1Cademy - Solving $$\frac{1}{c} + \frac{1}{m} = 1$$ for $$c$$

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Solving a Rational Equation for a Specific Variable

Example

Solving $\frac{1}{c} + \frac{1}{m} = 1$ for $c$

Solve the rational equation $\frac{1}{c} + \frac{1}{m} = 1$ for the variable $c$ . This example demonstrates the technique of factoring the target variable out when it appears in more than one term after clearing fractions, as well as the emergence of additional restrictions during the solving process.

Step 1 — Identify restricted values. Both $c$ and $m$ appear in denominators, so record $c \neq 0$ and $m \neq 0$ .

Step 2 — Clear the fractions. The denominators are $c$ , $m$ , and $1$ (implicit on the right). The LCD is $cm$ . Multiply every term on both sides by $cm$ :

$cm \cdot \frac{1}{c} + cm \cdot \frac{1}{m} = cm \cdot 1$

Step 3 — Simplify. Cancel the matching denominator factors in each product: $cm \cdot \frac{1}{c} = m$ , $cm \cdot \frac{1}{m} = c$ , and $cm \cdot 1 = cm$ . The fraction-free equation is:

$m + c = cm$

Step 4 — Isolate $c$ . The target variable $c$ now appears in two separate terms — once on the left side and once on the right side. Collect all terms containing $c$ on the right by subtracting $c$ from both sides:

$m = cm - c$

Factor $c$ out of the right side:

$m = c(m - 1)$

Divide both sides by $(m - 1)$ to isolate $c$ :

$c = \frac{m}{m - 1}$

Step 5 — State all restrictions. The original equation required $c \neq 0$ and $m \neq 0$ . However, the final expression has the denominator $(m - 1)$ , which equals zero when $m = 1$ . Therefore, the additional restriction $m \neq 1$ must also be stated.

This example highlights two important features of solving rational equations for a specific variable. First, when the target variable appears in more than one term after clearing fractions, all terms containing that variable must be gathered on one side and the variable factored out before dividing. Second, new restrictions can arise during the solving process beyond those identified from the original denominators — here, the restriction $m \neq 1$ comes from the denominator of the final answer, not from any denominator in the original equation.

Updated 2026-04-21

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OpenStax Elementary Algebra

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