Algebra, Equations and Inequalities

Substitute $x = -2$: $3(-2) - 5 = -6 - 5 = -11$. (Using $x = +2$ gives 1; treating $-5$ as $+5$ gives $-1$.)

Substitute $x = -3$: $(-3)^2 + 2(-3) = 9 + (-6) = 3$. Remember $(-3)^2 = 9$, not $-9$. (Treating it as $-9$ gives the wrong $-15$.)

Expanding multiplies every term inside the bracket by the 3: $3 \times 2x = 6x$ and $3 \times 5 = 15$, giving $6x + 15$. The student multiplied the first term but forgot the $+5$.

$6x$ means six lots of $x$ added together: $x + x + x + x + x + x$. ($2x \times 3x = 6x^2$, not $6x$; $6 + x$ and $x^6$ are different expressions.)

Perimeter $= 2(\text{length} + \text{width})$, so $2\big((x+4) + x\big) = 28$. This gives $2(2x + 4) = 28$, then $4x + 8 = 28$, so $4x = 20$ and $x = 5$. (The sides are then 9 cm and 5 cm, and $2(9 + 5) = 28$.)

Distribute the $-2$: $-2\times x = -2x$ and $-2\times(-4) = +8$, so $-2(x-4) = -2x + 8$. (Negative times negative is positive.)

Expand each bracket: $3(2x+1) = 6x + 3$ and $2(x-4) = 2x - 8$. Add them: $6x + 3 + 2x - 8 = 8x - 5$.

The HCF of 6 and 9 is 3: $\dfrac{6x}{3} = 2x$ and $\dfrac{9}{3} = 3$, so $6x + 9 = 3(2x + 3)$. (Check: $3\times 2x + 3\times 3 = 6x + 9$.)

The HCF of 12 and 18 is 6: $\dfrac{12a}{6} = 2a$ and $\dfrac{-18b}{6} = -3b$, so $12a - 18b = 6(2a - 3b)$. ($3(4a-6b)$ and $2(6a-9b)$ are only partly factorised.)

Subtract 5: $3x = 12$. Divide by 3: $x = \dfrac{12}{3} = 4$. (Check: $3(4) + 5 = 17$.)

Both $(+4)^2 = 16$ and $(-4)^2 = 16$, so $x = \pm 4$. (Giving only $x = 4$ misses the negative root; $x = 8$ doubles instead of square-rooting.)

Subtract $x$ from both sides: $2x + 5 = 11$. Subtract 5: $2x = 6$. Divide by 2: $x = \dfrac{6}{2} = 3$.

Angles on a straight line add to $180$°: $(2x + 10) + (x + 20) = 180$. Combine like terms: $3x + 30 = 180$, so $3x = 150$ and $x = 50$. (The two angles are then 110° and 70°.)

Area $= \text{length} \times \text{width}$, so $4(x + 3) = 32$. Divide both sides by 4: $x + 3 = 8$, so $x = 5$. (Check: length $= 8$ cm and $4 \times 8 = 32$.)

Multiply both sides by 2: $x - 4 = 10$. Add 4: $x = 14$. (Check: $\dfrac{14 - 4}{2} = 5$.)

Subtract 3: $2x > 6$. Divide by 2 (positive, so the sign stays): $x > 3$. (You only flip the sign when dividing by a negative.)

Substitute $C = 20$ into the formula: $F = 1.8(20) + 32 = 36 + 32 = 68$. (Computing $20 + 32$ and ignoring the $1.8$ gives 52; stopping at $36$ forgets to add 32.)

Divide both sides by $-2$ and flip the inequality: $x > -3$. (Check: $x = 0$ gives $-2(0) = 0 < 6$.) (Not flipping gives the wrong $x < -3$.)

The open (unfilled) circle at 2 means 2 is not included, so the inequality is strict. The arrow points right, meaning values larger than 2. Together: $x > 2$. (A closed circle would give $x \ge 2$; a left arrow would give $x < 2$.)

Expand: $2x + 6 - 4 = 12$, which simplifies to $2x + 2 = 12$. Subtract 2: $2x = 10$. Divide by 2: $x = 5$. (Check: $2(5+3) - 4 = 12$.)

Let $k$ be the kilometres: $5 + 2k = 23$. Subtract 5: $2k = 18$. Divide by 2: $k = \dfrac{18}{2} = 9$ km. (Dividing 23 by 2 forgets the base fee.)

Set up $50 + 8h \ge 130$. Subtract 50: $8h \ge 80$. Divide by 8: $h \ge \dfrac{80}{8} = 10$, so he must work at least 10 hours.

$2a = 8$, $-3b = -3(-2) = +6$, and $ab = 4\times(-2) = -8$. Sum: $8 + 6 - 8 = 6$. (Forgetting the $ab$ term gives 14.)

Expand each bracket: $4(2x-3) = 8x - 12$ and $-3(x-5) = -3x + 15$. Combine: $8x - 12 - 3x + 15 = 5x + 3$. (Forgetting to flip the sign of $-5$ gives the wrong $5x - 27$.)

Let the smallest be $n$. The sum is $n + (n+1) + (n+2) = 3n + 3$, and $3n + 3 > 30$, so $3n > 27$ and $n > 9$. Since the inequality is strict, $n = 9$ fails, so the smallest integer is $n = 10$. (Check: $10 + 11 + 12 = 33 > 30$.)