 6c = -1\) and \(-5a - 3c = 4\) and \(5a = 2 - 3c\) | ||||||||||
| For equations with two terms correct the signs can be ignored if the modulus of the numbers in the correct equation are unchanged eg For the correct equation \(5b - 7c = -1\) (so modulus 5, 7 and 1) equations with two terms correct include \(5b + 7c = 1\) and \(5b - 7c = 1\) and \(-5b - 7c = 1\) and \(-5b - 7c + 1 = 0\) | ||||||||||
| Up to M3 may be awarded for correct work with no, or incorrect answer, even if this is seen amongst multiple attempts | ||||||||||
| Elimination of variables may be seen from other approaches eg rearranges 1st equation to \(a = 4 - 3b + 2c\) and substitutes into the 2nd and 3rd equations | ||||||||||
| Correct values with no working | Zero | |||||||||
| Matrix method involving row reduction is equivalent to the methods in the mark scheme | ||||||||||
| Correct inverse matrix seen with three correct solutions | M3A2 |