Unit 7 · Lesson 5 of 5
Solving rational equations (and work/rate problems)
You'll be able to: solve equations with variables in the denominators, throw out answers that don't work, and set up "working together" and distance problems.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Answer from memory first, then check.
What is the LCD of \(\dfrac{3}{x}\) and \(\dfrac{5}{x^2 - 2x}\)?
Factor the denominators: \(x\) and \(x(x-2)\).
Use each different factor once (its most times in any one denominator): \(x(x-2)\).
Solve \(\dfrac{x}{3} + \dfrac{x}{4} = 7\).
Multiply every term by the LCD, 12: \(4x + 3x = 84\).
\(7x = 84\), so \(x = 12\).
Step 2 of 5The idea
The warm-up showed the trick: multiply every term by the LCD and the fractions disappear. That works even when the variable is in the denominator.
There is one catch. A denominator can never be 0, so values that make one 0 are excluded values. Sometimes the algebra hands you one of them anyway. That is an extraneous solution, and you throw it out.
Solving a rational equation.
- Factor the denominators and write the excluded values.
- Multiply every term on both sides by the LCD.
- Solve what is left (linear or quadratic).
- Throw out any answer that is an excluded value. If every answer is thrown out, there is no solution.
Working together. If one person does a job in \(a\) hours and another in \(b\) hours, together they take \(t\) hours, where \(\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{t}\). (\(\frac{1}{a}\) is the part of the job done in one hour.)
Distance problems. \(\text{time} = \dfrac{\text{distance}}{\text{rate}}\). "Same time" means you set two of these equal.
Why this shows up on the placement test: rational equations are on the list of equations you must solve, and test writers like to include an extraneous answer as a trap.
Step 3 of 5Worked examples
Example 1: \(\dfrac{x}{x-3} + \dfrac{1}{x} = \dfrac{9}{x^2 - 3x}\)
- Factor: \(x^2 - 3x = x(x-3)\). Excluded values: \(x \ne 0\) and \(x \ne 3\). The LCD is \(x(x-3)\).
- Multiply every term by \(x(x-3)\): \(x \cdot x + 1 \cdot (x-3) = 9\). (Each denominator cancels.)
- Simplify: \(x^2 + x - 3 = 9\), so \(x^2 + x - 12 = 0\).
- Factor: \((x+4)(x-3) = 0\), so \(x = -4\) or \(x = 3\).
- \(x = 3\) is excluded, so it is extraneous. The only solution is \(x = -4\). (Check: \(\frac{-4}{-7} + \frac{1}{-4} = \frac{9}{28}\), and \(\frac{9}{16 + 12} = \frac{9}{28}\).)
Example 2: working together
Ana can paint a fence in 3 hours. Ben can paint the same fence in 6 hours. How long does it take if they work together?
- In one hour Ana paints \(\frac{1}{3}\) of the fence and Ben paints \(\frac{1}{6}\). Together: \(\dfrac{1}{3} + \dfrac{1}{6} = \dfrac{1}{t}\).
- Multiply every term by the LCD, \(6t\): \(2t + t = 6\).
- \(3t = 6\), so \(t = 2\) hours. (Sense check: together must be faster than the faster person alone, 3 hours. It is.)
Step 4 of 5Practice
Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went. List every solution, separated by commas. If every answer is extraneous, type none.
Solve \(\dfrac{6}{x} + 1 = \dfrac{10}{x}\).
Excluded: \(x \ne 0\). Multiply every term by \(x\): \(6 + x = 10\).
\(x = 4\), which is allowed. Answer: \(x = 4\).
Solve \(\dfrac{5}{x+2} = \dfrac{3}{x}\).
Excluded: \(x \ne -2\), \(x \ne 0\). Multiply both sides by \(x(x+2)\): \(5x = 3(x+2)\).
\(5x = 3x + 6\), so \(2x = 6\) and \(x = 3\).
Check: \(\frac{5}{5} = 1\) and \(\frac{3}{3} = 1\). Answer: \(x = 3\).
