Moderate -0.5 This is a standard simultaneous equations question with one linear and one quadratic equation. The method is routine (substitute linear into quadratic, solve resulting quadratic, back-substitute), requiring only algebraic manipulation without novel insight. The 7 marks reflect working steps rather than conceptual difficulty, making it slightly easier than average.
Rearrange \(x - 2y - 1 = 0\) into \(x = ...\), or \(y = ...\), or \(2y = ..\) and attempt to fully substitute into 2nd equation.
M1
It does not need to be correct but a clear attempt must be made. Condone missing brackets \((2y+1)^2 + 4y^2 - 10(2y+1) + 9 = 0\)
\((2y+1)^2 + 4y^2 - 10(2y+1) + 9 = 0\)
\(x^2 + (x-1)^2 - 10x + 9 = 0\)
M1, A1
\(8y^2 - 16y = 0\)
\(2x^2 - 12x + 10 = 0\)
M1, A1
\(8y(y-2) = 0\) Alt \(y(8y-16) = 0\)
\(2(x-1)(x-5) = 0\) Alt \((2x-2)(x-5) = 0\)
M1
\(y = 0, y = 2\)
\(x = 1, x = 5\)
\(y = 0\) in \(x = 2y + 1 \Rightarrow x = 1\)
\(x = 1\) in \(y = \frac{x-1}{2} = 0\)
M1
\(y = 2\) in \(x = 2y + 1 \Rightarrow x = 5\)
\(x = 5\) in \(y = \frac{x-1}{2} = 2\)
\(x=1, y=0\) and \(x=5, y=2\)
\(x=1, y=0\) and \(x=5, y=2\)
A1, A1
(7 marks)
Special Cases where candidates write down answers with little or no working as can be awarded above.
- One correct solution – B2.
- Two correct solutions – B2, B2
To score all 7 marks candidates must prove that there are only two solutions. This could be justified by a sketch.
| $x = 2y + 1$ | $2y = x - 1$ |
Rearrange $x - 2y - 1 = 0$ into $x = ...$, or $y = ...$, or $2y = ..$ and attempt to fully substitute into 2nd equation. | M1 | It does not need to be correct but a clear attempt must be made. Condone missing brackets $(2y+1)^2 + 4y^2 - 10(2y+1) + 9 = 0$
| $(2y+1)^2 + 4y^2 - 10(2y+1) + 9 = 0$ | $x^2 + (x-1)^2 - 10x + 9 = 0$ | M1, A1 | Collect like terms to produce a quadratic equation in $x$ (or $y$) = 0. Either $A(y^2 - 2y) = 0$ or $B(x^2 - 6x + 5) = 0$
| $8y^2 - 16y = 0$ | $2x^2 - 12x + 10 = 0$ | M1, A1 |
| $8y(y-2) = 0$ Alt $y(8y-16) = 0$ | $2(x-1)(x-5) = 0$ Alt $(2x-2)(x-5) = 0$ | M1 |
| $y = 0, y = 2$ | $x = 1, x = 5$ | |
| $y = 0$ in $x = 2y + 1 \Rightarrow x = 1$ | $x = 1$ in $y = \frac{x-1}{2} = 0$ | M1 |
| $y = 2$ in $x = 2y + 1 \Rightarrow x = 5$ | $x = 5$ in $y = \frac{x-1}{2} = 2$ | |
| $x=1, y=0$ and $x=5, y=2$ | $x=1, y=0$ and $x=5, y=2$ | A1, A1 | Both $x$'s correct or both $y$'s correct or a correct matching pair. Accept as a coordinate. Do not accept correct answers that are obtained from incorrect equations.
| | **(7 marks)**
Special Cases where candidates write down answers with little or no working as can be awarded above.
- One correct solution – B2.
- Two correct solutions – B2, B2
To score all 7 marks candidates must prove that there are only two solutions. This could be justified by a sketch.
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