EQUATIONS

QUESTION 1 [6 marks] — Criterion A Medium

Solve each equation:

a. $3(x+4)=27$

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$3x+12=27 \Rightarrow 3x=15 \Rightarrow x=5$

b. $\dfrac{x-2}{5}=3$

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$x-2=15 \Rightarrow x=17$

c. $2(x-3)+4=16$

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$2x-6+4=16 \Rightarrow 2x-2=16 \Rightarrow 2x=18 \Rightarrow x=9$

QUESTION 2 [4 marks] — Criterion A Medium

Solve $5(2x-1)=3(x+8)$, showing every step.

a. Expand both sides.

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$10x-5=3x+24$

b. Collect $x$ terms on one side, constants on the other, and solve.

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$10x-3x=24+5 \Rightarrow 7x=29 \Rightarrow x=\frac{29}{7}\approx4.14$

QUESTION 3 [6 marks] — Criterion B Medium

Investigate whether the order of operations used to solve an equation with brackets affects the final answer.

a. Solve $4(x+3)=32$ by first EXPANDING the brackets.

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$4x+12=32 \Rightarrow 4x=20 \Rightarrow x=5$

b. Now solve the SAME equation by first DIVIDING both sides by 4 (before expanding).

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$x+3=8 \Rightarrow x=5$

c. Both methods give the same answer. Explain why either approach is valid, and suggest when one method might be more efficient than the other.

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Both are valid because they're just different (but equally legitimate) sequences of balanced operations — the equation stays true throughout either path. Dividing first tends to be more efficient when the bracket's coefficient divides evenly into the number on the other side (as here, 32÷4=8 is clean); expanding first is more reliable when it doesn't divide evenly.

QUESTION 4 [5 marks] — Criterion C Medium

A student solves $\dfrac{2x+6}{4}=5$ by cross-multiplying incorrectly, writing '$2x+6=5$' (forgetting to multiply the 5 by 4).

a. Explain the student's error, and give the correct first step and full solution.

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The student forgot that multiplying both sides by 4 affects the ENTIRE right side, not just leaving it as 5. Correct: $2x+6=5\times4=20 \Rightarrow 2x=14 \Rightarrow x=7$.

b. Verify $x=7$ by substituting into the original equation.

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$\frac{2(7)+6}{4}=\frac{20}{4}=5$ ?

QUESTION 5 [6 marks] — Criterion D Medium

A rectangular room has perimeter 34 m. The length is 3 m more than twice the width.

a. Let the width be $w$. Write an expression for the length, then write an equation for the perimeter using $P=2(\text{length}+\text{width})$.

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Length $=2w+3$. Perimeter equation: $2((2w+3)+w)=34$.

b. Solve your equation to find the width and length.

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$2(3w+3)=34 \Rightarrow 6w+6=34 \Rightarrow 6w=28 \Rightarrow w=\frac{28}{6}\approx4.67$ m. Length $=2(4.67)+3\approx12.33$ m.

QUESTION 6 [3 marks] — Criterion A Medium

Solve $4(2x-3)=5x+9$, showing every step.

a. Expand the left side.

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$8x-12=5x+9$

b. Collect $x$ terms and constants, then solve.

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$8x-5x=9+12 \Rightarrow 3x=21 \Rightarrow x=7$

QUESTION 7 [5 marks] — Criterion A Hard

Solve the equation involving fractions: $$\dfrac{x+3}{4}=\dfrac{2x-1}{6}$$

a. Eliminate the fractions by cross-multiplying (or finding a common denominator of 12).

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Cross-multiplying: $6(x+3)=4(2x-1) \Rightarrow 6x+18=8x-4$.

b. Solve for $x$.

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$18+4=8x-6x \Rightarrow 22=2x \Rightarrow x=11$

QUESTION 8 [6 marks] — Criterion B Hard

Investigate whether MULTIPLYING an entire equation by a NEGATIVE number changes anything important, using $-2x+6=10$.

a. Solve $-2x+6=10$ directly (using standard balance operations).

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$-2x=4 \Rightarrow x=-2$.

b. Now multiply EVERY term of the original equation by $-1$ first (giving $2x-6=-10$), then solve THIS new equation. Confirm you get the same solution, and explain why multiplying an entire equation by $-1$ (or any nonzero constant) never changes its solution.

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$2x-6=-10 \Rightarrow 2x=-4 \Rightarrow x=-2$ — matches. Multiplying EVERY term on BOTH sides by the same nonzero constant is a valid balanced operation (both sides are scaled identically), so it never changes which value of $x$ satisfies the equation — it only changes how the equation is temporarily written.

QUESTION 9 [5 marks] — Criterion C Medium

A student solving $\dfrac{2x}{3}=10$ multiplies both sides by 3, getting '$2x=30$', then INCORRECTLY divides by 3 again (instead of by 2) to finish, getting $x=10$.

a. Explain the error, and give the correct final answer.

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After correctly reaching $2x=30$, the student needed to divide by 2 (to undo the multiplication by 2 remaining in $2x$), NOT by 3 again — the '÷3' step was already used to clear the fraction, and shouldn't be repeated. Correct: $2x=30 \Rightarrow x=15$.

b. Verify $x=15$ using the ORIGINAL equation.

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$\frac{2(15)}{3}=\frac{30}{3}=10$ ?.

QUESTION 10 [5 marks] — Criterion C Medium

A student presents a complete, correct multi-step equation solution as a single unlabeled line of numbers: '3(x-2)=2x+7, 3x-6=2x+7, x=13'.

a. Rewrite this as a properly communicated solution, adding clear labels for each step (e.g. 'expand the bracket', 'collect like terms', etc.) so a reader can follow the LOGIC, not just the numbers.

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Solve $3(x-2)=2x+7$: Step 1 (expand the bracket): $3x-6=2x+7$. Step 2 (subtract $2x$ from both sides): $x-6=7$. Step 3 (add 6 to both sides): $x=13$.

b. Verify $x=13$ using the original equation.

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$3(13-2)=3(11)=33$. $2(13)+7=26+7=33$ ?.

QUESTION 11 [7 marks] — Criterion D Hard

A construction crew is laying rectangular paving. The length is $(2x+3)$m and width is $(x-1)$m. The perimeter is known to be 46m.

a. Write and solve an equation for $x$, using the perimeter formula $P=2(\text{length}+\text{width})$.

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$2[(2x+3)+(x-1)]=46 \Rightarrow 2(3x+2)=46 \Rightarrow 6x+4=46 \Rightarrow 6x=42 \Rightarrow x=7$.

b. Find the actual length and width, and verify the perimeter equals 46m.

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Length: $2(7)+3=17$m. Width: $7-1=6$m. Perimeter: $2(17+6)=2(23)=46$m ?.

Solving equations