Algebraic Fractions GCSE Questions

Answer $\dfrac{2x}{3}$

1. $\displaystyle \frac{6x^2}{9x} = \frac{6 \times x \times x}{9 \times x}$Write out the factors
2. $\displaystyle = \frac{2x}{3}$Cancel $3$ and $x$ from the top and bottom

• B1 $\dfrac{2x}{3}$ oe

Answer $3$

1. $\displaystyle \frac{3x + 6}{x + 2} = \frac{3(x + 2)}{x + 2}$Factorise the top
2. $\displaystyle = 3$Cancel the common factor $(x + 2)$

• M1 $3(x + 2)$
• A1 $3$

Answer $\dfrac{x}{x - 5}$

1. $\displaystyle \frac{x^2 + 5x}{x^2 - 25} = \frac{x(x + 5)}{(x + 5)(x - 5)}$Common factor on top; difference of two squares on the bottom
2. $\displaystyle = \frac{x}{x - 5}$Cancel $(x + 5)$

• M1 $x(x + 5)$
• M1 $(x + 5)(x - 5)$
• A1 $\dfrac{x}{x - 5}$

Answer $x + 2$

1. $\displaystyle \frac{x^2 + 3x + 2}{x + 1} = \frac{(x + 1)(x + 2)}{x + 1}$Factorise the quadratic on top
2. $\displaystyle = x + 2$

• M1 $(x + 1)(x + 2)$
• A1 $x + 2$

Answer $\dfrac{7x}{12}$

1. $\displaystyle \frac{x}{3} + \frac{x}{4} = \frac{4x}{12} + \frac{3x}{12}$Common denominator $12$
2. $\displaystyle = \frac{7x}{12}$

• M1 $\dfrac{4x}{12} + \dfrac{3x}{12}$
• A1 $\dfrac{7x}{12}$

Answer $\dfrac{x - 3}{x - 2}$

1. $\displaystyle \frac{x^2 - 9}{x^2 + x - 6} = \frac{(x + 3)(x - 3)}{(x + 3)(x - 2)}$Factorise both the top and the bottom
2. $\displaystyle = \frac{x - 3}{x - 2}$Cancel $(x + 3)$

• M1 $(x + 3)(x - 3)$
• M1 $(x + 3)(x - 2)$
• A1 $\dfrac{x - 3}{x - 2}$

Answer $\dfrac{5x + 2}{x(x + 1)}$

1. $\displaystyle \frac{2}{x} + \frac{3}{x + 1} = \frac{2(x + 1)}{x(x + 1)} + \frac{3x}{x(x + 1)}$Multiply each top by the other bottom
2. $\displaystyle = \frac{2x + 2 + 3x}{x(x + 1)}$
3. $\displaystyle = \frac{5x + 2}{x(x + 1)}$

• M1 Common denominator $x(x + 1)$
• M1 $2(x + 1) + 3x$
• A1 $\dfrac{5x + 2}{x(x + 1)}$ — allow $\dfrac{5x + 2}{x^2 + x}$

Answer $\dfrac{x + 3}{2x - 1}$

1. $\displaystyle 2x^2 + 7x + 3 = (2x + 1)(x + 3)$$ac = 6$: use $1$ and $6$
2. $\displaystyle 4x^2 - 1 = (2x + 1)(2x - 1)$Difference of two squares
3. $\displaystyle \frac{(2x + 1)(x + 3)}{(2x + 1)(2x - 1)} = \frac{x + 3}{2x - 1}$

• M1 $(2x + 1)(x + 3)$
• M1 $(2x + 1)(2x - 1)$
• A1 $\dfrac{x + 3}{2x - 1}$

Answer $\dfrac{x + 1}{x}$

1. $\displaystyle \frac{x + 2}{x - 1} \times \frac{(x + 1)(x - 1)}{x(x + 2)}$Factorise everything first
2. $\displaystyle = \frac{x + 1}{x}$Cancel $(x + 2)$ and $(x - 1)$

• M1 $(x + 1)(x - 1)$ or $x(x + 2)$
• M1 Both factorised and a correct cancellation seen
• A1 $\dfrac{x + 1}{x}$

Answer $\dfrac{3x + 11}{10}$

1. $\displaystyle \frac{5(x + 1)}{10} - \frac{2(x - 3)}{10}$
2. $\displaystyle = \frac{5x + 5 - 2x + 6}{10}$Careful: $-2(x - 3) = -2x + 6$
3. $\displaystyle = \frac{3x + 11}{10}$

