Linear Inequalities: Eduqas GCSE Higher June 2018 Paper 1 Q2

Higher June 2018 Paper 1 Q2

(a) Solve \(12x - 9 = 6 + 7x\). [2]

(b) Solve \(10(x + 2) - (2x - 9) = 30\). [3]

(c)

(i) Solve the inequality \(10x - 7 \leqslant 8\). [2]

(ii) Represent your answer to part (c)(i) on the number line below. [1]

(d) Gracie is trying to solve the equation \(x^2 - 5x + 6 = 0\).

Here is her work.

\(x^2 - 5x + 6 = 0\)

\((x - 3)(x - 2) = 0\)

\(x - 3 \qquad x - 2\)

\(x = -3, \quad x = -2\)

Is Gracie’s work correct?

Yes ☐ No ☐

Show clearly how you decide. [1]

Scheme	Marks	Comments
\(12x - 7x = 6 + 9\) or \(x = \dfrac{15}{5}\) or equivalent	B1	Seen or implied FT until 2nd error
\(x = 3\)	B1	Mark final answer; allow embedded answer for 2

Mark scheme (b)

Scheme	Marks	Comments
\(10x + 20 - (2x - 9) = 30\) or \(10(x + 2) - 2x + 9 = 30\) or better	B1	FT until 2nd error Seen or implied For expanding at least one pair of brackets correctly
\(10x - 2x = 30 - 9 - 20\) or better	B1	FT For collecting terms
\(x = \dfrac{1}{8}\) ISW	B1	FT; if FT the common error \(8x = 19\), then the answer must be 2.375 or \(\dfrac{19}{8}\), not rounded to e.g. 2.3

Mark scheme (c)

Scheme	Marks	Comments
(i) \(10x \leqslant 15\) or equivalent	M1
\(x \leqslant 1.5\) or equivalent	A1	Mark final answer No marks for use of “=”, unless finally replaced to give \(x \leqslant 1.5\) then award M1 A1.
(ii) Solid circle at 1.5 with arrow left	B1	STRICT FT ‘their (c)(i)’ provided an inequality Accept any unambiguous notation provided there is no termination (so not a line with a circle at each end)

Mark scheme (d)

Scheme	Marks	Comments
No with valid explanation. e.g. ‘The correct answers are \(x = 3\) and \(x = 2\).’ or ‘She has forgotten to put \(x - 3\) and \(x - 2\) equal to 0 and solve.’ or ‘When \(x = -3\) the value is 30.’ or ‘The correct solution is \(x = 3\) or \(x = 2\).’	E1	Do not accept e.g. ‘The values do not work’ without supporting evidence. Allow ‘No and (−3 −3 ) = −6 and (−2 −2 ) = −4’
	(9)