Algebraic fractions and rational equations

(a) $\dfrac{3x}{x - 2}$

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

• M1 Factorising the numerator to $3x(x + 2)$ and the denominator to $(x + 2)(x - 2)$.
• A1 $\frac{3x}{x - 2}$.

(b) $\dfrac{3}{2x}$

1. Multiply by the reciprocal.
   $$\frac{2x}{5} \times \frac{15}{4x^2} = \frac{30x}{20x^2} = \frac{3}{2x}$$

• M1 Multiplying by the reciprocal $\frac{15}{4x^2}$.
• A1 The correct answer, $\frac{3}{2x}$.

(a) $x - 4$

1. $$(x-4)(x+4)/(x+4)$$
2. Factorise the numerator using the difference of squares: $x^{2}$ $-$ 16 = (x $-$ 4)(x + 4).
3. Cancel the non-zero common factor x + 4 to obtain x $-$ 4.

• M1 Establishing $(x-4)(x+4)/(x+4)$ or an equivalent valid method.
• A1 Correct answer: $x - 4$

(a) $x = 4$ or $x = -1$

1. Multiply every term by $x(x + 2)$.
   $$2(x + 2) + 3x = x(x + 2)$$
2. Expand and rearrange.
   $$5x + 4 = x^2 + 2x \Rightarrow x^2 - 3x - 4 = 0$$
3. Factorise.
   $$(x - 4)(x + 1) = 0 \Rightarrow x = 4 \text{ or } x = -1$$

• M1 Multiplying through by $x(x + 2)$, including the right-hand side.
• M1 Reaching $x^2 - 3x - 4 = 0$.
• M1 Factorising (or using the formula) correctly.
• A1 $x = 4$ and $x = -1$.

(a) {(5 - $\sqrt{33}$)/2, (5 + $\sqrt{33}$)/2}

1. $$x(x+2)+3(x-2)=2(x^{2}-4)$$
2. $$x^{2}-5x-2=0$$
3. Multiply by (x $-$ 2)(x + 2): x(x + 2) + 3(x $-$ 2) = 2($x^{2}$ $-$ 4).
4. Expand and collect terms: $x^{2}$ + 5x $-$ 6 = $2x^{2}$ $-$ 8, so $x^{2}$ $-$ 5x $-$ 2 = 0.
5. Use the quadratic formula: x = (5 ± $\sqrt{33}$)/2.
6. Neither value is 2 or $-2$, so both are valid.

• M1 Establishing $x(x+2)+3(x-2)=2(x^{2}-4)$ or an equivalent valid method.
• M1 Establishing $x^{2}-5x-2=0$ or an equivalent valid method.
• A1 Correct answer: {(5 - $\sqrt{33}$)/2, (5 + $\sqrt{33}$)/2}

(a) $((x - 3)(x - 1))/((x - 2)(x + 3))$

1. $$(x-3)(x+3)/((x+3)(x-2))$$
2. $$\frac{x-1}{x+3}$$
3. Factorise: $x^{2}$ $-$ 9 = (x $-$ 3)(x + 3), and $x^{2}$ + x $-$ 6 = (x + 3)(x $-$ 2).
4. Dividing by (x + 3)/(x $-$ 1) means multiplying by (x $-$ 1)/(x + 3).
5. Cancel one common factor x + 3 to obtain (x $-$ 3)(x $-$ 1)/[(x $-$ 2)(x + 3)].

• M1 Establishing $(x-3)(x+3)/((x+3)(x-2))$ or an equivalent valid method.
• M1 Establishing $\frac{x-1}{x+3}$ or an equivalent valid method.
• A1 Correct answer: $((x - 3)(x - 1))/((x - 2)(x + 3))$

(a) $(x-3)/x$

1. Factorise the difference of squares before cancelling a whole factor.
   $$(x-3)(x+3)/(x(x+3))$$
2. Therefore $(x-3)/x$.

• M1 Factorise the difference of squares before cancelling a whole factor.
• A1 Correct answer: $(x-3)/x$

(b) $6$

1. Clear the denominator and solve the linear equation.
   $$2(x-3)=x$$
2. Therefore $6$.

• M1 Clear the denominator and solve the linear equation.
• A1 Correct answer: $6$

(a) $\frac{x}{x+7}$

1. Replace division by multiplication by the reciprocal.
   $$\frac{x-5}{x+7}\times\frac{x}{x-5}$$
2. So the answer is $\frac{x}{x+7}$.

• M1 Replace division by multiplication by the reciprocal.
• A1 Correct answer: $\frac{x}{x+7}$

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

1. Substitute $x = 14$ into $\frac{x}{x+7}$.
   $$\frac{14}{14+7}=\frac{14}{21}$$
2. So the value is $\frac{2}{3}$.

• M1 Substituting $x = 14$ into $\frac{x}{x+7}$.
• A1 Correct answer: $\frac{2}{3}$

(a) $\frac{x(x + 3) - 18}{(x - 3)(x + 3)} = \frac{(x + 6)(x - 3)}{(x - 3)(x + 3)} = \frac{x + 6}{x + 3}$

1. Factorise $x^2 - 9$ to see the common denominator.
   $$x^2 - 9 = (x - 3)(x + 3)$$
2. Combine.
   $$\frac{x(x + 3) - 18}{(x - 3)(x + 3)} = \frac{x^2 + 3x - 18}{(x - 3)(x + 3)}$$
3. Factorise the numerator and cancel.
   $$\frac{(x + 6)(x - 3)}{(x - 3)(x + 3)} = \frac{x + 6}{x + 3}$$

• P1 Using the common denominator $(x - 3)(x + 3)$.
• P1 Reaching the numerator $x^2 + 3x - 18$.
• M1 Factorising the numerator to $(x + 6)(x - 3)$.
• A1 Cancelling $(x - 3)$ to reach $\frac{x + 6}{x + 3}$.

(b) $x = 0$

1. $\frac{x + 6}{x + 3} = 2 \Rightarrow x + 6 = 2x + 6 \Rightarrow x = 0$.
2. Check in the original: $0 - \frac{18}{-9} = 2$.

• M1 Forming $x + 6 = 2(x + 3)$.
• A1 The correct answer, $x = 0$.