• M1 Common denominator $10$
• M1 $5(x + 1) - 2(x - 3)$
• A1 $\dfrac{3x + 11}{10}$

Answer $x = 4$

1. $\displaystyle 3(x + 2) + 2(x - 1) = 24$Multiply every term, including the $4$, by $6$
2. $\displaystyle 3x + 6 + 2x - 2 = 24$
3. $\displaystyle 5x + 4 = 24$
4. $\displaystyle x = 4$

• M1 Multiplies every term by $6$: $3(x + 2) + 2(x - 1) = 24$
• M1 $5x + 4 = 24$
• A1 $x = 4$

Answer $(2x + 1)(x - 3)$

1. $\displaystyle ac = -6;\quad (-6) \times 1 = -6,\ -6 + 1 = -5$
2. $\displaystyle 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3)$
3. $\displaystyle = (2x + 1)(x - 3)$

• M1 $(2x \pm 1)(x \pm 3)$
• A1 $(2x + 1)(x - 3)$

Answer $\dfrac{2x + 1}{x + 3}$

1. $\displaystyle \frac{(2x + 1)(x - 3)}{(x + 3)(x - 3)}$Use part (a) for the top
2. $\displaystyle = \frac{2x + 1}{x + 3}$Cancel $(x - 3)$

• M1 $(x + 3)(x - 3)$
• A1 $\dfrac{2x + 1}{x + 3}$ — ft their (a)

Answer $x = 3$ or $x = -2$

1. $\displaystyle 3(x + 1) - 2(x - 1) = (x - 1)(x + 1)$Multiply every term by $(x - 1)(x + 1)$
2. $\displaystyle 3x + 3 - 2x + 2 = x^2 - 1$
3. $\displaystyle x + 5 = x^2 - 1$
4. $\displaystyle x^2 - x - 6 = 0$
5. $\displaystyle (x - 3)(x + 2) = 0$
6. $\displaystyle x = 3 \quad\text{or}\quad x = -2$

• M1 Multiplies by $(x - 1)(x + 1)$: $3(x + 1) - 2(x - 1) = (x - 1)(x + 1)$
• M1 $x + 5 = x^2 - 1$
• M1 $x^2 - x - 6 = 0$ and $(x - 3)(x + 2) = 0$
• A1 $x = 3$ and $x = -2$

Answer $2x(x - 2)$

1. $\displaystyle \frac{x^2 - 4}{3x} \times \frac{6x^2}{x + 2}$Dividing: flip the second fraction and multiply
2. $\displaystyle = \frac{(x + 2)(x - 2)}{3x} \times \frac{6x^2}{x + 2}$
3. $\displaystyle = (x - 2) \times \frac{6x^2}{3x}$Cancel $(x + 2)$
4. $\displaystyle = 2x(x - 2)$$\dfrac{6x^2}{3x} = 2x$

• M1 $\dfrac{x^2 - 4}{3x} \times \dfrac{6x^2}{x + 2}$
• M1 $(x + 2)(x - 2)$
• A1 $2x(x - 2)$ — allow $2x^2 - 4x$

Answer $\dfrac{(x + 3) + 2(x + 1)}{(x + 1)(x + 3)} = \dfrac{3x + 5}{(x + 1)(x + 3)}$

1. $\displaystyle \frac{1(x + 3)}{(x + 1)(x + 3)} + \frac{2(x + 1)}{(x + 1)(x + 3)}$
2. $\displaystyle = \frac{x + 3 + 2x + 2}{(x + 1)(x + 3)}$
3. $\displaystyle = \frac{3x + 5}{(x + 1)(x + 3)}$"Show that": every line must be written down

• M1 Common denominator $(x + 1)(x + 3)$
• M1 $(x + 3) + 2(x + 1)$
• A1 Fully correct working to $\dfrac{3x + 5}{(x + 1)(x + 3)}$

Answer $2$

1. $\displaystyle \frac{x^2 + 2x - 15}{2x^2 - 7x + 3} \times \frac{4x - 2}{x + 5}$Dividing: flip and multiply
2. $\displaystyle = \frac{(x + 5)(x - 3)}{(2x - 1)(x - 3)} \times \frac{2(2x - 1)}{x + 5}$Factorise everything, including $4x - 2$
3. $\displaystyle = 2$$(x + 5)$, $(x - 3)$ and $(2x - 1)$ all cancel

