Algebraic Fractions

Use the common denominator x(x+1) and multiply each numerator by whatever its denominator was multiplied by.

1. Common denominator: x(x+1)
2. \dfrac{2}{x+1}=\dfrac{2x}{x(x+1)} and \dfrac{1}{x}=\dfrac{x+1}{x(x+1)}
3. Subtract the numerators: 2x-(x+1)=2x-x-1=x-1
4. Answer: \dfrac{x-1}{x(x+1)}
5. (The minus sign applies to the whole of (x+1), so -1 not +1.)

Find the lowest common multiple of 9y and 6y and use it as the denominator of both fractions.

1. The lowest common denominator of 9y and 6y is 18y
2. \dfrac{2}{9y}=\dfrac{4}{18y} and \dfrac{5}{6y}=\dfrac{15}{18y}
3. Add the numerators: 4+15=19
4. Answer: \dfrac{19}{18y}

Use the common denominator 15, and put the second numerator in a bracket so that the minus sign applies to both of its terms.

1. Common denominator: 15
2. \dfrac{2x+1}{3}=\dfrac{5(2x+1)}{15} and \dfrac{x-4}{5}=\dfrac{3(x-4)}{15}
3. Numerator: 5(2x+1)-3(x-4)=10x+5-3x+12
4. =7x+17
5. Answer: \dfrac{7x+17}{15}

Turn the second fraction upside down and multiply, then cancel the numbers and the letters separately.

1. Dividing by a fraction is the same as multiplying by its reciprocal
2. \dfrac{6p^2}{5q}\times\dfrac{10q^3}{9p}=\dfrac{60p^2q^3}{45pq}
3. Numbers: \dfrac{60}{45}=\dfrac43
4. Letters: \dfrac{p^2q^3}{pq}=pq^2
5. Answer: \dfrac{4pq^2}{3}

Factorise the top and the bottom, then look for a bracket that appears in both.

1. Top: 4x+12=4(x+3)
2. Bottom: x^2+x-6=(x+3)(x-2)
3. \dfrac{4(x+3)}{(x+3)(x-2)}
4. Cancel the common factor (x+3)
5. Answer: \dfrac{4}{x-2}

Use (x+4)(x-1) as the common denominator and multiply each numerator by the bracket its denominator is missing.

1. Common denominator: (x+4)(x-1)
2. \dfrac{x}{x+4}=\dfrac{x(x-1)}{(x+4)(x-1)} and \dfrac{2}{x-1}=\dfrac{2(x+4)}{(x+4)(x-1)}
3. Numerator: x(x-1)+2(x+4)=x^2-x+2x+8
4. =x^2+x+8 (this does not factorise, so nothing cancels)
5. Answer: \dfrac{x^2+x+8}{(x+4)(x-1)}

Factorise the top and the bottom completely before you cancel anything: the top has a common factor and then a difference of two squares.

1. Top: take out the common factor 3: 3x^2-12=3(x^2-4)
2. x^2-4 is a difference of two squares: 3(x^2-4)=3(x-2)(x+2)
3. Bottom: x^2+5x+6=(x+2)(x+3)
4. \dfrac{3(x-2)(x+2)}{(x+2)(x+3)}
5. Cancel the common factor (x+2)
6. Answer: \dfrac{3(x-2)}{x+3}

Take out the common factor x from the top and the bottom first: what is left on the top is a difference of two squares.

1. Top: 4x^3-x=x(4x^2-1)
2. 4x^2-1 is a difference of two squares: x(2x-1)(2x+1)
3. Bottom: 2x^2+x=x(2x+1)
4. \dfrac{x(2x-1)(2x+1)}{x(2x+1)}
5. Cancel x and (2x+1)
6. Answer: 2x-1

Both quadratics have a factor in common: factorise each one into two brackets and check by expanding.

1. Top: 3x^2+7x-6=(3x-2)(x+3)
2. Bottom: 4x^2+13x+3=(4x+1)(x+3)
3. \dfrac{(3x-2)(x+3)}{(4x+1)(x+3)}
4. Cancel the common factor (x+3)
5. Answer: \dfrac{3x-2}{4x+1}

Factorise the top as you would x^2-x-6, but with y in each bracket, and take out the common factor 2x from the bottom.

1. Top: x^2-xy-6y^2=(x-3y)(x+2y)
2. Bottom: 2x^2+4xy=2x(x+2y)
3. \dfrac{(x-3y)(x+2y)}{2x(x+2y)}
4. Cancel the common factor (x+2y)
5. Answer: \dfrac{x-3y}{2x}

Factorise 6-2x as -2(x-3) so that it matches one of the brackets on the top.

1. x^2+2x-15=(x+5)(x-3)
2. 6-2x=-2(x-3)
3. \dfrac{(x+5)(x-3)}{4x}\times\dfrac{8x^2}{-2(x-3)}
4. Cancel (x-3): \dfrac{8x^2(x+5)}{-8x}
5. \dfrac{8x^2}{-8x}=-x
6. Answer: -x(x+5)

Factorise all three expressions that can be factorised, then turn the second fraction upside down and multiply.

1. x^2-4=(x-2)(x+2) (difference of two squares)
2. 3x+6=3(x+2)
3. x^2-5x+6=(x-2)(x-3)
4. \dfrac{(x-2)(x+2)}{3(x+2)}\times\dfrac{9x}{(x-2)(x-3)}
5. Cancel (x+2) and (x-2): \dfrac{9x}{3(x-3)}
6. Answer: \dfrac{3x}{x-3}

Write the 1 as a fraction over the common denominator (x-2)(x+1) too.

