Question. Simplify $\dfrac{2m^2 - m - 1}{m + mn - 1 - n}$.

Solution.

Numerator (quadratic, cross method): $2m^2 - m - 1 = (2m + 1)(m - 1)$.

Denominator (4 terms, group 2-by-2):

$$m + mn - 1 - n = m(1 + n) - (1 + n) = (1 + n)(m - 1)$$

Combine:

$$\frac{(2m + 1)(m - 1)}{(1 + n)(m - 1)} = \frac{2m + 1}{1 + n}$$

Answer: $\dfrac{2m + 1}{1 + n}$.

Lesson: numerator and denominator may need different factorisation techniques. Always look at each independently.

Question. Simplify $\dfrac{a^2+1}{a^2-1} - \dfrac{2}{3(1-a)}$.

Solution.

Step 1 — Factorise denominators.

• $a^2 - 1 = (a-1)(a+1)$.
• $3(1 - a) = -3(a - 1)$, so $\dfrac{2}{3(1-a)} = -\dfrac{2}{3(a-1)}$.

Step 2 — Rewrite:

$$= \frac{a^2 + 1}{(a-1)(a+1)} \;+\; \frac{2}{3(a-1)}$$

(The minus became plus after flipping the sign.)

Step 3 — LCM $= 3(a-1)(a+1)$:

$$= \frac{3(a^2 + 1) + 2(a + 1)}{3(a-1)(a+1)}$$

Step 4 — Expand and simplify the numerator:

$$= \frac{3a^2 + 3 + 2a + 2}{3(a-1)(a+1)} = \frac{3a^2 + 2a + 5}{3(a-1)(a+1)}$$

Lesson: when a $(1 - a)$ vs $(a - 1)$ mismatch shows up, flip the sign before finding the LCM.

Question. Simplify $\dfrac{2x - 3}{3x + 1} \times \dfrac{9x^2 - 1}{5x - 2x^2 - 3}$.

Solution.

Factorise:

• $9x^2 - 1 = (3x - 1)(3x + 1)$ (DOS).
• $5x - 2x^2 - 3 = -(2x^2 - 5x + 3) = -(2x - 3)(x - 1) = (x - 1)(3 - 2x) \cdot ?$ — let's redo cleanly: $-2x^2 + 5x - 3 = -(2x - 3)(x - 1)$.

So:

$$= \frac{2x - 3}{3x + 1} \times \frac{(3x - 1)(3x + 1)}{-(2x - 3)(x - 1)}$$

Cancel $(2x - 3)$ and $(3x + 1)$:

$$= \frac{1}{1} \times \frac{(3x - 1)}{-(x - 1)} = -\frac{3x - 1}{x - 1}$$

Answer: $-\dfrac{3x - 1}{x - 1}$.

Lesson: factorise everything first; the minus sign from the sign-flip travels to the final answer.

Question. Solve $\dfrac{x}{x - 1} + 2 = \dfrac{4}{2 - x}$.

Solution.

Rewrite $2 - x = -(x - 2)$:

$$\frac{x}{x - 1} + 2 = -\frac{4}{x - 2}$$

LCM of denominators: $(x - 1)(x - 2)$. Multiply every term:

$$x(x - 2) + 2(x - 1)(x - 2) = -4(x - 1)$$

Expand:

$$x^2 - 2x + 2(x^2 - 3x + 2) = -4x + 4$$ $$x^2 - 2x + 2x^2 - 6x + 4 = -4x + 4$$ $$3x^2 - 8x + 4 = -4x + 4$$ $$3x^2 - 4x = 0$$ $$x(3x - 4) = 0$$

So $x = 0$ or $x = \dfrac{4}{3}$.

Check excluded values: $x \neq 1$ and $x \neq 2$. Both solutions are valid.

Answer: $x = 0$ or $x = \dfrac{4}{3}$.

Lesson: clear fractions by multiplying by the LCM, then solve the polynomial that drops out.

Question. Simplify $\left(\dfrac{1}{a} - \dfrac{1}{b}\right) \div \left(\dfrac{1}{a^2} - \dfrac{1}{b^2}\right)$.

Solution — combine each bracket, then divide.

Top bracket: $\dfrac{1}{a} - \dfrac{1}{b} = \dfrac{b - a}{ab}$.

Bottom bracket: $\dfrac{1}{a^2} - \dfrac{1}{b^2} = \dfrac{b^2 - a^2}{a^2 b^2}$.

Divide (multiply by reciprocal):

$$\frac{b - a}{ab} \times \frac{a^2 b^2}{b^2 - a^2}$$

Factorise $b^2 - a^2 = (b - a)(b + a)$ and cancel:

$$= \frac{(b-a)}{ab} \times \frac{a^2 b^2}{(b-a)(b+a)} = \frac{ab}{b + a}$$

Answer: $\dfrac{ab}{a + b}$.

Lesson: combine each nested piece into a single fraction first, then divide using reciprocals.

Question. Pump A fills a pool in $2x$ hours and pump B in $(3x - 4)$ hours. Working together they fill it in 5 hours. Form an equation in $x$ and solve.

Solution.

Step 1 — Rates per hour.

• Pump A rate: $\dfrac{1}{2x}$ pool/hour.
• Pump B rate: $\dfrac{1}{3x - 4}$ pool/hour.
• Combined: $\dfrac{1}{5}$ pool/hour.

Step 2 — Form the equation.

$$\frac{1}{2x} + \frac{1}{3x - 4} = \frac{1}{5}$$

Step 3 — Clear fractions. Multiply by $10x(3x - 4)$:

$$5(3x - 4) + 10x = 2x(3x - 4)$$ $$15x - 20 + 10x = 6x^2 - 8x$$ $$6x^2 - 33x + 20 = 0$$

Step 4 — Quadratic formula:

$$x = \frac{33 \pm \sqrt{33^2 - 4(6)(20)}}{12} = \frac{33 \pm \sqrt{1089 - 480}}{12} = \frac{33 \pm \sqrt{609}}{12}$$

$\sqrt{609} \approx 24.68$, so $x \approx 4.81$ or $x \approx 0.69$.

Step 5 — Reject unphysical roots. $x = 0.69 \Rightarrow 3x - 4 \approx -1.93 < 0$ (negative time). Reject.

Answer: $x \approx 4.81$ (to 2 d.p.).

Lesson: rates add, then clear fractions to get a quadratic. Always check the rejected root makes physical sense.