EQUATIONS

QUESTION 1 2 marks Criterion A

Medium

Consider the equation $3x+5=20$.

a. Test whether $x=5$ is a solution, by substituting it into the equation.

[1]

b. Test whether $x=6$ is a solution.

[1]

Show complete worked solution

(a)

$3(5)+5=15+5=20$ — yes, $x=5$ is a solution.

(b)

$3(6)+5=18+5=23\ne20$ — no, $x=6$ is NOT a solution.

QUESTION 2 4 marks Criterion A

Medium

For the equation $2x-7=x+3$, three students propose $x=8$, $x=10$, and $x=12$ as possible solutions.

a. Test all three proposed values, showing full working for each.

[3]

b. State which value is the correct solution.

[1]

Show complete worked solution

(a)

$x=8$: LHS$=2(8)-7=9$, RHS$=8+3=11$. Not equal. $x=10$: LHS$=2(10)-7=13$, RHS$=10+3=13$. Equal! $x=12$: LHS$=2(12)-7=17$, RHS$=12+3=15$. Not equal.

(b)

$x=10$

QUESTION 3 6 marks Criterion B

Medium

Investigate whether an equation can have MORE than one solution, using $x^2=9$.

a. Test $x=3$ and $x=-3$ in the equation $x^2=9$.

[2]

b. Now test $x=3$ and $x=-3$ in the equation $2x=6$ (a different, LINEAR equation).

[2]

c. Based on this investigation, explain why some equations (like $x^2=9$) have multiple solutions, while simple linear equations (like $2x=6$) typically have only one.

[2]

Show complete worked solution

(a)

$x=3$: $3^2=9$. $x=-3$: $(-3)^2=9$. Both work!

(b)

$x=3$: $2(3)=6$. $x=-3$: $2(-3)=-6\ne6$. Only $x=3$ works.

(c)

Squaring 'hides' the sign of the original number (since both a positive and its negative give the same square), so equations involving $x^2$ can have two solutions. Linear equations (just $x$, no power) don't have this sign-hiding effect, so they generally have exactly one solution.

QUESTION 4 4 marks Criterion C

Medium

A student says 'if I can find ONE value that works in an equation, there might be others too, so I should always keep guessing more values just in case.'

a. Explain why this approach (guessing and checking) is an inefficient and unreliable way to SOLVE an equation, even if it can verify a solution.

[2]

b. Explain why proper algebraic methods (like maintaining balance, covered elsewhere in this topic) are more reliable.

[2]

Show complete worked solution

(a)

Guessing and checking might miss the correct solution entirely if you don't happen to try the right number, and for equations with non-integer or unusual solutions, you could guess forever without finding it — it doesn't scale well and gives no certainty you've found ALL solutions.

(b)

Algebraic methods systematically isolate the variable through valid, reversible steps, guaranteeing you find the EXACT solution(s) without guessing — and the number of solutions an equation can have is determined by its structure (e.g. linear equations have exactly one), so you know when you're done.

QUESTION 5 4 marks Criterion D

Medium

A vending machine sells items for $x$ dollars each. A customer buys 4 items and receives \$2 change from a \$20 note, meaning $4x+2=20$.

a. Test whether $x=4.50$ is the correct price per item.

[2]

b. Explain, in context, why a solution like $x=-3$ would not make sense for this real-world equation, even if it satisfied the algebra.

[2]

Show complete worked solution

(a)

$4(4.50)+2=18+2=20$ — yes, $x=4.50$ satisfies the equation.

(b)

A negative price per item makes no real-world sense — you cannot pay a negative amount of money for a product. Even if $x=-3$ solved the equation mathematically (in this case it doesn't, but hypothetically), it would need to be rejected as an invalid answer in this real context.

QUESTION 6 5 marks Criterion A

Hard

For the equation $3x^2-5x-2=0$, three values are proposed as solutions: $x=2$, $x=-\frac{1}{3}$, $x=1$.

a. Test all three proposed values, showing full working, and identify which (if any) are genuine solutions.

[5]

Show complete worked solution

(a)

$x=2$: $3(4)-5(2)-2=12-10-2=0$ solution. $x=-\frac{1}{3}$: $3(\frac{1}{9})-5(-\frac{1}{3})-2=\frac{1}{3}+\frac{5}{3}-2=2-2=0$ solution. $x=1$: $3-5-2=-4\ne0$ not a solution. TWO of the three proposed values ($x=2$ and $x=-\frac{1}{3}$) are genuine solutions.

QUESTION 7 8 marks Criterion B

Hard

Investigate whether an equation can have DIFFERENT NUMBERS of solutions depending on its structure, comparing $x+5=12$, $x^2=16$, and $x^2+1=0$ (over the real numbers).

a. Solve each equation, stating how many real solutions each has.

[4]

b. Explain, in general terms, why a LINEAR equation (like the first) always has exactly one solution, while a QUADRATIC equation (like the second and third) can have 0, 1, or 2 real solutions depending on its specific structure.

[4]

Show complete worked solution

(a)

$x+5=12$: $x=7$ — exactly ONE solution. $x^2=16$: $x=4$ or $x=-4$ — exactly TWO solutions. $x^2+1=0$: $x^2=-1$, which has NO real solution (since a real number squared can never be negative) — ZERO real solutions.

(b)

A linear equation like $x+5=12$ involves $x$ to the power 1 only, so solving it always isolates $x$ to a single unique value via straightforward inverse operations. A quadratic equation involves $x^2$, which introduces the possibility of TWO different values of $x$ giving the same square (like $4^2=(-4)^2=16$) — but depending on the specific numbers involved, a quadratic might have two solutions (like $x^2=16$), exactly one repeated solution (a special case), or NO real solutions at all (like $x^2=-1$, since no real number squares to a negative).

QUESTION 8 6 marks Criterion B

Medium

Investigate whether EVERY value that makes an equation's LEFT side equal a certain number will also make an equivalent, REARRANGED version of the equation true.

a. Confirm $x=5$ is a solution to $2x-3=7$.

[2]

b. Rearrange $2x-3=7$ into the form $2x=10$ (adding 3 to both sides), then confirm $x=5$ ALSO satisfies THIS rearranged equation, and explain why any valid, balanced rearrangement of an equation must always share the exact same solution(s) as the original.

[4]

Show complete worked solution

(a)

$2(5)-3=10-3=7$.

