Higher June 2019 Paper 2 Q9
9 Solve the simultaneous equations
\[\begin{aligned} x + 2y &= -0.5 \\ 3x - y &= 16 \end{aligned}\]Show clear algebraic working.
(3)
| Scheme | Marks |
|---|---|
| Using elimination then substitution e.g. \(\begin{aligned} &x + 2y = -0.5 \\ +\; &6x - 2y = 32 \end{aligned}\) (\(7x = 31.5\)) or e.g. \(\begin{aligned} &3x + 6y = -1.5 \\ -\; &3x - y = 16 \end{aligned}\) (\(7y = -17.5\)) | M1 |
| e.g. ‘4.5’ + \(2y = -0.5\) or 3 × ‘4.5’ – \(y\) = 16 or e.g. \(x\) + 2 × ‘−2.5’ = −0.5 or \(3x\) – ‘−2.5’ = 16 | M1 |
| Working required Answer: \(x = 4.5\) \(y = -2.5\) | A1 |
| (3) | |
| (3 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 any one arithmetic error)
M1: (dep) for substituting their value found of one variable into one of the equations or
for repeating above method to find second variable
A1: (dep on first M1) for both solutions
| Scheme | Marks |
|---|---|
Using substitution \(3(-0.5 - 2y) - y = 16\) (\(7y = -17.5\)) or \(\dfrac{16 + y}{3} + 2y = -0.5\) (\(7y = -17.5\)) or \(3x - \left(\dfrac{-0.5 - x}{2}\right) = 16\) (\(7x = 31.5\)) or \(x + 2(3x - 16) = -0.5\) (\(7x = 31.5\)) | M1 |
e.g. \(x = -0.5 - 2\text{``}{-2.5}\text{''}\) or \(x = \dfrac{16 + \text{``}{-2.5}\text{''}}{3}\) or e.g. \(y = \dfrac{-0.5 - \text{``}{4.5}\text{''}}{2}\) or \(y = 3\text{``}{4.5}\text{''} - 16\) | M1 |
| Working required Answer: \(x = 4.5\) \(y = -2.5\) | A1 |
Notes
M1: for correctly writing \(x\) or \(y\) in terms of the other variable and correctly substituting
M1: (dep) for substituting their value found of one variable into one of the equations
A1: (dep on first M1) for both solutions