1Cademy - Subtracting $$\frac{3}{z+3}

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How to Add or Subtract Rational Expressions

Example

Subtracting $\frac{3}{z+3} - \frac{6z}{z^2-9}$

Subtract two rational expressions where finding the LCD involves factoring a difference of squares:

$\frac{3}{z + 3} - \frac{6z}{z^2 - 9}$

Step 1 — Find the LCD and rewrite each fraction. Factor the second denominator as a difference of squares: $z^2 - 9 = (z - 3)(z + 3)$ . The first denominator $(z + 3)$ is one of these factors. The LCD is $(z - 3)(z + 3)$ .

The second fraction already has the LCD. The first fraction is missing the factor $(z - 3)$ ; multiply its numerator and denominator by $(z - 3)$ :

$\frac{3(z - 3)}{(z - 3)(z + 3)} - \frac{6z}{(z - 3)(z + 3)}$

Distribute in the first numerator: $3(z - 3) = 3z - 9$ .

$\frac{3z - 9}{(z - 3)(z + 3)} - \frac{6z}{(z - 3)(z + 3)}$

Step 2 — Subtract the rational expressions. Subtract the numerators over the common denominator:

$\frac{3z - 9 - 6z}{(z - 3)(z + 3)}$

Combine like terms: $3z - 6z = -3z$ :

$\frac{-3z - 9}{(z - 3)(z + 3)}$

Step 3 — Simplify, if possible. Factor the numerator by extracting the greatest common factor, $-3$ :

$\frac{-3(z + 3)}{(z - 3)(z + 3)}$

Cancel the shared factor $(z + 3)$ :

$\frac{-3}{z - 3}$

This example highlights a subtraction problem where the first denominator is a factor of the second. Once the fractions are rewritten with the LCD and their numerators are combined, factoring the resulting numerator reveals a shared factor with the denominator, allowing the expression to be fully simplified.

Updated 2026-04-30

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OpenStax Intermediate Algebra

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