• M1 Flips the second fraction and multiplies
• M1 $(x + 5)(x - 3)$ and $(2x - 1)(x - 3)$
• M1 $4x - 2 = 2(2x - 1)$ and correct cancelling
• A1 $2$

Answer $x = -5$

1. $\displaystyle x(x - 1) + 4(x + 2) = (x + 2)(x - 1)$Multiply every term by $(x + 2)(x - 1)$
2. $\displaystyle x^2 - x + 4x + 8 = x^2 + x - 2$
3. $\displaystyle x^2 + 3x + 8 = x^2 + x - 2$
4. $\displaystyle 2x = -10$The $x^2$ terms cancel, so this is linear
5. $\displaystyle x = -5$Check: $\tfrac{-5}{-3} + \tfrac{4}{-6} = \tfrac{5}{3} - \tfrac{2}{3} = 1$

• M1 Multiplies by $(x + 2)(x - 1)$: $x(x - 1) + 4(x + 2) = (x + 2)(x - 1)$
• M1 $x^2 + 3x + 8 = x^2 + x - 2$
• M1 $2x = -10$
• A1 $x = -5$

Answer $\dfrac{4(2x - 1)}{(x - 2)(x + 2)}$

1. $\displaystyle \frac{(x + 1)(x + 2) - (x - 3)(x - 2)}{(x - 2)(x + 2)}$
2. $\displaystyle = \frac{x^2 + 3x + 2 - (x^2 - 5x + 6)}{(x - 2)(x + 2)}$Keep the bracket: the minus applies to every term
3. $\displaystyle = \frac{8x - 4}{(x - 2)(x + 2)}$
4. $\displaystyle = \frac{4(2x - 1)}{(x - 2)(x + 2)}$

• M1 Common denominator $(x - 2)(x + 2)$
• M1 $(x + 1)(x + 2) - (x - 3)(x - 2)$
• M1 $x^2 + 3x + 2 - (x^2 - 5x + 6)$
• A1 $\dfrac{8x - 4}{(x - 2)(x + 2)}$ or $\dfrac{4(2x - 1)}{x^2 - 4}$ oe

Answer $a = 1$

1. $\displaystyle x^2 - 9 = (x + 3)(x - 3)$Factorise the bottom
2. $\displaystyle \frac{x^2 + ax - 12}{(x + 3)(x - 3)} = \frac{x + 4}{x + 3}$So $(x - 3)$ must have cancelled from the top
3. $\displaystyle x^2 + ax - 12 = (x + 4)(x - 3) = x^2 + x - 12$
4. $\displaystyle a = 1$

• M1 $x^2 - 9 = (x + 3)(x - 3)$
• M1 The top must be $(x + 4)(x - 3)$
• A1 $a = 1$

Answer $5(x + 3) + 2(x - 1) = (x - 1)(x + 3)$, so $7x + 13 = x^2 + 2x - 3$, so $x^2 - 5x - 16 = 0$.

1. $\displaystyle 5(x + 3) + 2(x - 1) = (x - 1)(x + 3)$Multiply every term by $(x - 1)(x + 3)$
2. $\displaystyle 7x + 13 = x^2 + 2x - 3$
3. $\displaystyle 0 = x^2 - 5x - 16$

• M1 Multiplies by $(x - 1)(x + 3)$: $5(x + 3) + 2(x - 1) = (x - 1)(x + 3)$
• M1 $7x + 13 = x^2 + 2x - 3$
• A1 Correct rearrangement to $x^2 - 5x - 16 = 0$

Answer $x = 7.22$ or $x = -2.22$

1. $\displaystyle x = \frac{5 \pm \sqrt{(-5)^2 - 4 \times 1 \times (-16)}}{2} = \frac{5 \pm \sqrt{89}}{2}$It doesn't factorise, so use the formula
2. $\displaystyle x = 7.216\ldots \quad\text{or}\quad x = -2.216\ldots$
3. $\displaystyle x = 7.22 \quad\text{or}\quad x = -2.22$

• M1 Correct substitution: $\dfrac{5 \pm \sqrt{25 + 64}}{2}$
• A1 $7.22$
• A1 $-2.22$