Moderate -0.5 This is a straightforward simultaneous equations problem requiring substitution of a linear equation into a circle equation, then solving a quadratic. It's slightly easier than average because the algebra is clean (coefficients are small integers) and it's a standard textbook exercise with no conceptual challenges, though it does require multiple steps and careful algebraic manipulation.
Substitute for \(x/y\) or valid attempt to eliminate one variable
\(10x^2 + 24x - 18 = 0\), \(5x^2 + 12x - 9 = 0\)
A1
Correct three term quadratic in solvable form
\((5x-3)(x+3) = 0\)
M1dep*
Attempt to solve resulting three term quadratic
\(x = \frac{3}{5},\ x = -3\)
A1
Correct \(x\) values
\(y = \frac{29}{5},\ y = -5\)
A1
Correct \(y\) values
If \(x\) eliminated: \(10y^2 - 8y + 290 = 0\); \(5y^2 - 4y + 145 = 0\); \((5y-29)(y+5)=0\). Award A1 A0 for one pair correctly found from correct quadratic. Spotted solutions: If M0 DM0: SC B1 \(x=\frac{3}{5}, y=\frac{29}{5}\) www; SC B1 \(x=-3, y=-5\) www. Must show on both line and curve.
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 + (3x+4)^2 = 34$ | M1* | Substitute for $x/y$ or valid attempt to eliminate one variable |
| $10x^2 + 24x - 18 = 0$, $5x^2 + 12x - 9 = 0$ | A1 | Correct three term quadratic in solvable form |
| $(5x-3)(x+3) = 0$ | M1dep* | Attempt to solve resulting three term quadratic |
| $x = \frac{3}{5},\ x = -3$ | A1 | Correct $x$ values |
| $y = \frac{29}{5},\ y = -5$ | A1 | Correct $y$ values |
**If $x$ eliminated:** $10y^2 - 8y + 290 = 0$; $5y^2 - 4y + 145 = 0$; $(5y-29)(y+5)=0$. Award A1 A0 for one pair correctly found from correct quadratic. **Spotted solutions:** If M0 DM0: SC B1 $x=\frac{3}{5}, y=\frac{29}{5}$ www; SC B1 $x=-3, y=-5$ www. Must show on both line and curve.
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