Example

Try It 7.54: Simplifying 13c+41c+4+c3\frac{1-\frac{3}{c+4}}{\frac{1}{c+4}+\frac{c}{3}} by Writing it as Division

To simplify the complex rational expression 13c+41c+4+c3\frac{1-\frac{3}{c+4}}{\frac{1}{c+4}+\frac{c}{3}} by writing it as division, perform the following steps:

Step 1. Simplify the numerator and denominator. In the numerator, the common denominator is c+4c+4. Rewrite 11 as c+4c+4\frac{c+4}{c+4}. Subtracting yields c+43c+4=c+1c+4\frac{c+4-3}{c+4} = \frac{c+1}{c+4}. In the denominator, the common denominator is 3(c+4)3(c+4). Rewrite as 33(c+4)+c(c+4)3(c+4)=3+c2+4c3(c+4)=c2+4c+33(c+4)\frac{3}{3(c+4)} + \frac{c(c+4)}{3(c+4)} = \frac{3+c^2+4c}{3(c+4)} = \frac{c^2+4c+3}{3(c+4)}. The complex fraction becomes: c+1c+4c2+4c+33(c+4)\frac{\frac{c+1}{c+4}}{\frac{c^2+4c+3}{3(c+4)}}.

Step 2. Rewrite as fraction division. Replace the main fraction bar with division: c+1c+4÷c2+4c+33(c+4)\frac{c+1}{c+4} \div \frac{c^2+4c+3}{3(c+4)}

Step 3. Multiply by the reciprocal and simplify. Rewrite as multiplication by the reciprocal: c+1c+43(c+4)c2+4c+3\frac{c+1}{c+4} \cdot \frac{3(c+4)}{c^2+4c+3} Factor the denominator c2+4c+3c^2+4c+3 to (c+1)(c+3)(c+1)(c+3). (c+1)3(c+4)(c+4)(c+1)(c+3)\frac{(c+1) \cdot 3(c+4)}{(c+4)(c+1)(c+3)} Divide out the common factors (c+1)(c+1) and (c+4)(c+4). The final simplified expression is: 3c+3\frac{3}{c+3}

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Updated 2026-04-30

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Ch.7 Rational Expressions and Functions - Intermediate Algebra @ OpenStax

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