Higher November 2019 Paper 3 Q5
5 Solve the simultaneous equations [3 marks]
\(7x + 2y = 36\)
\(3x + 2y = 16\)
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(7x - 3x = 36 - 16\) | M1 | oe elimination of one variable implied by \(4x = n\), where \(n < 36\) and \(n \ne 16\) |
| \(4x = 20\) or \(x = 5\) | A1 | oe |
| \(y = 0.5\) | A1 | oe |
| Alternative method 2 | ||
| \(7 \times 2y - 3 \times 2y = 7 \times 16 - 3 \times 36\) or \(14y - 6y = 112 - 108\) | M1 | oe elimination of one variable implied by \(21x + 14y = 112\) and \(21x + 6y = 108\) followed by \(8y = n\), where \(n < 112\) and \(n \ne 36\), 16 or 20 |
| \(8y = 4\) or \(y = 0.5\) | A1 | oe |
| \(x = 5\) | A1 | |
| Alternative method 3 | ||
| \(36 - 7x = 16 - 3x\) or \(\dfrac{36 - 2y}{7} = \dfrac{16 - 2y}{3}\) | M1 | oe elimination of one variable |
| \(4x = 20\) or \(x = 5\) or \(8y = 4\) or \(y = 0.5\) | A1 | oe collects terms oe |
| \(x = 5\) and \(y = 0.5\) | A1 | oe |
Additional guidance
| \(x = 5\) and \(y = 0.5\) | M1A1A1 |
| One correct value with one incorrect value (or no second value) and no working eg \(x = 5\) and \(y = 2\) or eg \(x = 5\) | M1A1A0 |
| Embedded, correct values in both equations eg \(7 \times 5 + 2 \times 0.5 = 36\) and \(3 \times 5 + 2 \times 0.5 = 16\) | M1A1A0 |
| Embedded, correct values in one equation only eg \(7 \times 5 + 2 \times 0.5 = 36\) | M1A0A0 |