Moderate -0.3 This is a straightforward simultaneous equations problem requiring substitution of a linear equation into a quadratic, then solving the resulting quadratic equation. While it involves a difference of squares rather than the more common circle/parabola, the method is standard C1 fare with no conceptual challenges—slightly easier than average due to the clean algebraic manipulation required.
Substitute for \(x/y\) or valid attempt to eliminate one of the variables
\(3x^2-20x+28=0\)
A1
Three term quadratic in solvable form
\((3x-14)(x-2)=0\)
M1dep
Correct method to solve three term quadratic – see appendix 1
\(x=\frac{14}{3}, x=2\)
A1
Both \(x\) values correct. Spotted solutions: If M*0: SC B1 \(x=2, y=1\) www; SC B1 \(x=\frac{14}{3}, y=-\frac{13}{3}\) www. Must show on both line and curve
\(y=-\frac{13}{3}, y=1\)
A1
Both \(y\) values correct. Allow 1 A mark for one correct pair of \(x\) and \(y\) from correct factorisation. (Can then get 5/5 if both found www and exactly two solutions justified)
If \(y\) eliminated: \(3y^2+10y-13=0\); \((3y+13)(x-1)=0\)
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2-(5-2x)^2=3$ | M1* | Substitute for $x/y$ or valid attempt to eliminate one of the variables |
| $3x^2-20x+28=0$ | A1 | Three term quadratic in solvable form |
| $(3x-14)(x-2)=0$ | M1dep | Correct method to solve three term quadratic – see appendix 1 |
| $x=\frac{14}{3}, x=2$ | A1 | Both $x$ values correct. **Spotted solutions: If M*0:** SC B1 $x=2, y=1$ www; SC B1 $x=\frac{14}{3}, y=-\frac{13}{3}$ www. Must show on both line and curve |
| $y=-\frac{13}{3}, y=1$ | A1 | Both $y$ values correct. **Allow 1 A mark for one correct pair of $x$ and $y$ from correct factorisation.** (Can then get 5/5 if both found www and exactly two solutions justified) |
**If $y$ eliminated:** $3y^2+10y-13=0$; $(3y+13)(x-1)=0$
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