(b)

$2(5)=10$, confirming $x=5$ satisfies the rearranged form too. This must ALWAYS be true because a valid rearrangement only involves performing the SAME operation on both sides of the equation (maintaining balance) — this never changes which values of $x$ make the two sides equal, it only changes how the equation LOOKS, not what it fundamentally represents.

QUESTION 9 4 marks Criterion C

Medium

A student tests $x=4$ in the equation $\dfrac{2x+6}{x-1}=\dfrac{14}{3}$ by cross-multiplying INCORRECTLY as '$3(2x+6)=14(x-1)$ becomes $3(14)=14(3)$' (substituting too early, before finishing the cross-multiplication setup).

a. Explain the correct order of steps for testing a proposed solution in an equation involving fractions: should you substitute FIRST and then simplify each side, or attempt to rearrange the general equation using the specific number partway through? Demonstrate the CORRECT way to test $x=4$.

[4]

Show complete worked solution

(a)

The safest, clearest approach is to substitute $x=4$ FIRST into the original equation, then simplify each side SEPARATELY and compare. Testing $x=4$: LHS $=\dfrac{2(4)+6}{4-1}=\dfrac{14}{3}$. RHS $=\dfrac{14}{3}$. Since LHS = RHS, $x=4$ IS a valid solution — substituting first and simplifying each side independently avoids the confusion of mixing algebraic rearrangement with a specific numerical test value.

QUESTION 10 7 marks Criterion D

Medium

A rectangular garden's area (in m²) is given by $A=x(x+8)$, where $x$ is the width. The garden's owner wants to know if a width of $x=6$m gives an area of at least 80m².

a. Test $x=6$ in the area formula, and determine whether it meets the 80m² requirement.

[3]

b. The owner also wants to know if $x=5$m would work. Test this value, and if it does NOT meet the requirement, find (by testing nearby whole-number values) the SMALLEST whole-number width that DOES satisfy the 80m² requirement.

[4]

Show complete worked solution

(a)

$A=6(6+8)=6(14)=84$ m². Since $84\ge80$, YES, $x=6$m meets the requirement.

(b)

$x=5$: $A=5(13)=65$m² — does NOT meet 80m². Testing further: $x=5.5$: not whole number, skip. Since $x=5$ gives 65 (too small) and $x=6$ gives 84 (sufficient), the smallest WHOLE-number width meeting the requirement is $x=6$m.

QUESTION 1 5 marks Criterion A

Medium

Solve each linear equation:

a. $x+9=17$

[1]

b. $5x=35$

[1]

c. $\dfrac{x}{4}=6$

[1]

d. $3x-2=13$

[2]

Show complete worked solution

(a)

$$x=8$$

(b)

$$x=7$$

(c)

$$x=24$$

(d)

$$3x=15 \Rightarrow x=5$$

QUESTION 2 4 marks Criterion A

Medium

Solve $4x+7=2x+19$, showing every step.

a. Move the $x$ terms to one side of the equation.

[2]

b. Complete the solution.

[2]

Show complete worked solution

(a)

$4x-2x+7=19 \Rightarrow 2x+7=19$

(b)

$2x=12 \Rightarrow x=6$

QUESTION 3 5 marks Criterion B

Medium

Investigate what happens to the solution of a linear equation when every term is multiplied by the same constant.

a. Solve $x+3=8$.

[1]

b. Now solve $2x+6=16$ (every term in the first equation multiplied by 2).

[2]

c. Compare the two solutions, and explain why multiplying every term of an equation by the same nonzero constant doesn't change the solution.

[2]

Show complete worked solution

(a)

$x=5$

(b)

$2x=10 \Rightarrow x=5$

(c)

Both give $x=5$. Multiplying every term (on both sides) by the same constant is a 'balanced' operation — whatever you do to one side, you do to the other, so the equality (and therefore the solution) is preserved.

QUESTION 4 4 marks Criterion C

Medium

A student solves $2x+5=15$ by writing '$2x=15-5=10$, so $x=10-2=8$' (subtracting instead of dividing in the last step).

a. Identify the error, and give the correct solution.

[2]

b. Verify the correct solution by substituting back into the original equation.

[2]

Show complete worked solution

(a)

The student correctly found $2x=10$, but then subtracted 2 instead of DIVIDING by 2 to isolate $x$. Correct: $2x=10 \Rightarrow x=10\div2=5$.

(b)

$2(5)+5=10+5=15$, confirming $x=5$ is correct (while the student's $x=8$ gives $2(8)+5=21\ne15$).

QUESTION 5 5 marks Criterion D

Medium

A phone repair shop charges a \$25 call-out fee plus \$15 per hour of labour. A customer's total bill was \$85.

a. Write a linear equation for the number of hours $h$ worked, and solve it.

[3]

b. The technician actually recorded working 4 hours and 15 minutes. Explain the discrepancy, and state what the CORRECT bill should have been.

[2]

Show complete worked solution

(a)

$$25+15h=85 \Rightarrow 15h=60 \Rightarrow h=4 \text{ hours}$$

(b)

4 hours 15 minutes $=4.25$ hours, giving a correct bill of $25+15(4.25)=25+63.75=\$88.75$ — the original \$85 bill undercharged the customer by \$3.75, likely due to rounding down the time worked.

QUESTION 6 7 marks Criterion A

Medium

Solve each equation:

a. $4x-9=23$

[2]

b. $\dfrac{x}{3}+7=15$

[2]

c. $\dfrac{x+4}{5}=9$

[3]

Show complete worked solution

(a)

$4x=32 \Rightarrow x=8$

(b)

$\dfrac{x}{3}=8 \Rightarrow x=24$

(c)

$x+4=45 \Rightarrow x=41$

QUESTION 7 6 marks Criterion A

Hard

A cinema's total ticket revenue for one screening was \$1,244. Adult tickets cost \$14 each and child tickets cost \$8 each. There were 100 tickets sold in total.

a. Let $a$ be the number of adult tickets sold. Write the number of child tickets in terms of $a$, then form a linear equation for the revenue.

[3]

b. Solve your equation to find the number of adult and child tickets sold.

[3]

Show complete worked solution

(a)

Child tickets: $100-a$.

Revenue equation: $$14a+8(100-a)=1244.$$

(b)

$$14a+8(100-a)=1244$$ $$14a+800-8a=1244$$ $$6a=444$$ $$a=74.$$

Therefore 74 adult tickets and $100-74=26$ child tickets were sold. Check: $$14(74)+8(26)=1036+208=\boxed{1244}.$$

QUESTION 8 4 marks Criterion B

Medium

Investigate what happens to the SOLUTION of a linear equation $ax+b=c$ when you double BOTH $b$ and $c$ (but keep $a$ and the equation's basic structure the same).

a. Solve $3x+4=19$.

