Moderate -0.3 This is a standard C1 simultaneous equations question requiring substitution of a linear equation into a quadratic, then solving the resulting quadratic equation. While it involves multiple steps (substitution, expansion, factorization/formula, back-substitution), these are routine techniques with no conceptual challenges, making it slightly easier than the average A-level question.
Attempts to make \(y\) the subject of the linear equation and substitutes into the other equation. Allow slips e.g. substituting \(y = -4x + 1\)
\(21x^2 + 10x + 1 = 0\)
A1
Correct 3 term quadratic (terms do not need to be all on the same side). The "\(= 0\)" may be implied by subsequent work
\((7x+1)(3x+1) = 0 \Rightarrow x = -\frac{1}{7}, -\frac{1}{3}\)
dM1 A1
dM1: Solves a 3 term quadratic by usual rules to give at least one value for \(x\). Dependent on first method mark. A1: Two separate correct exact answers. Allow exact equivalents e.g. \(x = -\frac{6}{42}, -\frac{14}{42}\)
\(y = -\frac{3}{7}, \frac{1}{3}\)
M1 A1
M1: Substitutes to find at least one \(y\) value. A1: \(y = -\frac{3}{7}, \frac{1}{3}\) (two correct exact answers). Allow exact equivalents e.g. \(y = -\frac{18}{42}, \frac{14}{42}\)
Coordinates do not need to be paired
Note: if the linear equation is explicitly rearranged to \(y = 4x + 1\), this gives correct answers for \(x\) and possibly for \(y\). If not already lost, deduct the final A1.
Correct 3 term quadratic. "\(= 0\)" may be implied by subsequent work
\((7y+3)(3y-1) = 0 \Rightarrow y = -\frac{3}{7}, \frac{1}{3}\)
dM1 A1
dM1: Solves a 3 term quadratic to give at least one value for \(y\). Dependent on first method mark. A1: Two separate correct exact answers. Allow e.g. \(y = -\frac{18}{42}, \frac{14}{42}\)
\(x = -\frac{1}{7}, -\frac{1}{3}\)
M1 A1
M1: Substitutes to find at least one \(x\) value. A1: \(x = -\frac{1}{7}, -\frac{1}{3}\) (two correct exact answers). Allow e.g. \(x = -\frac{6}{42}, -\frac{14}{42}\)
Coordinates do not need to be paired
Note: if the linear equation is explicitly rearranged to \(x = (y+1)/4\), this gives correct answers for \(y\) and possibly for \(x\). If not already lost, deduct the final A1.
[6 marks total]
# Question 5:
## Way 1
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = -4x - 1 \Rightarrow (-4x-1)^2 + 5x^2 + 2x = 0$ | M1 | Attempts to make $y$ the subject of the linear equation and substitutes into the other equation. Allow slips e.g. substituting $y = -4x + 1$ |
| $21x^2 + 10x + 1 = 0$ | A1 | Correct 3 term quadratic (terms do not need to be all on the same side). The "$= 0$" may be implied by subsequent work |
| $(7x+1)(3x+1) = 0 \Rightarrow x = -\frac{1}{7}, -\frac{1}{3}$ | dM1 A1 | dM1: Solves a **3 term** quadratic by usual rules to give at least one value for $x$. **Dependent on first method mark.** A1: Two separate correct exact answers. Allow exact equivalents e.g. $x = -\frac{6}{42}, -\frac{14}{42}$ |
| $y = -\frac{3}{7}, \frac{1}{3}$ | M1 A1 | M1: Substitutes to find at least one $y$ value. A1: $y = -\frac{3}{7}, \frac{1}{3}$ (two correct exact answers). Allow exact equivalents e.g. $y = -\frac{18}{42}, \frac{14}{42}$ |
**Coordinates do not need to be paired**
**Note: if the linear equation is explicitly rearranged to $y = 4x + 1$, this gives correct answers for $x$ and possibly for $y$. If not already lost, deduct the final A1.**
**[6]**
## Way 2
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = -\frac{1}{4}y - \frac{1}{4} \Rightarrow y^2 + 5(-\frac{1}{4}y - \frac{1}{4})^2 + 2(-\frac{1}{4}y - \frac{1}{4}) = 0$ | M1 | Attempts to make $x$ the subject and substitutes into the other equation. Allow slips |
| $\frac{21}{16}y^2 + \frac{1}{8}y - \frac{3}{16} = 0 \;\; (21y^2 + 2y - 3 = 0)$ | A1 | Correct 3 term quadratic. "$= 0$" may be implied by subsequent work |
| $(7y+3)(3y-1) = 0 \Rightarrow y = -\frac{3}{7}, \frac{1}{3}$ | dM1 A1 | dM1: Solves a **3 term** quadratic to give at least one value for $y$. **Dependent on first method mark.** A1: Two separate correct exact answers. Allow e.g. $y = -\frac{18}{42}, \frac{14}{42}$ |
| $x = -\frac{1}{7}, -\frac{1}{3}$ | M1 A1 | M1: Substitutes to find at least one $x$ value. A1: $x = -\frac{1}{7}, -\frac{1}{3}$ (two correct exact answers). Allow e.g. $x = -\frac{6}{42}, -\frac{14}{42}$ |
**Coordinates do not need to be paired**
**Note: if the linear equation is explicitly rearranged to $x = (y+1)/4$, this gives correct answers for $y$ and possibly for $x$. If not already lost, deduct the final A1.**
**[6 marks total]**
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