Foundation June 2025 Paper 2R Q19
19 Solve the simultaneous equations
\[\begin{aligned} 3x + 2y &= 10 \\ 3x - 4y &= 16 \end{aligned}\]Show clear algebraic working.
(3)
| Scheme | Marks |
|---|---|
eg \(\begin{aligned} &3x + 2y = 10 \\ -\;&3x - 4y = 16 \end{aligned}\) \((6y = -6)\) or eg \(3\left(\dfrac{10 - 2y}{3}\right) - 4y = 16\) or \(3\left(\dfrac{16 + 4y}{3}\right) + 2y = 10\) or \(6y = -6\) oe or Eg \(\begin{aligned} &6x + 4y = 20 \\ +\;&3x - 4y = 16 \end{aligned}\) \((9x = 36)\) or eg \(3x - 4\left(\dfrac{10 - 3x}{2}\right) = 16\) or \(3x + 2\left(\dfrac{3x - 16}{4}\right) = 10\) or \(9x = 36\) oe | M1 |
\(3x + 2 \times \text{``}{-1}\text{''} = 10\) or \(3x - 4 \times \text{``}{-1}\text{''} = 16\) or \(x = \dfrac{10 - 2 \times \text{``}{-1}\text{''}}{3}\) or \(x = \dfrac{16 + 4 \times \text{``}{-1}\text{''}}{3}\) or \(3 \times \text{``}{4}\text{''} + 2y = 10\) or \(3 \times \text{``}{4}\text{''} - 4y = 16\) or \(y = \dfrac{10 - 3 \times \text{``}{4}\text{''}}{2}\) or \(y = \dfrac{3 \times \text{``}{4}\text{''} - 16}{4}\) | M1 |
| Working required Answer: \(x = 4\) \(y = -1\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: a correct method to eliminate \(x\) or \(y\): coefficients of \(x\) or \(y\) are the same and the correct operation to eliminate is selected; if operator not written, the correct operation can be implied by 2 out of 3 terms correct
Allow one arithmetic error if multiplying to equate coefficients
or
for a correct substitution of one variable into the other equation
NB: the mark is for the method and not for the result of the method. However, if the correct result of this method is seen, the mark can be awarded
M1: dep on M1
a correct substitution to find the value of the second variable using their value or for starting again with elimination or substitution (as above)
A1: dep on M1