[1]

b. Now solve $3x+8=38$ (both $b=4\to8$ and $c=19\to38$ have been doubled). Compare the new solution to the original.

[3]

Show complete worked solution

(a)

$3x=15 \Rightarrow x=5$

(b)

$3x=30 \Rightarrow x=10$ — exactly DOUBLE the original solution ($5\to10$). Doubling both $b$ and $c$ (while keeping $a$ fixed) doubles the solution $x$, since the equation $3x=30$ is simply the original $3x=15$ scaled by 2 throughout.

QUESTION 9 5 marks Criterion C

Medium

A student solves $5x-8=2x+13$ by writing '$5x-2x=13-8$', getting $3x=5$, then $x=\frac{5}{3}$.

a. Verify whether $x=\frac{5}{3}$ is actually correct by substituting it back into BOTH sides of the original equation.

[3]

b. Identify the student's error, and find the correct solution.

[2]

Show complete worked solution

(a)

LHS: $5(\frac{5}{3})-8=\frac{25}{3}-8=\frac{25-24}{3}=\frac{1}{3}$. RHS: $2(\frac{5}{3})+13=\frac{10}{3}+13=\frac{10+39}{3}=\frac{49}{3}$. Since $\frac{1}{3}\ne\frac{49}{3}$, $x=\frac{5}{3}$ is INCORRECT.

(b)

The student correctly moved the $x$ terms ($5x-2x=3x$) but incorrectly moved the constants — it should be $13+8$ (not $13-8$), since $-8$ moves to the other side by ADDING 8. Correct: $3x=21 \Rightarrow x=7$.

QUESTION 10 7 marks Criterion D

Hard

A rental car company charges a \$45 base fee plus \$0.35 per km driven. A customer's total bill was \$122.50.

a. Write and solve a linear equation for the distance driven.

[3]

b. The company rounds the final bill to the nearest \$0.50. Find the whole-number distances that could produce a displayed bill of \$122.50.

[4]

Show complete worked solution

(a)

Let $k$ be the distance in kilometres.

$$45+0.35k=122.50$$ $$0.35k=77.50$$ $$k=221.428\ldots.$$

If the bill were exact, the distance would be approximately $\boxed{221.43\text{ km}}$.

(b)

A bill rounds to \$122.50 when $$122.25\le45+0.35k<122.75.$$

Thus $$220.714\ldots\le k<222.142\ldots.$$

For a whole-number distance, $$\boxed{k=221\text{ km or }k=222\text{ km}}.$$

QUESTION 1 3 marks Criterion A

Medium

Solve $x-8=15$, explaining the balance operation used at each step.

a. State what operation must be performed to BOTH sides to isolate $x$, and why.

[2]

b. Complete the solution.

[1]

Show complete worked solution

(a)

Add 8 to both sides. This is needed because the equation must stay 'balanced' — whatever is done to one side must be done to the other, to keep both sides equal.

(b)

$$x-8+8=15+8 \Rightarrow x=23$$

QUESTION 2 3 marks Criterion A

Medium

Solve $\dfrac{x}{5}+3=9$, showing the balance operation used at each of the two steps.

a. Perform the first balancing step (removing the $+3$).

[2]

b. Perform the second balancing step to fully isolate $x$.

[1]

Show complete worked solution

(a)

Subtract 3 from both sides: $$\frac{x}{5}=6$$

(b)

Multiply both sides by 5: $$x=30$$

QUESTION 3 5 marks Criterion B

Medium

Investigate what happens to an equation's balance if you perform an operation on only ONE side.

a. Start with the true equation $10=10$. Add 5 to ONLY the left side. Is the resulting statement still true?

[2]

b. Now add 5 to BOTH sides of $10=10$. Is this still true?

[1]

c. Explain why this simple demonstration justifies the rule 'always do the same operation to both sides' when solving equations.

[2]

Show complete worked solution

(a)

$10+5=10 \Rightarrow 15=10$ — this is FALSE. Adding to only one side breaks the equality.

(b)

$10+5=10+5 \Rightarrow 15=15$ — TRUE, the balance is maintained.

(c)

Since an equation represents a true balance between two equal quantities, any operation applied to only one side will generally destroy that balance (as shown), while applying the SAME operation to both sides preserves it — this is why balance must always be maintained on both sides throughout solving.

QUESTION 4 5 marks Criterion C

Medium

A student solving $x+6=14$ writes: 'I'll just move the 6 to the other side and change its sign, getting $x=14-6=8$' without mentioning balance at all.

a. Explain how the student's 'move it and change the sign' shortcut is actually just a quick way of describing the balance operation (subtracting 6 from both sides).

[3]

b. Explain why understanding the BALANCE reasoning (not just the shortcut) becomes important for more complex equations later on.

[2]

Show complete worked solution

(a)

Subtracting 6 from BOTH sides gives $x+6-6=14-6$, which simplifies to $x=8$. The '6' effectively disappears from the left side and reappears as '$-6$' on the right — this is exactly what 'moving it and changing the sign' describes, just using balance-operation language instead of a shortcut phrase.

(b)

The shortcut can become confusing or lead to errors in more complex equations (e.g. those with fractions, brackets, or the variable on both sides), where it's not always obvious what should 'move'. Understanding the underlying balance principle ensures the method can be applied correctly and confidently to ANY equation, not just simple ones.

QUESTION 5 4 marks Criterion D

Medium

A see-saw is perfectly balanced with a 40 kg weight on one side and an unknown combination of weights ($x$ kg plus a fixed 15 kg weight) on the other side, so $x+15=40$.

a. Using the balance analogy directly (removing equal weight from both sides), find $x$.

[2]

b. Explain how this physical see-saw scenario directly mirrors the algebraic rule of 'doing the same operation to both sides' of an equation.

[2]

Show complete worked solution

(a)

Removing 15 kg from both sides keeps the see-saw balanced: $x=40-15=25$ kg.

(b)

Just as removing the same amount of weight from both sides of a physically balanced see-saw keeps it balanced, performing the same mathematical operation on both sides of an equation keeps the equality true — the see-saw is a physical illustration of exactly the same balance principle used in algebra.

