EQUATIONS

QUESTION 1 [5 marks] — Criterion A Medium

Solve each linear equation:

a. $x+9=17$

[1]

Show Solution

$$x=8$$

b. $5x=35$

[1]

Show Solution

$$x=7$$

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

[1]

Show Solution

$$x=24$$

d. $3x-2=13$

[2]

Show Solution

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

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]

Show Solution

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

b. Complete the solution.

[2]

Show Solution

$2x=12 \Rightarrow x=6$

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]

Show Solution

$x=5$

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

[2]

Show Solution

$2x=10 \Rightarrow x=5$

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 Solution

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]

Show Solution

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

[2]

Show Solution

$2(5)+5=10+5=15$ ?, 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]

Show Solution

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

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 Solution

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]

Show Solution

$4x=32 \Rightarrow x=8$

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

[2]

Show Solution

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

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

[3]

Show Solution

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

QUESTION 7 [6 marks] — Criterion A Hard

A cinema's total ticket revenue for one screening was \$1,240, from a mix of adult tickets (\$14 each) and child tickets (\$8 each). There were 100 tickets sold in total.

a. Let $a$ be the number of adult tickets sold. Write an expression for the number of child tickets sold (in terms of $a$), then write a linear equation for the total revenue.

[3]

Show Solution

Child tickets: $100-a$. Revenue equation: $$14a+8(100-a)=1240$$

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

[3]

Show Solution

$14a+800-8a=1240 \Rightarrow 6a=440 \Rightarrow a\approx73.3$ — since ticket counts must be WHOLE numbers, this suggests the given revenue figure (\$1,240) may not correspond to an exact whole-number ticket split; testing $a=73$: revenue $=14(73)+8(27)=1022+216=1238$ (close but not exact), while $a=74$: $14(74)+8(26)=1036+208=1244$ — neither gives EXACTLY \$1,240, indicating the scenario's numbers don't perfectly align with whole-ticket sales.

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]

Show Solution

$3x=15 \Rightarrow x=5$

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 Solution

$3x=30 \Rightarrow x=10$ — 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]

Show Solution

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

[2]

Show Solution

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 to find the number of km driven.

[3]

Show Solution

$45+0.35k=122.50 \Rightarrow 0.35k=77.50 \Rightarrow k=221.43$km (approximately).

b. The company rounds the FINAL bill to the nearest \$0.50 for simplicity (not the km figure). Given this, the true bill might not be EXACTLY \$122.50. Find the RANGE of possible km values (to the nearest km) that could have resulted in a bill rounding to \$122.50, assuming rounding to the nearest \$0.50.

[4]

Show Solution

Bill rounds to \$122.50 if the TRUE bill is between \$122.25 and \$122.75. Lower km: $45+0.35k=122.25 \Rightarrow k\approx220.7$km. Upper km: $45+0.35k=122.75 \Rightarrow k\approx222.1$km. So the true distance was likely between approximately 221 and 222 km.

Linear equations