Algebraic fraction sums

Rename the fractions to a common unit

For 1/(x-2) + 2/(x-3), the original requires x ≠ 2,3. Use denominator (x-2)(x-3):

Each numerator is multiplied by the missing denominator factor. Adding the denominators themselves would not preserve the fractions.

For subtraction, keep the numerator in brackets: 1/(x-2)-2/(x-3) = [(x-3)-2(x-2)]/[(x-2)(x-3)] = (1-x)/[(x-2)(x-3)].

Factor a quadratic denominator first

1/(x²-9) + 2/(x-3) has x²-9 = (x-3)(x + 3) and excludes x = 3,-3. Its sum is [1 + 2(x + 3)]/[(x-3)(x + 3)] = (2x + 7)/[(x-3)(x + 3)].

For repeated factors, use the highest needed power: 1/(x-3) + 2/((x-3)²) = [(x-3) + 2]/(x-3)² = (x-1)/(x-3)², x ≠ 3. Multiplying both denominators in full would work but creates an unnecessarily large expression.

Check your understanding 1 A cancelled factor does not remove its exclusion

Simplify 1/(x-1)-2/(x²-1) and state the original restrictions.

Your answer or drawing on paper

Show answer

For x ≠ 1,-1, use denominator (x-1)(x + 1). The numerator is (x + 1)-2 = x-1, giving 1/(x + 1) on that original domain. The cancelled value x = 1 remains excluded.

Compare with a numerical calculation

At x = 4, check the first sum using the original fractions and the combined expression.