Higher June 2023 Paper 3 Q17
17 Solve \(\quad \dfrac{x + 8}{2} + \dfrac{9 - x}{5} = 4\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – multiplies through by 10 or common denominator of 10 | ||
| \(5(x + 8) + 2(9 - x)\) or \(5x + 40 + 18 - 2x\) | M1 | oe numerator on the left-hand side if written as a fraction allow one error or omission in the expansion if brackets not seen eg \(5x + 18 - 2x\) |
| \(3x + 58\) | A1 | may be implied by eg \(3x + 18 = 0\) or \(3x = -18\) |
| their \((3x + 58) = 4 \times\) (their 10) or their \((3x + 58) = 40\) or \(3x + 18 = 0\) or \(3x = -18\) | M1 | oe allow an unsimplified expression for their \((3x + 58)\) equation may be implied by answer |
| \(-6\) | A1ft | ft M1A0M1 |
| Alternative method 2 – collects terms with fractions | ||
| \(\dfrac{x}{2} + 4 + \dfrac{9}{5} - \dfrac{x}{5}\) | M1 | oe eg \(0.5x + 4 + 1.8 - 0.2x\) allow one error |
| \(\dfrac{3}{10}x + \dfrac{29}{5}\) | A1 | oe eg \(0.3x + 5.8\) |
| \(\dfrac{3}{10}x = \dfrac{20}{5} - \dfrac{29}{5}\) or \(\dfrac{3}{10}x = -\dfrac{9}{5}\) | M1 | oe eg \(0.3x = -1.8\) terms must be collected |
| \(-6\) | A1ft | ft M1A0M1 |
Additional guidance
| Accept decimal answers for follow through correct to 1 dp or better | |
| Apply the principles of alt 1 for any use of other common denominators eg common denominator of 20 (or multiplication through by 20) | |
| \(10(x + 8) + 4(9 - x) = 6x + 116\) | M1A1 |
| \(6x + 116 = 80 \qquad x = -6\) | M1A1 |
| An incorrect simplification of \(5x + 40 + 18 - 2x\) may still gain the third and fourth marks | |
| eg \(5x + 40 + 18 - 2x = 3x + 68\) followed by \(3x + 68 = 40\) and \(x = -\dfrac{28}{3}\) | M1A0M1A1ft |
| eg \(5x + 40 + 18 - 2x = 2x + 68\) followed by \(2x + 68 = 40\) and \(x = -14\) | M1A0M1A1ft |
| An incorrect denominator may still gain the third and fourth marks \(\dfrac{5x + 40 + 18 - 2x}{7}\) followed by \(5x + 40 + 18 - 2x = 28\) and \(x = -10\) | M1A0M1A1ft |
| Denominator not processed \(3x + 58 = 4\) followed by \(3x = -54\) and \(x = -18\) | M1A1M0A0 |
| \((x + 8) + (9 - x) = 40\) | M0A0M1A0 |
| Two errors in the expansion but with brackets seen may go on to get the third and fourth marks \(5(x + 8) + 2(9 - x) = 5x + 8 + 18 - x\) | 1st M1A0 |
| Two errors in the expansion and no brackets seen, no follow through allowed \(5x + 8 + 18 - x\) followed by \(4x + 26 = 40\) and \(x = \dfrac{14}{4}\) | M0A0M1A0 |