QUESTION 6 3 marks Criterion A

Easy

Solve $\dfrac{x}{4}-3=6$, explaining the balance operation at each of the two steps.

a. Perform the first balancing step.

[2]

b. Perform the second balancing step to fully isolate $x$.

[1]

Show complete worked solution

(a)

Add 3 to both sides: $$\frac{x}{4}=9$$

(b)

Multiply both sides by 4: $$x=36$$

QUESTION 7 5 marks Criterion B

Medium

Investigate whether the ORDER of balance operations matters when solving $2x+6=20$ — specifically, does it matter whether you SUBTRACT 6 first, or DIVIDE by 2 first?

a. Solve by subtracting 6 from both sides FIRST, then dividing by 2.

[2]

b. Now solve by dividing EVERY term by 2 first (giving $x+3=10$), then subtracting 3. Confirm you reach the same answer, and explain why BOTH valid orderings of balance operations must always lead to the same correct solution.

[3]

Show complete worked solution

(a)

$2x=14 \Rightarrow x=7$.

(b)

$x+3=10 \Rightarrow x=7$ — matches. Both orderings work because each step, regardless of order, is a VALID balanced operation (applied equally to both sides) — as long as every step maintains the equality, the final isolated value of $x$ must be the same, since balanced operations never change WHICH value of $x$ makes the equation true, only how the equation is temporarily expressed along the way.

QUESTION 8 6 marks Criterion C

Medium

A student solving $5x-2=3x+8$ says: 'I'll just move everything to one side by adding $2-3x-8$ to both sides all at once, since I can do multiple things in one step.'

a. Explain why combining MULTIPLE balance operations into a single step, while mathematically valid IN PRINCIPLE, can be RISKY for someone still learning to solve equations, and suggest what could go wrong.

[3]

b. Demonstrate the SAME solution using clear, SEPARATE balance steps instead, to show a safer approach.

[3]

Show complete worked solution

(a)

While it IS valid to perform multiple balanced operations at once (as long as they're applied identically to both sides), doing so increases the risk of ARITHMETIC ERRORS — combining several sign changes and term movements in one step makes it harder to spot mistakes, and harder for a reader (or the student themselves, checking their own work) to verify each individual step was done correctly. Breaking the process into smaller, clearly labelled steps reduces error risk, especially while still building confidence with the method.

(b)

Subtract $3x$ from both sides: $$2x-2=8.$$

Add 2: $$2x=10.$$

Divide by 2: $$\boxed{x=5}.$$

QUESTION 9 5 marks Criterion C

Medium

A student solving $2(x+5)=3(x-1)$ writes only: '2x+10=3x-3, x=13' with no intermediate steps shown for HOW they moved from the expanded equation to the final answer.

a. Fill in the MISSING intermediate balance steps between '$2x+10=3x-3$' and 'x=13', showing each operation clearly.

[3]

b. Verify $x=13$ is correct by substituting into the ORIGINAL (unexpanded) equation.

[2]

Show complete worked solution

(a)

$2x+10=3x-3$. Subtract $2x$ from both sides: $10=x-3$. Add 3 to both sides: $13=x$, i.e. $x=13$.

(b)

$2(13+5)=2(18)=36$. $3(13-1)=3(12)=36$ — both sides equal 36.

QUESTION 10 6 marks Criterion D

Medium

A weighing scale used in a pharmacy is balanced with a container of unknown TOTAL mass (container + contents) $x$ grams plus a fixed 15g weight on one side, and a 47g reference weight on the other side, so $x+15=47$.

a. Using the balance principle directly (removing equal mass from both sides), find $x$, the total mass of the container and its contents.

[2]

b. The empty container alone (with no contents) is known to weigh 9g. Find the mass of the contents ALONE, and determine whether it meets the prescription requirement of exactly 25g.

[4]

Show complete worked solution

(a)

Removing 15g from both sides: $x=47-15=32$g.

(b)

Contents mass $=x-\text{empty container mass}=32-9=23$g. Since $23\text{g}\ne25\text{g}$ (and is actually LESS), the contents do NOT meet the required 25g — the pharmacist would need to add approximately 2 more grams of contents.

QUESTION 1 4 marks Criterion A

Medium

State the inverse operation needed to undo each of the following, then use it to solve for $x$:

a. $x+12=30$

[1]

b. $x-9=4$

[1]

c. $7x=63$

[1]

d. $\dfrac{x}{6}=3$

[1]

Show complete worked solution

(a)

Inverse of $+12$ is $-12$: $$x=30-12=18$$

(b)

Inverse of $-9$ is $+9$: $$x=4+9=13$$

(c)

Inverse of $\times7$ is $\div7$: $$x=63\div7=9$$

(d)

Inverse of $\div6$ is $\times6$: $$x=3\times6=18$$

QUESTION 2 4 marks Criterion A

Medium

Solve $\sqrt{x}=7$ and $x^2=49$ (for $x>0$), using inverse operations.

a. State the inverse of 'square root' and use it to solve $\sqrt{x}=7$.

[2]

b. State the inverse of 'squaring' and use it to solve $x^2=49$ for $x>0$.

[2]

Show complete worked solution

(a)

Inverse of $\sqrt{\ }$ is squaring: $$x=7^2=49$$

(b)

Inverse of squaring is square-rooting: $$x=\sqrt{49}=7$$

QUESTION 3 5 marks Criterion B

Medium

Investigate why squaring and square-rooting are inverse operations, and why order matters when applying two inverse operations in sequence.

a. Start with $x=5$. Square it, then take the square root of the result. What do you get?

[2]

b. Now start with $x=5$ again. Take the square root FIRST (of 5, giving an irrational number), then square that result. Do you still get back to 5?

[2]

c. Explain why applying an operation and then its EXACT inverse always 'undoes' the original change, regardless of the order.

[1]

Show complete worked solution

(a)

$5^2=25$, then $\sqrt{25}=5$ — back to the original 5.

(b)

$$\left(\sqrt5\right)^2=5.$$ This order returns the starting value because the square root is defined as the nonnegative value and the input $5$ lies in its real domain.

(c)

The order is valid only on the appropriate domain. For $x\ge0$, $\left(\sqrt{x}\right)^2=x$. For every real $x$, however, $$\sqrt{x^2}=|x|,$$ so squaring and then taking the principal square root returns $|x|$, not $x$, when $x<0$.

