Operations with Algebraic Fractions

The denominators are already the same, so only the numerators need adding.

1. The denominators are equal, so add the numerators.
2. (x+2)+(2x-1)=3x+1.
3. The fraction is \frac{3x+1}{5}.

The denominators are already the same.

1. Keep the common denominator x.
2. The numerator is 3+2=5.
3. The answer is \frac{5}{x}.

In a product, cancel common factors across numerators and denominators.

1. Multiply numerators and denominators to obtain \frac{9x}{6x}.
2. Cancel the nonzero factor x, then reduce \frac{9}{6}.
3. The result is \frac{3}{2}.

Keep the first fraction and multiply by the reciprocal of the second.

1. Dividing by a fraction means multiplying by its reciprocal.
2. Write \frac{x+4}{x}\times\frac{x}{2}.
3. Cancel the nonzero factor x to obtain \frac{x+4}{2}.

Use 3x as the common denominator.

1. The denominators x and 3 have no common factor, so use 3x as the common denominator.
2. Multiply the top and bottom of \frac{1}{x} by 3, and of \frac{1}{3} by x: \frac{3}{3x}+\frac{x}{3x}.
3. Add the numerators: \frac{x+3}{3x}.

The lowest common denominator is x^2, not x^3.

1. x^2 is a multiple of x, so the lowest common denominator is x^2.
2. Multiply the top and bottom of \frac{3}{x} by x: \frac{3x}{x^2}.
3. Subtract the numerators: \frac{3x}{x^2}-\frac{2}{x^2}=\frac{3x-2}{x^2}.

Use 12 as the common denominator, and keep each numerator in a bracket.

1. Use the common denominator 12.
2. The numerator is 3(x+1)-2(x-3).
3. Expand carefully: 3x+3-2x+6=x+9.
4. The result is \frac{x+9}{12}.

Look for an entire bracket appearing in both a numerator and a denominator.

1. Write the product as \frac{(x+2)(x-3)}{(x-3)(x+5)}.
2. Cancel the nonzero factor x-3.
3. The result is \frac{x+2}{x+5}.

Use (x+1)(x-1) as the common denominator.

1. The denominators have no common factor, so use (x+1)(x-1) as the common denominator.
2. Multiply the top and bottom of each fraction by the other denominator: \frac{2(x-1)+3(x+1)}{(x+1)(x-1)}.
3. Expand the numerator: 2x-2+3x+3=5x+1.
4. The answer is \frac{5x+1}{(x+1)(x-1)}, which can also be written as \frac{5x+1}{x^2-1}.

Write 1 as a fraction with denominator x+2.

1. Write 1 as \frac{x+2}{x+2}.
2. The new numerator is x-(x+2)=-2.
3. The answer is \frac{-2}{x+2}.

Use 2x(x+4) as the common denominator.

1. The denominators 2x and x+4 have no common factor, so the common denominator is 2x(x+4).
2. Rewrite each fraction: \frac{3}{2x}=\frac{3(x+4)}{2x(x+4)} and \frac{1}{x+4}=\frac{2x}{2x(x+4)}.
3. Subtract the numerators: 3(x+4)-2x=3x+12-2x=x+12.
4. The answer is \frac{x+12}{2x(x+4)}. The numerator has no factor in common with the denominator, so this is its simplest form.

Factorise both quadratics before multiplying the fractions.

1. Factorise each quadratic. x^2-1=(x-1)(x+1) is a difference of two squares, and x^2+4x+3=(x+1)(x+3).
2. The product becomes \frac{(x-1)(x+1)}{(x+1)(x+3)}\times\frac{x+3}{x-1}.
3. Every bracket on the top, (x-1), (x+1) and (x+3), also appears on the bottom, so they all cancel.
4. The expression simplifies to 1.

Invert the divisor and factorise the difference of two squares.

