Foundation June 2025 Paper 2 Q25
25 Solve the simultaneous equations
\[\begin{aligned} 3x + 5y &= 8 \\ 4x + y &= -3.5 \end{aligned}\]Show clear algebraic working.
(3)
| Scheme | Marks |
|---|---|
eg \(3x + 5y = 8\) \(20x + 5y = -17.5\) Subtracting (\(3x - 20x = 8 - -17.5\) or \(-17x = 25.5\)) or \(3x + 5(-3.5 - 4x) = 8\) or \(4x + \dfrac{8 - 3x}{5} = -3.5\) or eg \(12x + 20y = 32\) \(12x + 3y = -10.5\) Subtracting (\(20y - 3y = 32 - -10.5\) or \(17y = 42.5\)) or \(3\left(\dfrac{-3.5 - y}{4}\right) + 5y = 8\) or \(4\left(\dfrac{8 - 5y}{3}\right) + y = -3.5\) | M1 |
\(3 \times \text{``}{-1.5}\text{''} + 5y = 8\) or \(4 \times \text{``}{-1.5}\text{''} + y = -3.5\) or \(y = -3.5 - 4 \times \text{``}{-1.5}\text{''}\) or \(y = \dfrac{8 - 3 \times \text{``}{-1.5}\text{''}}{5}\) or \(3x + 5 \times \text{``}{2.5}\text{''} = 8\) or \(4x + \text{``}{2.5}\text{''} = -3.5\) or \(x = \dfrac{-3.5 - \text{``}{2.5}\text{''}}{4}\) or \(x = \dfrac{8 - 5 \times \text{``}{2.5}\text{''}}{3}\) | M1 |
| Working required Answer: \(x = -1.5\) \(y = 2.5\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for a correct method to eliminate \(x\) or \(y\): coefficients of \(x\) or \(y\) the same and correct operator to eliminate selected variable (condone any one arithmetic error in multiplication)
or
writing \(x\) or \(y\) in terms of the other variable and correctly substituting (condone missing brackets)
NB The mark is for the method and not for the result of the method. However, if the correct result of the method is seen, the mark can be awarded.
M1: dep on first M1 for a correct method to find other variable by substitution of found variable into one equation
or
for repeating the above method to find the second variable.
A1: oe dep on M1