QUESTION 4 4 marks Criterion C

Medium

A student solving $x-4=10$ uses the WRONG inverse operation, writing '$x=10-4=6$' (subtracting again instead of adding).

a. Explain the student's error, and identify the CORRECT inverse operation needed.

[2]

b. Verify the correct answer, and show why the student's answer ($x=6$) fails when substituted back into the original equation.

[2]

Show complete worked solution

(a)

The equation has $-4$ applied to $x$; to undo (invert) subtraction, you must ADD, not subtract again. Correct: $x=10+4=14$.

(b)

Correct check: $14-4=10$. Student's check: $6-4=2\ne10$ — confirms $x=6$ is wrong.

QUESTION 5 4 marks Criterion D

Medium

A hot air balloon is at height $h$ metres. After descending 35 m, it is at 60 m.

a. Write an equation for this situation ($h-35=60$), and use the correct inverse operation to solve for $h$.

[2]

b. The balloon then needs to rise back to its original height of 95 m from 60 m, at a constant rate of 5 m per minute. Using inverse operations (or otherwise), find how long this will take.

[2]

Show complete worked solution

(a)

Inverse of $-35$ is $+35$: $$h=60+35=95 \text{ m}$$

(b)

Height still needed: $95-60=35$ m. Time $=35\div5=7$ minutes.

QUESTION 6 4 marks Criterion A

Medium

Use inverse operations to solve each equation:

a. $x^3=125$

[2]

b. $\sqrt{x}=8$

[2]

Show complete worked solution

(a)

Inverse of cubing is cube-rooting: $$x=\sqrt[3]{125}=5$$

(b)

Inverse of square-rooting is squaring: $$x=8^2=64$$

QUESTION 7 4 marks Criterion A

Medium

Solve $2\sqrt{x}+3=15$, using inverse operations in the CORRECT order.

a. Isolate the square root term first (undo the $+3$, then undo the $\times2$).

[3]

b. Complete the solution by undoing the square root.

[1]

Show complete worked solution

(a)

$2\sqrt{x}=12 \Rightarrow \sqrt{x}=6$.

(b)

$$x=6^2=36$$

QUESTION 8 5 marks Criterion B

Hard

Investigate whether 'squaring then square-rooting' and 'square-rooting then squaring' ALWAYS return you to the original number, testing with a NEGATIVE starting value, $x=-4$.

a. Square $-4$, then take the square root of the result. Do you return to $-4$?

[2]

b. Explain WHY this happens, referring to the fact that the square root symbol $\sqrt{\ }$ conventionally refers to the POSITIVE root only, even when the original number being squared was negative.

[3]

Show complete worked solution

(a)

$(-4)^2=16$. $\sqrt{16}=4$ — this does NOT return to $-4$; it gives $+4$ instead.

(b)

The square root operation, by mathematical CONVENTION, always returns the non-negative (principal) root — so $\sqrt{16}=4$ specifically, never $-4$, even though BOTH $4^2$ and $(-4)^2$ equal 16. This means 'square then square-root' does NOT perfectly reverse for negative starting numbers — the true inverse relationship only holds cleanly for non-negative starting values, which is an important subtlety when using square-rooting as an 'inverse' of squaring.

QUESTION 9 3 marks Criterion C

Medium

A student solving $x^2=49$ using inverse operations writes only '$x=7$', missing a second valid solution.

a. Explain the student's error, and state BOTH correct solutions.

[3]

Show complete worked solution

(a)

Squaring is NOT a one-to-one operation — both $7^2=49$ AND $(-7)^2=49$, so 'undoing' a square must consider BOTH the positive and negative square roots. Correct solutions: $x=7$ or $x=-7$ (i.e. $x=\pm7$).

QUESTION 10 6 marks Criterion D

Hard

The relationship between a cube's volume $V$ and its side length $s$ is $V=s^3$. A storage cube has volume $V=343$ cm³.

a. Use the inverse operation (cube root) to find the side length.

[2]

b. A larger storage cube has EXACTLY DOUBLE the volume of this one (686 cm³). Find its side length (to 2 decimal places), and determine whether DOUBLING the volume also DOUBLES the side length (compare the new side length to double the original 7cm).

[4]

Show complete worked solution

(a)

$$s=\sqrt[3]{343}=7\text{ cm}$$

(b)

$s=\sqrt[3]{686}\approx8.82$cm. This is NOT double the original side length (which would be $14$cm) — doubling the VOLUME of a cube does NOT double its side length, since volume scales with the CUBE of the side length, not linearly; a much smaller increase in side length (from 7cm to about 8.82cm) is enough to double the volume.

QUESTION 11 6 marks Criterion D

Hard

A physics formula relates a pendulum's period $T$ (seconds) to its length $L$ (metres): $T=2\pi\sqrt{\dfrac{L}{9.8}}$. A pendulum has a measured period of $T=2$ seconds.

a. Rearrange the formula (using inverse operations step by step) to solve for $L$ in terms of $T$.

[4]

b. Substitute $T=2$ to find the pendulum's length, correct to 2 decimal places.

[2]

Show complete worked solution

(a)

$\dfrac{T}{2\pi}=\sqrt{\dfrac{L}{9.8}}$ (divide by $2\pi$). $\left(\dfrac{T}{2\pi}\right)^2=\dfrac{L}{9.8}$ (square both sides). $$L=9.8\left(\frac{T}{2\pi}\right)^2$$

(b)

$$L=9.8\left(\frac{2}{2\pi}\right)^2=9.8\times\left(\frac{1}{\pi}\right)^2\approx0.99\text{ m}$$

QUESTION 1 3 marks Criterion A

Medium

The flowchart shows an unknown $x$ having two operations applied, resulting in 23.

a. Write the equation shown by the flowchart.

[1]

b. Solve the flowchart in REVERSE (starting from 23, undoing each operation in opposite order) to find $x$.

[2]

Show complete worked solution

(a)

$$4x+3=23$$

(b)

Reverse: $23-3=20$, then $20\div4=5$. So $x=5$.

QUESTION 2 4 marks Criterion A

Medium

The flowchart shows two operations applied to $x$, resulting in 18.

a. Write the equation shown, and solve it forward (using algebra, not the flowchart) to check your understanding.

[2]

b. Now solve using the REVERSE flowchart method (undo $\times2$ first, then undo $-6$), and confirm you get the same answer.

[2]

Show complete worked solution

(a)

Equation: $2(x-6)=18$. Solving: $x-6=9 \Rightarrow x=15$.

