Moderate -0.3 This is a standard P1 simultaneous equations question requiring substitution of a linear equation into a quadratic (circle equation), then solving the resulting quadratic. While it involves multiple steps (substitution, expansion, collecting terms, quadratic formula or factoring, back-substitution), these are all routine techniques with no conceptual difficulty or novel insight required. Slightly easier than average due to being a well-practiced question type.
4. Use algebra to solve the simultaneous equations
$$\begin{array} { r }
y - 3 x = 4 \\
x ^ { 2 } + y ^ { 2 } + 6 x - 4 y = 4
\end{array}$$
You must show all stages of your working.
\(y=3x+4 \Rightarrow x^2+(3x+4)^2+6x-4(3x+4)=4\) or \(x=\frac{y-4}{3}\Rightarrow\left(\frac{y-4}{3}\right)^2+y^2+6\left(\frac{y-4}{3}\right)-4y=4\)
M1
Rearrange linear equation to \(y=\ldots\) or \(x=\ldots\) and fully substitute into second equation
\(5x^2+9x-2(=0)\) or \(5y^2-13y-46(=0)\)
M1A1
Collect terms to produce 2 or 3 term quadratic \(=0\); condone slips; may be multiples e.g. \(10x^2+18x-4=0\)
\((5x-1)(x+2)=0 \Rightarrow x=\ldots\) or \((5y-23)(y+2)=0 \Rightarrow y=\ldots\)
dM1
Attempt to factorise/solve or use quadratic formula; dependent on both previous M marks; factored expression must equal their quadratic
\(x=0.2,\ x=-2\) or \(y=4.6,\ y=-2\)
B1
Correct answers for both values of \(x\) or both values of \(y\); only scored if correct 3TQ achieved
Substitutes \(x\) into \(y=3x+4\) / substitutes \(y\) into \(x=\frac{y-4}{3}\)
M1
Substitute at least one value to find other variable; may be implied by final answers
\(x=0.2\left(\text{or }\frac{1}{5}\right),\ y=4.6\left(\text{or }4\frac{3}{5}\text{ or }\frac{23}{5}\right)\) and \(x=-2,\ y=-2\)
A1
Fully correct solution with all previous marks awarded; both pairs found and simplified; condone recovery of invisible brackets
(7 marks)
# Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=3x+4 \Rightarrow x^2+(3x+4)^2+6x-4(3x+4)=4$ **or** $x=\frac{y-4}{3}\Rightarrow\left(\frac{y-4}{3}\right)^2+y^2+6\left(\frac{y-4}{3}\right)-4y=4$ | M1 | Rearrange linear equation to $y=\ldots$ or $x=\ldots$ and fully substitute into second equation |
| $5x^2+9x-2(=0)$ or $5y^2-13y-46(=0)$ | M1A1 | Collect terms to produce 2 or 3 term quadratic $=0$; condone slips; may be multiples e.g. $10x^2+18x-4=0$ |
| $(5x-1)(x+2)=0 \Rightarrow x=\ldots$ or $(5y-23)(y+2)=0 \Rightarrow y=\ldots$ | dM1 | Attempt to factorise/solve or use quadratic formula; dependent on both previous M marks; factored expression must equal their quadratic |
| $x=0.2,\ x=-2$ or $y=4.6,\ y=-2$ | B1 | Correct answers for both values of $x$ or both values of $y$; only scored if correct 3TQ achieved |
| Substitutes $x$ into $y=3x+4$ / substitutes $y$ into $x=\frac{y-4}{3}$ | M1 | Substitute at least one value to find other variable; may be implied by final answers |
| $x=0.2\left(\text{or }\frac{1}{5}\right),\ y=4.6\left(\text{or }4\frac{3}{5}\text{ or }\frac{23}{5}\right)$ **and** $x=-2,\ y=-2$ | A1 | Fully correct solution with all previous marks awarded; both pairs found and simplified; condone recovery of invisible brackets |
**(7 marks)**
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4. Use algebra to solve the simultaneous equations
$$\begin{array} { r }
y - 3 x = 4 \\
x ^ { 2 } + y ^ { 2 } + 6 x - 4 y = 4
\end{array}$$
You must show all stages of your working.\\
\hfill \mbox{\textit{Edexcel P1 2020 Q4 [7]}}