ALGEBRAIC OPERATIONS

QUESTION 1 [3 marks] — Criterion A Medium

Simplify each algebraic fraction fully:

a. $\dfrac{6x}{9}$

[1]

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$$\frac{2x}{3}$$

b. $\dfrac{8ab}{12a}$

[1]

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$$\frac{2b}{3}$$

c. $\dfrac{15x^2}{5x}$

[1]

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$$3x$$

QUESTION 2 [3 marks] — Criterion A Medium

Simplify $\dfrac{12x^2y}{18xy^2}$, showing full working.

a. Find the HCF of the numerical coefficients (12 and 18), and simplify the numerical part.

[1]

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HCF$(12,18)=6$, so $\frac{12}{18}=\frac{2}{3}$

b. Simplify the $x$ terms and the $y$ terms separately (using index laws), then combine everything into the final simplified fraction.

[2]

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$x$ terms: $\frac{x^2}{x}=x$. $y$ terms: $\frac{y}{y^2}=\frac{1}{y}$. Combining: $$\frac{12x^2y}{18xy^2} = \frac{2x}{3y}$$

QUESTION 3 [5 marks] — Criterion B Medium

Investigate what happens to an algebraic fraction when the SAME factor is present in both the numerator and denominator.

a. Simplify $\dfrac{x(x+3)}{2(x+3)}$, explaining which factor cancels.

[2]

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The factor $(x+3)$ appears in both the numerator and denominator, so it cancels (as long as $x\ne-3$): $$\frac{x(x+3)}{2(x+3)} = \frac{x}{2}$$

b. Now try to simplify $\dfrac{x+3}{2+3}$. Does anything cancel here? Explain the difference between this and part (a).

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Nothing cancels here — $\frac{x+3}{2+3}=\frac{x+3}{5}$, which cannot be simplified further. The key difference is that in part (a), $(x+3)$ was a FACTOR (multiplied) in both parts, while here, 3 is just a TERM being added — you can only cancel common FACTORS, not individual terms within a sum.

QUESTION 4 [4 marks] — Criterion C Medium

A student simplifies $\dfrac{x+4}{x}$ by 'cancelling the $x$', writing the answer as $4$.

a. Explain why this cancellation is incorrect.

[2]

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You can only cancel a factor that multiplies the ENTIRE numerator and the entire denominator. Here, $x$ is only part of a SUM in the numerator ($x+4$), not a factor of the whole numerator, so it cannot be cancelled with the $x$ in the denominator.

b. Verify the error by evaluating both $\dfrac{x+4}{x}$ and $4$ at $x=2$.

[2]

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$\dfrac{x+4}{x}=\dfrac{6}{2}=3$. The claimed 'simplified' answer of 4 does not match (3 vs 4), confirming the cancellation was invalid.

QUESTION 5 [4 marks] — Criterion D Medium

A recipe requires $\dfrac{3x}{4}$ cups of flour per batch, where $x$ represents the number of people being served.

a. Simplify the expression for the amount of flour needed per person (i.e. divide by $x$), and interpret what this simplified fraction means.

[2]

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$$\dfrac{3x}{4} \div x = \dfrac{3x}{4x} = \dfrac{3}{4}$$ This means each person requires $\frac{3}{4}$ cup of flour, regardless of how many people are being served — a constant amount per person.

b. A caterer is serving 24 people. Use the ORIGINAL expression (not the per-person one) to find the total flour needed.

[2]

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$$\dfrac{3(24)}{4} = \dfrac{72}{4}=18 \text{ cups}$$

QUESTION 6 [7 marks] — Criterion A Hard

Simplify each algebraic fraction, factoring first where needed:

a. $\dfrac{x^2+5x}{x}$

[2]

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$$\frac{x(x+5)}{x}=x+5$$

b. $\dfrac{x^2-9}{x-3}$

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$$\frac{(x-3)(x+3)}{x-3}=x+3$$

c. $\dfrac{2x^2+6x}{4x}$

[2]

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$$\frac{2x(x+3)}{4x}=\frac{x+3}{2}$$

QUESTION 7 [5 marks] — Criterion B Hard

Investigate what happens when you simplify $\dfrac{x^2-a^2}{x-a}$ for different values of $a$.

a. Simplify $\dfrac{x^2-9}{x-3}$ (i.e. $a=3$) by factoring the numerator as a difference of squares.

