\[\begin{aligned} 2x + y &= 8 \\ x - y &= 1 \end{aligned}\]
(c)
Represent the inequality \(-2 \leqslant x \leqslant 3\) on the number line below. [1]
(d) Solve \(\dfrac{2x}{3} \lt 4\). [2]
Mark scheme (a)
Scheme
Marks
Comments
\(2x = 5\)
B1
\(x = \dfrac{5}{2}\) oe, ISW
B1
FT from ‘their \(ax = b\)’ provided \(a \neq b\) or 0 or 1 and \(b \neq 0\); accept \(\dfrac{b}{a}\) but if on FT \(\dfrac{b}{a}\) simplifies to an integer the answer must be given as an integer.
‘\(x\) =’ can be omitted but must not be wrong if there.
Correct answer implies first B1.
Mark scheme (b)
Scheme
Marks
Comments
\(x = 3\)
B1
\(y = 2\)
B1
Mark scheme (c)
Scheme
Marks
Comments
Line with solid circles at both ends starting at \(-2\) and ending at 3
B1
Mark scheme (d)
Scheme
Marks
Comments
\(2x \lt 4 \times 3\)
M1
\(x \lt 6\)
A1
No marks for use of “=”, unless finally replaced to give \(x \lt 6\) then award M1 A1. \(x\) <= 6 is A0
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