Higher June 2022 Paper 1R Q14
14 Solve the simultaneous equations
\[3x - 5y = 25\] \[4x + 3y = 14\]Show clear algebraic working.
(4)
| Scheme | Marks |
|---|---|
Elimination eg \(9x - 15y = 75\) \(20x + 15y = 70\) + \((29x = 145)\) or \(12x - 20y = 100\) \(12x + 9y = 42\) − \((-29y = 58)\) or Substitution eg \(4\left(\dfrac{25 + 5y}{3}\right) + 3y = 14\) or \(4x + 3\left(\dfrac{25 - 3x}{-5}\right) = 14\) or \(3\left(\dfrac{14 - 3y}{4}\right) - 5y = 25\) or \(3x - 5\left(\dfrac{14 - 4x}{3}\right) = 25\) | M1 |
| A1 | |
| eg \(3x - 5 \times \text{``}{-2}\text{''} = 25\) or \(4x + 3 \times \text{``}{-2}\text{''} = 14\) or \(3 \times \text{``}{5}\text{''} - 5y = 25\) or \(4 \times \text{``}{5}\text{''} + 3y = 14\) | M1 |
| Working required Answer: \(x = 5\) \(y = -2\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for a correct method to eliminate \(x\) or \(y\): coefficients of \(x\) or \(y\) the same and correct operation to eliminate selected variable (condone 1 arithmetical error)
or
for correctly writing \(x\) or \(y\) in terms of the other variable and correctly substituting
A1: dep on M1 for \(x = 5\) or \(y = -2\)
M1: dep on M1 for substitution of found variable
or
repeating the steps in first M1 for the second variable
A1: cao, dep on M1
a correct answer without working scores no marks