1. Common denominator: (x-2)(x+1)
2. 1=\dfrac{(x-2)(x+1)}{(x-2)(x+1)}
3. Numerator: (x-2)(x+1)+4(x+1)-3(x-2)
4. =x^2-x-2+4x+4-3x+6
5. =x^2+8
6. Answer: \dfrac{x^2+8}{(x-2)(x+1)}

Factorise x^2+4x-12 to find the common denominator, then multiply every term in the bracket when you work out the first numerator.

1. x^2+4x-12=(x+6)(x-2)
2. \dfrac{5}{x-2}=\dfrac{5(x+6)}{(x+6)(x-2)}
3. Numerator: 5(x+6)-(5x+7)=5x+30-5x-7
4. =23
5. Answer: k=23

Multiply every term by x(x+1) to clear the fractions, then rearrange into a quadratic equal to zero.

1. Multiply every term by x(x+1): 4x+3(x+1)=2x(x+1)
2. 7x+3=2x^2+2x
3. 2x^2-5x-3=0
4. (2x+1)(x-3)=0
5. Neither value makes a denominator zero, so both are valid.
6. Answer: x=3 or x=-\dfrac12

Write each bracket as a single fraction first, and notice that y^2-x^2 is a difference of two squares.

1. First bracket: \dfrac1x-\dfrac1y=\dfrac{y-x}{xy}
2. Second bracket: \dfrac{1}{x^2}-\dfrac{1}{y^2}=\dfrac{y^2-x^2}{x^2y^2}
3. y^2-x^2=(y-x)(y+x)
4. Flip and multiply: \dfrac{y-x}{xy}\times\dfrac{x^2y^2}{(y-x)(y+x)}
5. Cancel (y-x) and xy
6. Answer: \dfrac{xy}{x+y}

Add the fractions of the tank that each pipe fills in one hour. Answer part (a) before attempting part (b).

Part (a)

1. In one hour pipe A fills \dfrac1x of the tank
2. Pipe B takes x+5 hours, so in one hour it fills \dfrac{1}{x+5} of the tank
3. Together they fill the tank in 6 hours, so \dfrac16 of it in one hour
4. Answer: \dfrac1x+\dfrac{1}{x+5}=\dfrac16

Part (b)

1. Multiply every term by 6x(x+5): 6(x+5)+6x=x(x+5)
2. 12x+30=x^2+5x
3. x^2-7x-30=0
4. (x-10)(x+3)=0
5. Answer: x=10 or x=-3

Part (c)

1. A time cannot be negative, so x=10
2. Pipe B takes x+5 hours
3. Answer: 15 hours

Multiply both sides by the denominator from part (a), which is positive because x>0. Answer part (a) before attempting part (b).

Part (a)

1. Common denominator: x(x+1)
2. Numerator: 6(x+1)-4x=6x+6-4x
3. =2x+6
4. Answer: \dfrac{2x+6}{x(x+1)}

Part (b)

1. Use part (a): \dfrac{2x+6}{x(x+1)}\geqslant1
2. x>0, so x(x+1) is positive and you can multiply both sides by it without reversing the sign
3. 2x+6\geqslant x^2+x
4. x^2-x-6\leqslant0
5. (x-3)(x+2)\leqslant0, so -2\leqslant x\leqslant3
6. Combine with x>0
7. Answer: 0<x\leqslant3

Split the fraction into two terms and write each as a power of x before you differentiate. Answer part (a) before attempting part (b).

Part (a)

1. Split the fraction: y=\dfrac{x^4}{2x^2}+\dfrac{16}{2x^2}
2. y=\dfrac12x^2+8x^{-2}
3. Differentiate each term: \dfrac12\times2x=x and 8\times(-2)x^{-3}=-16x^{-3}
4. Answer: \dfrac{\mathrm{d}y}{\mathrm{d}x}=x-16x^{-3}

Part (b)

1. At a stationary point \dfrac{\mathrm{d}y}{\mathrm{d}x}=0: x-16x^{-3}=0
2. Multiply by x^3: x^4-16=0
3. x^4=16
4. Answer: x=2 or x=-2

Factorise all four expressions first, then turn the division into a multiplication by flipping the second fraction.

Part (a)

1. 3x^2-10x-8=(3x+2)(x-4)
2. x^2-6x+8=(x-2)(x-4)
3. 9x^2-4=(3x-2)(3x+2) (difference of two squares)
4. 6x^2-4x=2x(3x-2)
5. Flip the second fraction and multiply: \dfrac{(3x+2)(x-4)}{(x-2)(x-4)}\times\dfrac{2x(3x-2)}{(3x-2)(3x+2)}
6. Cancel (x-4), (3x+2) and (3x-2)
7. Answer: \dfrac{2x}{x-2}

Part (b)

1. Use part (a): \dfrac{2x}{x-2}=x-3
2. Multiply both sides by (x-2): 2x=(x-3)(x-2)
3. 2x=x^2-5x+6
4. x^2-7x+6=0
5. (x-1)(x-6)=0
6. Neither value makes a denominator zero, so both are valid.
7. Answer: x=1 or x=6