(b)

Reverse: $18\div2=9$, then $9+6=15$. Matches, $x=15$.

QUESTION 3 5 marks Criterion B

Medium

Investigate why the REVERSE flowchart method requires undoing operations in the OPPOSITE order to how they were applied.

a. Consider $x\to(\times3)\to(+7)\to25$. If you tried to undo in the SAME order (subtract 7 first is correct, but what if you divided by 3 first instead)? Show what happens if you incorrectly divide by 3 before subtracting 7.

[3]

b. Explain, like unpacking nested boxes, why operations must be undone in reverse (last-applied, first-undone) order.

[2]

Show complete worked solution

(a)

Incorrect order: $25\div3\approx8.33$, then $8.33-7\approx1.33$ — this does NOT match the actual solution. Correct order: $25-7=18$, then $18\div3=6$, giving $x=6$ (verify: $3(6)+7=25$).

(b)

The LAST operation applied when building forward is the 'outermost' one, so it must be the FIRST one removed when working backward — like taking off your shoes before your socks, even though socks went on first. Undoing in the wrong order doesn't correctly peel back each layer.

QUESTION 4 4 marks Criterion C

Medium

A student builds a flowchart for $\dfrac{x+4}{3}=7$ as $x\to(+4)\to(\div3)\to7$, but then reverses it incorrectly as 'undo $+4$ first, then undo $\div3$' (same order, not reversed).

a. Explain the student's error, and perform the CORRECT reverse process.

[3]

b. Verify $x=17$ by substituting into the original equation.

[1]

Show complete worked solution

(a)

The student forgot to REVERSE the order — since $\div3$ was applied last (forward), it must be undone FIRST (backward). Correct: $7\times3=21$ (undo $\div3$), then $21-4=17$ (undo $+4$). So $x=17$.

(b)

$\frac{17+4}{3}=\frac{21}{3}=7$

QUESTION 5 4 marks Criterion D

Medium

A taxi fare is modelled by a flowchart: take the time in hours $t$, multiply by 12, then add a base fee of 45, giving the total fare in dollars.

a. If a fare came to \$165, use the reverse flowchart method to find the trip's duration $t$.

[2]

b. A 10-hour taxi trip seems unusually long. Suggest what the flowchart's numbers might actually represent instead of 'hours' and 'dollars per hour', to make the scenario more realistic (e.g. a different time unit or fare structure).

[2]

Show complete worked solution

(a)

Reverse: $165-45=120$, then $120\div12=10$. So $t=10$ hours.

(b)

A more realistic interpretation might be that $t$ represents distance in KILOMETRES (not hours), with \$12 per km plus a \$45 base/booking fee — a 10 km trip costing \$165 total is far more typical for a taxi fare than a 10-hour trip.

QUESTION 6 3 marks Criterion A

Medium

The flowchart shows two operations applied to $x$, giving 27.

a. Write the equation shown, and solve it using the reverse flowchart method.

[3]

Show complete worked solution

(a)

Equation: $5x-8=27$. Reverse: $27+8=35$, then $35\div5=7$. So $x=7$.

QUESTION 7 7 marks Criterion B

Hard

Investigate how a THREE-operation flowchart is reversed, using $x \to (+6) \to (\times3) \to (-4) \to 41$.

a. Write the equation this flowchart represents.

[2]

b. Reverse the flowchart (undoing all THREE operations in the correct opposite order), showing each step.

[3]

c. Verify $x=9$ by substituting into the original equation.

[2]

Show complete worked solution

(a)

$$3(x+6)-4=41$$

(b)

Reverse: undo $-4$ first: $41+4=45$. Undo $\times3$: $45\div3=15$. Undo $+6$: $15-6=9$. So $x=9$.

(c)

$3(9+6)-4=3(15)-4=45-4=41$.

QUESTION 8 5 marks Criterion B

Medium

Investigate why a flowchart involving DIVISION must be reversed using MULTIPLICATION, using the flowchart shown.

a. Write the equation, and identify the TWO operations shown (in order).

[2]

b. Reverse the flowchart correctly, explaining specifically why the LAST operation (÷4, working backward from division) must be undone using MULTIPLICATION, not further division.

[3]

Show complete worked solution

(a)

Equation: $\frac{x}{4}+9=20$. Operations: divide by 4, then add 9.

(b)

Reverse: undo $+9$: $20-9=11$. Undo $\div4$: $11\times4=44$. So $x=44$. Division and multiplication are INVERSE operations — since the forward flowchart used $\div4$, reversing it requires the OPPOSITE (inverse) operation, $\times4$, to correctly 'undo' the division and recover the original value.

QUESTION 9 4 marks Criterion C

Medium

A student building a flowchart for $\dfrac{3x-5}{2}=8$ draws it as $x\to(\times3)\to(-5)\to(\div2)\to8$, then reverses it by undoing $\times3$ FIRST (in the same order as building it), instead of reversing the order.

a. Explain the student's error, and perform the CORRECT reverse process, step by step.

[4]

Show complete worked solution

(a)

The student forgot that reversing a flowchart requires undoing the operations in OPPOSITE order to how they were built (last-applied, first-undone) — undoing $\times3$ first (the FIRST operation applied) is incorrect. Correct reverse order: undo $\div2$ first: $8\times2=16$. Undo $-5$: $16+5=21$. Undo $\times3$: $21\div3=7$. So $x=7$.

QUESTION 10 7 marks Criterion D

Hard

A cleaning service charges according to a flowchart: take the number of rooms cleaned $t$, multiply by 18, then add a fixed \$65 travel fee.

a. If a bill came to \$191, use the reverse flowchart method to find the number of rooms cleaned.

[3]

b. The cleaning service is considering changing their pricing to a flowchart with the operations REVERSED in order (add \$65 first, THEN multiply by 18), keeping the same numbers. Determine the NEW bill for 7 rooms under this alternative structure, and explain why changing the ORDER of operations in the flowchart (not just the numbers) produces a genuinely different pricing formula.

[4]

Show complete worked solution

(a)

Reverse: $191-65=126$, then $126\div18=7$. So $t=7$ rooms.

(b)

New formula: $18(t+65)$. For $t=7$: $18(72)=1296$ — a MASSIVELY higher bill (\$1,296 vs \$191). Changing the order fundamentally changes the formula's structure: the original $(18t+65)$ applies the \$65 fee ONCE regardless of room count, while the reversed version $18(t+65)$ effectively multiplies the \$65 fee by 18 too, making it scale disproportionately — this shows that in a flowchart, the ORDER of operations is just as important as the individual operations themselves.

