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Algebra · Algebraic fractions

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Adding and subtracting algebraic fractions

Why algebraic fractions need a common denominator exactly like numeric ones, how the product of the denominators always works, and the bracket-and-sign discipline of the combined numerator.

Algebra · Algebraic fractions

Adding and subtracting algebraic fractions

Why algebraic fractions need a common denominator exactly like numeric ones, how the product of the denominators always works, and the bracket-and-sign discipline of the combined numerator.

Why it works

2x\frac{2}{x} and 3x+1\frac{3}{x+1} are different-sized pieces — xx-ths and (x+1)(x+1)-ths — so, exactly as with 13+14\frac{1}{3} + \frac{1}{4}, they must be recut to a common size before the counts can add. The product of the denominators always serves: here x(x+1)x(x + 1).

2x+3x+1=2(x+1)x(x+1)+3xx(x+1)=2(x+1)+3xx(x+1)=5x+2x(x+1).\frac{2}{x} + \frac{3}{x+1} = \frac{2(x+1)}{x(x+1)} + \frac{3x}{x(x+1)} = \frac{2(x+1) + 3x}{x(x+1)} = \frac{5x + 2}{x(x+1)}.

Each numerator is multiplied by the other fraction's denominator — in a BRACKET, expanded carefully. The denominators themselves never add: 52x+1\frac{5}{2x+1}-style answers fail the number test instantly.

Subtraction: the minus hits the WHOLE second numerator.

3x+2−1x−1=3(x−1)−(x+2)(x+2)(x−1)=3x−3−x−2(x+2)(x−1)=2x−5(x+2)(x−1).\frac{3}{x+2} - \frac{1}{x-1} = \frac{3(x-1) - (x+2)}{(x+2)(x-1)} = \frac{3x - 3 - x - 2}{(x+2)(x-1)} = \frac{2x - 5}{(x+2)(x-1)}.

The step −(x+2)=−x−2-(x + 2) = -x - 2 is where this topic's marks die — write the bracket, then push the minus through.

Numeric denominators: use the LCM as usual. x4+x+13=3x+4(x+1)12=7x+412\frac{x}{4} + \frac{x+1}{3} = \frac{3x + 4(x+1)}{12} = \frac{7x + 4}{12}; and with a shared letter, 52x+34x=104x+34x=134x\frac{5}{2x} + \frac{3}{4x} = \frac{10}{4x} + \frac{3}{4x} = \frac{13}{4x} — the LCM 4x4x beats the blunt product 8x28x^2 (which works but then needs simplifying).

A whole number is a fraction over 1. To combine 2+3x−12 + \frac{3}{x-1}, recut the 2:

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

Leave the denominator factorised. (x+2)(x−1)(x+2)(x-1) is a better final form than x2+x−2x^2 + x - 2: it shows the structure and is what mark schemes print. Expand only the top.