Moderate -0.8 This is a straightforward simultaneous equations question requiring substitution of the linear equation into the quadratic, expanding to get a quadratic equation, and solving. It's a standard C1 exercise with clear structure and routine algebraic manipulation, making it easier than average but not trivial since students must correctly expand and solve the resulting quadratic.
Correct 3 term quadratic (not necessarily all in one side)
\((2x-9)(x+2)(=0)\)
M1
Correct method to solve quadratic
\(x = \frac{9}{2}, x = -2\)
A1
x values correct
\(y = \frac{25}{2}, y = 32\)
A1, 5
y values correct
SR If A0 A0, one correct pair of values, spotted or from correct factorisation www
B1
$2x^2 - 8x + 8 = 26 - 3x$ | M1 | Attempt to eliminate x or y | Must be a clear attempt to reduce to one variable. Condone poor algebra for first mark.
$2x^2 - 5x - 18(=0)$ | A1 | Correct 3 term quadratic (not necessarily all in one side) | If x eliminated: $y = 2(\frac{26-y}{3} - 2)^2$ leading to $2y^2 - 89y + 800 = 0$ or $(2y-25)(y-32)=0$ etc.
$(2x-9)(x+2)(=0)$ | M1 | Correct method to solve quadratic |
$x = \frac{9}{2}, x = -2$ | A1 | x values correct |
$y = \frac{25}{2}, y = 32$ | A1, 5 | y values correct |
SR If A0 A0, one correct pair of values, spotted or from correct factorisation www | B1 |