1. Dividing by \frac{x+4}{6x} is the same as multiplying by its reciprocal, \frac{6x}{x+4}.
2. Factorise the difference of two squares: x^2-16=(x-4)(x+4).
3. The expression becomes \frac{(x-4)(x+4)}{3x}\times\frac{6x}{x+4}.
4. Cancel x+4 and x, and simplify \frac{6}{3}=2. This leaves 2(x-4), which is 2x-8.

Factorise the second denominator first.

1. Factorise the second denominator: 2x-2=2(x-1), so 2(x-1) is a common denominator.
2. Multiply the top and bottom of the first fraction by 2: \frac{2}{x-1}=\frac{4}{2(x-1)}.
3. Add the numerators: \frac{4+1}{2(x-1)}=\frac{5}{2(x-1)}.

For the difference use the shared denominator; for the quotient invert the second fraction.

Part (a)

1. The denominator is shared.
2. The numerator is (x+3)-(x-2)=5.
3. Thus P-Q=\frac{5}{x-1}.

Part (b)

1. Write the quotient as \frac{x+3}{x-1}\times\frac{x-1}{x-2}.
2. Cancel x-1 to obtain \frac{x+3}{x-2}.
3. The exclusion x=2 is needed because Q must not be zero.

Factorise x^2-4 first; it contains the other denominator.

1. Factorise the difference of two squares: x^2-4=(x-2)(x+2). This is a common denominator for both fractions.
2. Multiply the top and bottom of \frac{3}{x+2} by x-2: \frac{3(x-2)}{(x-2)(x+2)}.
3. Subtract the numerators, keeping a bracket round the second one: 3(x-2)-(x-1)=3x-6-x+1=2x-5.
4. The answer is \frac{2x-5}{(x-2)(x+2)}, which can also be written as \frac{2x-5}{x^2-4}.

Factorise 3x^2+5x-2, x^2+4x+4 and 6x-2 before you divide.

1. Dividing by a fraction is the same as multiplying by its reciprocal: \frac{3x^2+5x-2}{x^2+4x+4}\times\frac{x}{6x-2}.
2. Factorise each part: 3x^2+5x-2=(3x-1)(x+2), x^2+4x+4=(x+2)^2 and 6x-2=2(3x-1).
3. The product is \frac{(3x-1)(x+2)\times x}{(x+2)(x+2)\times2(3x-1)}.
4. Cancel the common factors 3x-1 and x+2. This leaves \frac{x}{2(x+2)}, which can also be written as \frac{x}{2x+4}.

Factorise every part, then compare the brackets x-3 and 3-x.

1. Factorise: x^2-3x=x(x-3), and 9-x^2=(3-x)(3+x) is a difference of two squares.
2. The product is \frac{x(x-3)(x+3)}{(3-x)(3+x)(x+1)}.
3. Cancel x+3, which is the same as 3+x.
4. x-3 and 3-x are not the same: x-3=-(3-x), so \frac{x-3}{3-x}=-1.
5. This leaves \frac{-x}{x+1}, which can also be written as -\frac{x}{x+1}.

Complete the addition inside the bracket before carrying out the division.

1. Work out the bracket first. Its common denominator is (x-2)(x+2): \frac{x(x+2)+2(x-2)}{(x-2)(x+2)}.
2. Expand the numerator: x^2+2x+2x-4=x^2+4x-4.
3. Dividing by \frac{1}{x+2} is the same as multiplying by x+2: \frac{x^2+4x-4}{(x-2)(x+2)}\times(x+2).
4. Cancel x+2. The numerator x^2+4x-4 does not factorise, so the answer is \frac{x^2+4x-4}{x-2}.

Factorise x^2-1; it is a common denominator for all three fractions.

1. x^2-1=(x-1)(x+1), so (x-1)(x+1) is a common denominator for all three fractions.
2. Rewrite each fraction over it: \frac{x+1}{(x-1)(x+1)}+\frac{x-1}{(x-1)(x+1)}-\frac{2}{(x-1)(x+1)}.
3. Combine the numerators: (x+1)+(x-1)-2=2x-2.
4. Factorise the numerator as 2(x-1) and cancel x-1: \frac{2(x-1)}{(x-1)(x+1)}=\frac{2}{x+1}.