Foundation June 2019 Paper 2 Q20
20 Solve the simultaneous equations
\(\begin{aligned} x + 2y &= -0.5 \\ 3x - y &= 16 \end{aligned}\)
Show clear algebraic working.
(3)
Using elimination then substitution
| Scheme | Marks |
|---|---|
| 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. \(\text{``}{4.5}\text{''} + 2y = -0.5\) or \(3 \times \text{``}{4.5}\text{''} - y = 16\) or e.g. \(x + 2 \times \text{``}{-2.5}\text{''} = -0.5\) or \(3x - \text{``}{-2.5}\text{''} = 16\) | M1 |
| \(x = 4.5\), \(y = -2.5\) | 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
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
Alternative scheme: using substitution
| Scheme | Marks |
|---|---|
\(3(-0.5 - 2y) - y = 16\) (\(7y = -17.5\)) 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 |
| \(x = 4.5\), \(y = -2.5\) | A1 |
Notes
M1: for correctly writing \(x\) or \(y\) in terms of the other variable and correctly substituting
(condone any one arithmetic error)
M1: (dep) for substituting their value found of one variable into one of the equations
A1: (dep on first M1) for both solutions