At the start of each shift, a delivery driver records \(x\), the number of parcels being carried above or below the standard load for that route, so \(x\) c...
Assessment:Mathematics Specification A 4MA1 | Paper 3 Mock 01 | Structured / Short AnswerSubject:Mathematics Specification A - 4MA1
At the start of each shift, a delivery driver records \(x\), the number of parcels being carried above or below the standard load for that route, so \(x\) can be negative. The load satisfies \(-7 \le 4x-3 \lt 9\). Solve this inequality for \(x\), and represent your solution on the number line shown. (3)
This question solves a compound (three-part) inequality by applying the same operation to all three parts at once, and then represents the resulting range on a number line.
Starting from \(-7 \le 4x-3 \lt 9\), add \(3\) to all three parts:
\[-4 \le 4x \lt 12\]
[1 mark]
Dividing all three parts by \(4\) (positive, so the inequality directions are unchanged):
\[-1 \le x \lt 3\]
[1 mark]
Representing this on a number line, with a filled circle at \(x=-1\) (included) and an open circle at \(x=3\) (not included), joined by a thick line:
[1 mark]
The filled circle at \(-1\) shows that \(x=-1\) is a valid load reading, since the original inequality used \(\le\) there; the open circle at \(3\) shows \(x=3\) is not valid, since the original inequality used the strict \(\lt\) at that end.
This question solves a compound (three-part) inequality by applying the same operation to all three parts at once, and then represents the resulting range on a number line.
Starting from \(-7 \le 4x-3 \lt 9\), add \(3\) to all three parts:
\[-4 \le 4x \lt 12\]
[1 mark]
Dividing all three parts by \(4\) (positive, so the inequality directions are unchanged):
\[-1 \le x \lt 3\]
[1 mark]
Representing this on a number line, with a filled circle at \(x=-1\) (included) and an open circle at \(x=3\) (not included), joined by a thick line:
[1 mark]
The filled circle at \(-1\) shows that \(x=-1\) is a valid load reading, since the original inequality used \(\le\) there; the open circle at \(3\) shows \(x=3\) is not valid, since the original inequality used the strict \(\lt\) at that end.