QUESTION 1 6 marks Criterion A

Medium

Solve each equation:

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

[2]

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

[2]

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

[2]

Show complete worked solution

(a)

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

(b)

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

(c)

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

QUESTION 2 4 marks Criterion A

Medium

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

a. Expand both sides.

[2]

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

[2]

Show complete worked solution

(a)

$10x-5=3x+24$

(b)

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

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.

[2]

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

[2]

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.

[2]

Show complete worked solution

(a)

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

(b)

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

(c)

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.

[3]

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

[2]

Show complete worked solution

(a)

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)

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

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 $P=2(\text{length}+\text{width})$.

[3]

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

[3]

Show complete worked solution

(a)

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

(b)

$$2(3w+3)=34$$ $$6w+6=34$$ $$6w=28$$ $$w=\frac{14}{3}\text{ m}\approx4.67\text{ m}.$$ Retain the exact width when finding the length: $$L=2\left(\frac{14}{3}\right)+3=\frac{37}{3}\text{ m}\approx\boxed{12.33\text{ m}}.$$ Thus the width is $\boxed{\frac{14}{3}\text{ m}}$ and the length is $\boxed{\frac{37}{3}\text{ m}}$.

QUESTION 6 3 marks Criterion A

Medium

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

a. Expand the left side.

[1]

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

[2]

Show complete worked solution

(a)

$8x-12=5x+9$

(b)

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

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).

[3]

b. Solve for $x$.

[2]

Show complete worked solution

(a)

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

(b)

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

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).

[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.

[4]

Show complete worked solution

(a)

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

(b)

$2x-6=-10 \Rightarrow 2x=-4 \Rightarrow x=-2$ — 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.

[3]

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

[2]

Show complete worked solution

(a)

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)

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

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.

[3]

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

[2]

Show complete worked solution

(a)

Expand: $$3x-6=2x+7.$$

Subtract $2x$: $$x-6=7.$$

Add 6: $$\boxed{x=13}.$$

(b)

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

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 $P=2(\text{length}+\text{width})$.

[4]

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

[3]

Show complete worked solution

(a)

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

(b)

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

QUESTION 1 6 marks Criterion A

Medium

Solve each equation, where $x$ appears more than once:

a. $5x+3=2x+18$

[2]

b. $7x-4=3x+20$

[2]

c. $9-2x=x+3$

[2]

Show complete worked solution

(a)

$3x=15 \Rightarrow x=5$

(b)

$4x=24 \Rightarrow x=6$

(c)

$6=3x \Rightarrow x=2$

QUESTION 2 3 marks Criterion A

Medium

Solve $3(x+2)=x+14$, where $x$ appears on both sides (one inside brackets).

a. Expand the left side first.

[1]

b. Collect $x$ terms and solve.

[2]

Show complete worked solution

(a)

$3x+6=x+14$

(b)

$3x-x=14-6 \Rightarrow 2x=8 \Rightarrow x=4$

QUESTION 3 5 marks Criterion B

Medium

Investigate whether it matters WHICH side you move the $x$ terms to when solving an equation with $x$ on both sides.

a. Solve $6x+2=2x+18$ by moving the $x$ terms to the LEFT side (subtracting $2x$ from both sides).

[2]

b. Now solve the SAME equation by instead moving the $x$ terms to the RIGHT side (subtracting $6x$ from both sides).

[2]

c. Compare your two answers, and explain why the choice of which side to collect $x$ terms on doesn't affect the final solution.

[1]

Show complete worked solution

(a)

$4x+2=18 \Rightarrow 4x=16 \Rightarrow x=4$

(b)

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

(c)

Both give $x=4$ — the choice is just a matter of preference/convenience (usually choosing the side that avoids negative coefficients), since either path represents valid, balanced operations leading to the same true solution.

QUESTION 4 5 marks Criterion C

Medium

A student solving $8x-3=3x+22$ moves terms around and writes '$8x-3x=22-3$', getting $5x=19$.

a. Identify the student's error (a sign mistake when moving terms), and give the correct working.

[3]

b. Verify $x=5$ by substituting into the original equation.

[2]

Show complete worked solution

(a)

When moving $-3$ from the left side to the right, it should become $+3$ (not stay as $-3$ to subtract), and when moving $3x$ from the right to the left, it becomes $-3x$. Correct: $8x-3x=22+3 \Rightarrow 5x=25 \Rightarrow x=5$.

(b)

$8(5)-3=37$ and $3(5)+22=37$ — both sides equal 37.

QUESTION 5 5 marks Criterion D

Medium

Two mobile phone plans are being compared. Plan A costs \$20 flat plus \$0.10 per minute. Plan B costs \$35 flat plus \$0.05 per minute.

a. Write an equation setting the two plans' total costs equal (using $m$ for minutes), and solve to find the number of minutes at which both plans cost the same.

[3]

b. For a customer who typically uses 500 minutes per month, which plan is cheaper? Justify your answer using your equation's result.

[2]

Show complete worked solution

(a)

$20+0.10m=35+0.05m \Rightarrow 0.05m=15 \Rightarrow m=300$ minutes.

(b)

Since the break-even point is 300 minutes, and 500 minutes is MORE than this, the plan with the lower per-minute rate (Plan B) becomes cheaper beyond 300 minutes — so Plan B is the better choice for a 500-minute user.

QUESTION 6 3 marks Criterion A

Medium

Solve $9x-4=5x+20$, where $x$ appears on both sides.

a. Collect the $x$ terms and constants, then solve.

[3]

Show complete worked solution

(a)

$9x-5x=20+4 \Rightarrow 4x=24 \Rightarrow x=6$

QUESTION 7 7 marks Criterion B

Hard

Investigate what happens when solving an equation with $x$ on both sides leads to the $x$ terms CANCELLING OUT completely, using $6x+12=2x+12$ vs $4x+7=4x+9$.

a. Solve $6x+12=2x+12$ using standard balance operations, and interpret the result.

[3]

b. Now attempt to solve $4x+7=4x+9$. What happens when you try to collect the $x$ terms, and what does this tell you about how many solutions this equation has?

[4]

Show complete worked solution

(a)

$6x-2x=12-12 \Rightarrow 4x=0 \Rightarrow x=0$. This has exactly ONE valid solution ($x=0$).

