Moderate -0.3 This is a standard C1 simultaneous equations question requiring substitution of the linear equation into the quadratic, expanding to form a quadratic in one variable, and solving. The arithmetic is straightforward and the method is routine textbook material, making it slightly easier than average but still requiring careful algebraic manipulation.
\(y=2x+4 \Rightarrow 4x^2+(2x+4)^2+20x=0\) or \(2x=y-4 \Rightarrow (y-4)^2+y^2+10(y-4)=0\)
M1
Attempts to rearrange linear equation to \(y=\ldots\) or \(x=\ldots\) or \(2x=\ldots\) and attempts to fully substitute into second equation
\(8x^2+36x+16=0\) or \(2y^2+2y-24=0\)
M1 A1
M1: Collects terms to produce quadratic \(=0\). A1: Correct three term quadratic in \(x\) or \(y\)
\((4)(2x+1)(x+4)=0 \Rightarrow x=\ldots\) or \((2)(y+4)(y-3)=0 \Rightarrow y=\ldots\)
M1
Attempt to factorise and solve or complete the square or use correct quadratic formula for a 3 term quadratic
\(x=-0.5,\ x=-4\) or \(y=-4,\ y=3\)
A1 cso
Correct answers for either both values of \(x\) or both values of \(y\)
Sub into \(y=2x+4\) or sub into \(x=\frac{y-4}{2}\)
M1
Substitutes at least one value of \(x\) into a correct equation as far as \(y=\ldots\) or substitutes at least one value of \(y\) into a correct equation as far as \(x=\ldots\)
\(y=3,\ y=-4\) and \(x=-4,\ x=-0.5\)
A1
Fully correct solutions and simplified. Pairing not required. If extra values of \(x\) or \(y\), score A0
# Question 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| $y=2x+4 \Rightarrow 4x^2+(2x+4)^2+20x=0$ **or** $2x=y-4 \Rightarrow (y-4)^2+y^2+10(y-4)=0$ | M1 | Attempts to rearrange linear equation to $y=\ldots$ or $x=\ldots$ or $2x=\ldots$ and attempts to **fully** substitute into second equation |
| $8x^2+36x+16=0$ **or** $2y^2+2y-24=0$ | M1 A1 | M1: Collects terms to produce quadratic $=0$. A1: Correct three term quadratic in $x$ or $y$ |
| $(4)(2x+1)(x+4)=0 \Rightarrow x=\ldots$ **or** $(2)(y+4)(y-3)=0 \Rightarrow y=\ldots$ | M1 | Attempt to factorise and solve or complete the square or use correct quadratic formula **for a 3 term quadratic** |
| $x=-0.5,\ x=-4$ **or** $y=-4,\ y=3$ | A1 cso | Correct answers for either both values of $x$ or both values of $y$ |
| Sub into $y=2x+4$ **or** sub into $x=\frac{y-4}{2}$ | M1 | Substitutes at least one value of $x$ into a **correct** equation as far as $y=\ldots$ or substitutes at least one value of $y$ into a **correct** equation as far as $x=\ldots$ |
| $y=3,\ y=-4$ and $x=-4,\ x=-0.5$ | A1 | Fully correct solutions and simplified. **Pairing not required.** If extra values of $x$ or $y$, score A0 |
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