Solve \(\dfrac{2}{x-1} = \dfrac{3}{x+2}\).
Excluded: \(x \ne 1\), \(x \ne -2\). Multiply both sides by \((x-1)(x+2)\): \(2(x+2) = 3(x-1)\).
\(2x + 4 = 3x - 3\), so \(x = 7\).
Check: \(\frac{2}{6} = \frac{1}{3}\) and \(\frac{3}{9} = \frac{1}{3}\). Answer: \(x = 7\).
Solve \(\dfrac{x}{x-5} = \dfrac{5}{x-5} + 2\).
Excluded: \(x \ne 5\). Multiply every term by \(x - 5\): \(x = 5 + 2(x-5)\).
\(x = 5 + 2x - 10 = 2x - 5\), so \(x = 5\).
But 5 is excluded (it makes the denominators 0). It is extraneous, so there is no solution.
Solve \(\dfrac{3}{x} + \dfrac{x}{x+2} = \dfrac{4}{x^2 + 2x}\).
Factor: \(x^2 + 2x = x(x+2)\). Excluded: \(x \ne 0\), \(x \ne -2\). The LCD is \(x(x+2)\).
Multiply every term by it: \(3(x+2) + x \cdot x = 4\), so \(x^2 + 3x + 6 = 4\) and \(x^2 + 3x + 2 = 0\).
\((x+1)(x+2) = 0\) gives \(x = -1\) or \(x = -2\). But \(-2\) is excluded.
Answer: \(x = -1\).
Solve \(\dfrac{1}{x} + \dfrac{1}{x+2} = \dfrac{4}{3}\).
Excluded: \(x \ne 0\), \(x \ne -2\). The LCD is \(3x(x+2)\).
Multiply every term by it: \(3(x+2) + 3x = 4x(x+2)\), so \(6x + 6 = 4x^2 + 8x\).
Move everything to one side: \(0 = 4x^2 + 2x - 6 = 2(2x^2 + x - 3) = 2(2x+3)(x-1)\).
\(x = 1\) or \(x = -\frac{3}{2}\). Neither is excluded, so both are solutions.
Maria can clean the office alone in 5 hours. When Jo helps, they finish together in 3 hours. How many hours would Jo take to clean the office alone?
Let \(j\) be Jo's time alone. Together: \(\dfrac{1}{5} + \dfrac{1}{j} = \dfrac{1}{3}\).
Multiply every term by \(15j\): \(3j + 15 = 5j\), so \(15 = 2j\).
\(j = \frac{15}{2} = 7.5\) hours. Check: \(\frac{1}{5} + \frac{2}{15} = \frac{3}{15} + \frac{2}{15} = \frac{1}{3}\).
Kim drives 150 miles in the same time that Lee drives 120 miles. Kim drives 10 miles per hour faster than Lee. What is Lee's speed, in miles per hour?
Let \(r\) be Lee's speed, so Kim's is \(r + 10\). Time is distance over rate, and the times are equal: \(\dfrac{150}{r + 10} = \dfrac{120}{r}\).
Multiply both sides by \(r(r+10)\): \(150r = 120(r + 10) = 120r + 1200\).
\(30r = 1200\), so \(r = 40\) mph. Check: Kim takes \(150 \div 50 = 3\) hours and Lee takes \(120 \div 40 = 3\) hours.
Step 5 of 5Wrap-up
You can now solve rational equations: write the excluded values, multiply every term by the LCD, solve, and throw out any extraneous answer. You can also set up "working together" and "same time" distance problems.
Read or watch next: OpenStax Intermediate Algebra 2e, 7.4 Solve Rational Equations and 7.5 Solve Applications with Rational Equations (free textbook, with more practice at the end of each section).
Also good: Khan Academy: Rational equations (short videos). Quick rules: Rational expressions cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "Why can solving an equation give me an answer that doesn't work?" It knows this course and can explain things another way.