Easy -1.8 This is a straightforward linear inequality requiring only basic algebraic manipulation: multiply through by 30 to clear fractions, collect like terms, and solve. It's a routine C1 exercise with no conceptual difficulty beyond fraction handling, making it significantly easier than average A-level questions.
For multiplying up correctly or for correct use of a common denominator; first three Ms may be earned with an equality; condone omission of brackets only if then expanded as if brackets present
\(12x+6 < 15x+20\) or ft
M1
For expanding brackets correctly; for combined first two steps with one error, such as \(12x+6 < 15x+4\), allow M1M0; e.g. \(\frac{12x+6}{30} < \frac{15x+20}{30}\) oe earns M1M1
\(-14 < 3x\) or \(-3x < 14\) or ft
M1
For collecting terms correctly; ft from two \(x\) terms and two constants
\(x > -\frac{14}{3}\) oe or ft isw
M1
For final division of their inequality with \(ax\) on one side, \(a \neq 1\) or \(0\), and non-zero number on the other; allow working with equality and making correct decision at end; e.g. allow last M1 for \(x > \frac{14}{-3}\) or \(\frac{-14}{3} < x\) isw; reminder: \((-14/3, \infty)\) is acceptable notation; allow SC3 for \(-14/3\) found without correct inequality symbol(s)
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6(2x+1) < 5(3x+4)$ | M1 | For multiplying up correctly or for correct use of a common denominator; first three Ms may be earned with an equality; condone omission of brackets only if then expanded as if brackets present |
| $12x+6 < 15x+20$ or ft | M1 | For expanding brackets correctly; for combined first two steps with one error, such as $12x+6 < 15x+4$, allow M1M0; e.g. $\frac{12x+6}{30} < \frac{15x+20}{30}$ oe earns M1M1 |
| $-14 < 3x$ or $-3x < 14$ or ft | M1 | For collecting terms correctly; ft from two $x$ terms and two constants |
| $x > -\frac{14}{3}$ oe or ft isw | M1 | For final division of their inequality with $ax$ on one side, $a \neq 1$ or $0$, and non-zero number on the other; allow working with equality and making correct decision at end; e.g. allow last M1 for $x > \frac{14}{-3}$ or $\frac{-14}{3} < x$ isw; reminder: $(-14/3, \infty)$ is acceptable notation; allow SC3 for $-14/3$ found without correct inequality symbol(s) |
| **Or:** $\frac{1}{5} - \frac{4}{6} < \frac{3x}{6} - \frac{2x}{5}$ oe | or M1 | |
| $\frac{-7}{15} < \frac{3x}{30}$ oe or ft | M2 | M1 for one side correct ft |
| $x > -\frac{14}{3}$ oe or ft isw | M1 | As in previous method |
| **[4]** | | |
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