(a) Write down the inequality shown on the number line. (2)
(b) Solve the inequality \(7a - 5 \leqslant 3a + 28\) Show clear algebraic working. (2)
Mark scheme (a)
Scheme
Marks
\(-2 \lt x \leqslant 1\)
B2
(2)
Notes
B2: accept \(1 \geqslant x \gt -2\) or \(x \gt -2, x \leqslant 1\) if not B2 then B1 for \(-2 \lt x\) or \(x \leqslant 1\) or \(-2 \leqslant x \lt 1\) or \(-2 \leqslant x \leqslant 1\) or \(-2 \lt x \lt 1\) Condone use of a variable other than \(x\) but not 0
Mark scheme (b)
Scheme
Marks
\(7a - 3a \leqslant 28 + 5\) or \(4a \leqslant 33\) or \(-5 - 28 \leqslant 3a - 7a\) or \(-33 \leqslant -4a\)
M1
Working required Answer: \(a \leqslant 8.25\)
A1
(2)
(4 marks)
Notes
M1: for \(a\) terms on one side and numbers on the other. Condone = rather than \(\leqslant\) or any other sign for this mark.
A1: (dep on M1) oe eg \(a \leqslant \dfrac{33}{4}\) or \(a \leqslant 8\dfrac{1}{4}\) or \(8.25 \geqslant a\)
must have correct sign on answer line
(sight of correct answer in working space and just 8.25 on answer line gains M1 only).
Correct answer scores full marks (unless from obvious incorrect working) Answer: \((x + 8)(x - 3)\)
A1
(2)
(ii) Answer: –8 and 3
B1
(1)
Notes
M1: for \((x \pm 8)(x \pm 3)\) or \((x + a)(x + b)\) where \(ab = -24\) or \(a + b = 5\) and, \(a\) and \(b\) are integers
A1: for \((x + 8)(x - 3)\) Allow any letter for \(x\)
Must be in the form \((x + a)(x + b)\) where \(a\) and \(b\) are integers
B1: must ft from their answer in (a)(i) ft from their incorrect factors in the form \((x + a)(x + b)\) Award B0 for –8 and 3 if no marks scored in (a)(i)
Mark scheme (b)
Scheme
Marks
\(3y - 7y \gt -10 - 5\) or \(5 + 10 \gt 7y - 3y\)
M1
\(-4y \gt -15\) or \(15 \gt 4y\) or \(y = \dfrac{15}{4}\) oe or
M1
Working required
Answer: \(y \lt \dfrac{15}{4}\)
A1
(3)
(6 marks)
Notes
M1: allow use of = or condone incorrect inequality sign
M1: allow use of = or condone incorrect inequality sign
A1: dep on M1
oe eg \(y \lt 3.75\) or \(\dfrac{15}{4} \gt y\) or \(3.75 \gt y\)
Must have correct sign on answer line
NB Sight of correct answer in working space and just \((y =)\;\dfrac{15}{4}\) oe on answer line gains M2 only
(a) On the grid, draw the straight line with equation
(i) \(x = 3\) (ii) \(y = 1\) (iii) \(x + y = 7\) Label each line with its equation. (3)
(b) Show, by shading on the grid, the region that satisfies all three of the inequalities \(x \geqslant 3 \qquad y \geqslant 1 \qquad x + y \leqslant 7\) Label the region R(1)
Mark scheme (a)
Scheme
Marks
(i)
B1
(ii)
B1
(iii) Line length 2 cm + but shaded area must be enclosed for the mark in (b)
If unlabelled, award: \(x = 3\) and \(y = 3\) B1 B0 \(y = 1\) and \(x = 1\) B0 B1 \(x = 3\) and \(x = 1\) and \(y = 1\) B0 B1 \(x = 3\) and \(y = 1\) and \(y = 3\) B1 B0
\(x = 3\) and \(x = 1\), \(y = 1\) and \(y = 3\) B0 B0
B1
(3)
Notes
B1: \(x = 3\) drawn
B1: \(y = 1\) drawn
B1: \(x + y = 7\) drawn
Allow dashed lines or solid lines for graphs of minimum length 2 squares condone lack of labels if unambiguous
Mark scheme (b)
Scheme
Marks
B1
(1)
(4 marks)
Notes
B1: correct region shaded – shaded in or out – labelled R or clear intention to be the required region (ft only for one vertical line (not \(x = 0\)), one horizontal line (not \(y = 0\)) and one line with a negative gradient eg \(x = 1\), \(y = 3\) and \(x + y = 7\))
