Foundation June 2021 Paper 2 Q22
22 Solve the simultaneous equations
\(5a + 2c = 10\)
\(2a - 4c = 7\)
Show clear algebraic working.
(3)
| Scheme | Marks |
|---|---|
eg \(10a + 4c = 20\) \(+\ 2a - 4c = 7\) eg [\(c = \dfrac{10 - 5a}{2}\)] oe \(2a - 4\left(\dfrac{10 - 5a}{2}\right) = 7\) oe or eg \(10a + 4c = 20\) \(-\ 10a - 20c = 35\) eg [\(a = \dfrac{7 + 4c}{2}\)] oe \(5\left(\dfrac{7 + 4c}{2}\right) + 2c = 10\) oe | M1 |
| eg \(5 \times \text{“}2.25\text{”} + 2c = 10\) or \(2 \times \text{“}2.25\text{”} - 4c = 7\) or eg \(5a + 2 \times \text{“}{-0.625}\text{”} = 10\) or \(2a - 4 \times \text{“}{-0.625}\text{”} = 7\) | M1 |
| Working required Answer: \(a = 2.25\) \(c = -0.625\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: multiplication of one or both equation(s) with correct operation selected (allow one arithmetic error) (if + or – is not shown then assume it is the operation that at least 2 of the 3 terms have been calculated for)
or
correct rearrangement of one equation with substitution into second
M1: (dep on previous M1 but not on a correct first value) correct method to find second unknown – this could be a correct substitution into one of the equations given or calculated or starting again with the same style of working as for the first method mark
A1: oe eg \(a = \dfrac{9}{4}\), \(c = -\dfrac{5}{8}\) for both solutions
dependent on first M1