(b)

Subtracting $4x$ from both sides: $7=9$ — this is a FALSE statement (7 does not equal 9), regardless of what $x$ is. This means the equation $4x+7=4x+9$ has NO SOLUTION at all — there is no value of $x$ that could ever make both sides equal, since the $x$ terms cancel identically but leave behind two DIFFERENT constants.

QUESTION 8 6 marks Criterion B

Hard

Investigate the equation $3(x+2)=3x+6$, where the $x$ terms ALSO cancel — but compare the result to the 'no solution' case above.

a. Expand the left side, then attempt to collect $x$ terms as usual.

[3]

b. Explain what this result means: how many solutions does $3(x+2)=3x+6$ actually have, and why is this a fundamentally DIFFERENT outcome from the 'no solution' case investigated earlier, even though both involved the $x$ terms cancelling?

[3]

Show complete worked solution

(a)

$3x+6=3x+6$. Subtracting $3x$ from both sides: $6=6$ — this is a TRUE statement, for any value of $x$.

(b)

This equation has INFINITELY MANY solutions — EVERY value of $x$ satisfies it, since $3(x+2)$ is simply the EXPANDED (equivalent) form of $3x+6$; they're not two different expressions that happen to be equal, they're the SAME expression written two ways. This is fundamentally different from the 'no solution' case: there, the $x$ terms cancelled leaving a FALSE constant statement (impossible for any $x$); here, they cancel leaving a TRUE constant statement (true for every $x$) — the key difference is whether the leftover constant equation is true or false.

QUESTION 9 3 marks Criterion C

Medium

A student solving $7x-3=7x+5$ doesn't notice the $x$ terms are identical, and after subtracting $7x$ from both sides gets '$-3=5$', then writes 'x = -3/5 or something, I'm confused' without recognizing what this actually means.

a. Explain to the confused student what '$-3=5$' (a statement with NO variable left at all) actually indicates about the ORIGINAL equation's solution set, correcting their attempt to still solve for a numerical $x$.

[3]

Show complete worked solution

(a)

When ALL the variable terms cancel and you're left with a FALSE numerical statement (like $-3=5$, which is never true), this means the original equation has NO SOLUTION whatsoever — there is no value of $x$, no matter how it's calculated, that could make the original equation true. The student's instinct to 'solve for $x$' from this point is misguided, since there's no $x$ left in the statement to solve for; the correct response is to conclude 'no solution', not attempt further algebra.

QUESTION 10 3 marks Criterion C

Medium

A classmate claims: 'if solving an equation ever gives a statement with no $x$ left in it, that ALWAYS means there's no solution.'

a. Explain why this claim is only PARTIALLY correct, using the earlier finding that $3(x+2)=3x+6$ (which also loses its $x$ terms during solving) actually has INFINITELY MANY solutions, not zero.

[3]

Show complete worked solution

(a)

The classmate's claim is incomplete — losing the $x$ terms during solving can lead to TWO very different outcomes: if the remaining constant statement is FALSE (like $-3=5$), there's no solution; but if the remaining constant statement is TRUE (like $6=6$, from $3(x+2)=3x+6$), the equation actually has INFINITELY MANY solutions (every value of $x$ works). The key isn't just 'the $x$ disappeared' — it's whether the LEFTOVER constant statement is true or false that determines the outcome.

QUESTION 11 6 marks Criterion D

Hard

Two ride-share drivers compare pay structures. Driver A earns a flat \$80 per shift plus \$1.50 per km driven. Driver B earns a flat \$50 per shift plus \$2.10 per km driven.

a. Write an equation setting the two drivers' total pay equal (using $k$ for km driven), and solve to find the break-even distance.

[3]

b. For a shift involving 80km of driving, determine which driver earns MORE, and by how much, justifying your answer using the break-even point found in part (a).

[3]

Show complete worked solution

(a)

$80+1.50k=50+2.10k \Rightarrow 30=0.60k \Rightarrow k=50$km.

(b)

Since 80km EXCEEDS the 50km break-even point, and Driver B has the HIGHER per-km rate (\$2.10 vs \$1.50), Driver B earns more beyond the break-even distance. Driver A: $80+1.50(80)=80+120=\$200$. Driver B: $50+2.10(80)=50+168=\$218$. Driver B earns \$18 more.

QUESTION 12 7 marks Criterion D

Hard

A gym offers two membership plans. Plan X: \$30/month plus \$8 per visit. Plan Y: \$60/month plus \$4 per visit.

a. Find the number of visits per month at which BOTH plans cost the same.

[3]

b. Since visits must be a WHOLE number, explain what this non-whole break-even point (7.5) means practically for someone deciding between the two plans — specifically, which plan is cheaper for someone visiting 7 times per month, and which is cheaper for 8 times per month?

[4]

Show complete worked solution

(a)

$30+8v=60+4v \Rightarrow 4v=30 \Rightarrow v=7.5$ visits.

(b)

At 7 visits: Plan X $=30+56=\$86$; Plan Y $=60+28=\$88$ — Plan X is cheaper. At 8 visits: Plan X $=30+64=\$94$; Plan Y $=60+32=\$92$ — Plan Y is cheaper. The non-whole break-even (7.5) means the CHEAPER plan actually SWITCHES between 7 and 8 visits — Plan X is better for LIGHTER gym use (7 or fewer visits per month), while Plan Y becomes better once visits reach 8 or more per month.

QUESTION 13 7 marks Criterion D

Hard

Two internet providers offer plans. Provider A: \$600 setup fee plus \$9 per month. Provider B: \$900 setup fee plus \$15 per month, but includes a free premium router (value irrelevant to the cost comparison here).

a. Write an equation setting the total costs equal after $m$ months, and solve for $m$.

[3]

b. Explain what a NEGATIVE 'break-even time' means in this context, and what it tells a customer about which plan is cheaper for EVERY realistic (positive) number of months.

[4]

Show complete worked solution

(a)

$$600+9m=900+15m$$ Solving: $600-900=15m-9m \Rightarrow -300=6m \Rightarrow m=-50$ — a NEGATIVE result.

(b)

A negative break-even point means the two costs NEVER become equal for any realistic (positive) number of months — since Provider A has BOTH a lower setup fee AND a lower monthly rate, Provider A is cheaper for every possible positive value of $m$, with the cost gap only growing larger over time. The negative solution is a mathematical signal that the 'equal cost' scenario would only occur in a nonsensical negative-time scenario, confirming Provider A dominates entirely in this case.

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