Moderate -0.3 This is a straightforward simultaneous equations problem requiring substitution of the linear equation into the quadratic curve, then solving the resulting quadratic. It's slightly easier than average because the linear equation is already in a convenient form for substitution and the algebra is manageable, though students must be careful with the quadratic manipulation.
Attempt substitution from (ii) into (i) or (i) into (ii) and obtain equation in one letter
\(12y - 6y^2 + 16 - 16y + 4y^2 = -14\)
\(2y^2 + 4y - 30 = 0\)
Answer
Marks
Guidance
\(y^2 + 2y - 15 = 0 \Rightarrow (y+5)(y-3) = 0\)
A1 (1.1)
Obtain correct 3-term quadratic equation e.g. \(x^2 - 12x - 28 = 0\); method may not be seen
\(y = -5\) or \(3\)
A1 (1.1)
cao; or \(x = 14\) or \(-2\)
e.g. \(x + 2(-5) = 4\) and \(x + 2\times3 = 4\)
M1 (2.2a)
Substitute their \(y\) values into either equation; or their \(x\) values
Points of intersection are \((14, -5)\) & \((-2, 3)\)
A1 (1.1)
or \(x=14, y=-5\); \(x=-2, y=3\). Must be clearly paired cao
# Question 7:
$3(4-2y)y + (4-2y)^2 = -14$ | M1 (3.1a) | Attempt substitution from (ii) into (i) or (i) into (ii) and obtain equation in one letter
$12y - 6y^2 + 16 - 16y + 4y^2 = -14$
$2y^2 + 4y - 30 = 0$
$y^2 + 2y - 15 = 0 \Rightarrow (y+5)(y-3) = 0$ | A1 (1.1) | Obtain correct 3-term quadratic equation e.g. $x^2 - 12x - 28 = 0$; method may not be seen
$y = -5$ or $3$ | A1 (1.1) | cao; or $x = 14$ or $-2$
e.g. $x + 2(-5) = 4$ and $x + 2\times3 = 4$ | M1 (2.2a) | Substitute their $y$ values into either equation; or their $x$ values
Points of intersection are $(14, -5)$ & $(-2, 3)$ | A1 (1.1) | or $x=14, y=-5$; $x=-2, y=3$. Must be clearly paired cao
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