Example

Subtracting 5c+4c23\frac{5c+4}{c-2} - 3

Subtract a whole number from a rational expression by first rewriting the whole number as a fraction with denominator 11: 5c+4c23\frac{5c + 4}{c - 2} - 3

Step 1 — Write the whole number as a fraction. Express 33 as 31\frac{3}{1} so that both terms are rational expressions: 5c+4c231\frac{5c + 4}{c - 2} - \frac{3}{1}

Step 2 — Find the LCD and rewrite. The denominators are c2c - 2 and 11. Since any expression divided by 11 is itself, the LCD is simply c2c - 2. The first fraction already has the LCD. Multiply the numerator and denominator of the second fraction by c2c - 2: 5c+4c23(c2)1(c2)\frac{5c + 4}{c - 2} - \frac{3(c - 2)}{1 \cdot (c - 2)} Distribute in the second numerator, 3(c2)=3c63(c - 2) = 3c - 6: 5c+4c23c6c2\frac{5c + 4}{c - 2} - \frac{3c - 6}{c - 2}

Step 3 — Subtract the numerators over the common denominator. Place the second numerator in parentheses and distribute the negative sign: 5c+4(3c6)c2=5c+43c+6c2=2c+10c2\frac{5c + 4 - (3c - 6)}{c - 2} = \frac{5c + 4 - 3c + 6}{c - 2} = \frac{2c + 10}{c - 2}

Step 4 — Factor to check for common factors. Extract the GCF from the numerator, 2c+10=2(c+5)2c + 10 = 2(c + 5): 2(c+5)c2\frac{2(c + 5)}{c - 2} Neither 22 nor c+5c + 5 matches the denominator factor c2c - 2, so the expression is already fully simplified.

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Updated 2026-06-25

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