[2]

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$$\frac{(x-3)(x+3)}{x-3}=x+3$$

b. Repeat for $\dfrac{x^2-16}{x-4}$ ($a=4$) and $\dfrac{x^2-49}{x-7}$ ($a=7$). State the general pattern for $\dfrac{x^2-a^2}{x-a}$.

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$\dfrac{x^2-16}{x-4}=x+4$. $\dfrac{x^2-49}{x-7}=x+7$. General pattern: $$\frac{x^2-a^2}{x-a}=x+a$$

QUESTION 8 [3 marks] — Criterion C Hard

A student simplifies $\dfrac{x+7}{7}$ by 'cancelling the 7', writing the answer as $x$.

a. Explain why this cancellation is invalid, and verify using $x=14$ that the student's answer is wrong.

[3]

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The 7 in the numerator is part of a SUM ($x+7$), not a FACTOR of the entire numerator — you can only cancel factors that multiply the WHOLE numerator and WHOLE denominator, not terms being added. Verify: $\dfrac{14+7}{7}=\dfrac{21}{7}=3$, but the student's 'simplified' answer would give $x=14$ — clearly different (3 vs 14), confirming the error.

QUESTION 9 [5 marks] — Criterion C Hard

A student simplifies $\dfrac{2x+4}{2}$ as $x+4$ (only dividing the FIRST term by 2, forgetting the second).

a. Explain the error, and give the correct simplification, making clear that BOTH terms in the numerator must be divided.

[3]

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$\dfrac{2x+4}{2} = \dfrac{2x}{2}+\dfrac{4}{2} = x+2$ — EVERY term in the numerator must be divided by the denominator, not just the first one. The student's answer ($x+4$) incorrectly left the second term unchanged.

b. Verify the correct answer using $x=5$.

[2]

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$\dfrac{2(5)+4}{2}=\dfrac{14}{2}=7$. Correct formula: $5+2=7$ ? (student's wrong version would give $5+4=9\ne7$).

QUESTION 10 [6 marks] — Criterion D Hard

A water tank drains according to the formula: time to drain $=\dfrac{V}{r}$ minutes, where $V$ is volume in litres and $r$ is drain rate in litres/minute. A tank has volume $V=3x^2+9x$ litres and drains at rate $r=3x$ litres/minute.

a. Simplify the expression for draining time, factoring the numerator first.

[3]

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$$\frac{3x^2+9x}{3x} = \frac{3x(x+3)}{3x} = x+3 \text{ minutes}$$

b. If $x=8$, find the actual draining time, and verify by calculating $V$ and $r$ separately first, then dividing directly.

[3]

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Using simplified formula: $8+3=11$ minutes. Verify: $V=3(64)+9(8)=192+72=264$L. $r=3(8)=24$L/min. Direct: $264\div24=11$ minutes ?.

QUESTION 11 [6 marks] — Criterion D Hard

A charity divides a fundraising total of $\$(5x^2+15x)$ equally among $5x$ volunteer teams for their operating budgets.

a. Simplify the expression for the amount each team receives.

[2]

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$$\frac{5x^2+15x}{5x} = \frac{5x(x+3)}{5x} = x+3$$

b. If there are 40 volunteer teams (i.e. $5x=40$), find the amount each team receives, and the TOTAL fundraising amount that was originally divided.

[4]

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$5x=40 \Rightarrow x=8$. Amount per team: $x+3=11$ (i.e. \$11 thousand, or whatever unit $x$ represents — treating as \$(x+3)\times1000 if $x$ is in thousands, but taking the numbers at face value: \$11 per team). Total: $5x^2+15x=5(64)+15(8)=320+120=\$440$ (matching $40\text{ teams}\times\$11=\$440$ ?).

Algebraic fractions