Applications of Rational Expressions and Equations

Overview

Rational expressions and equations can be applied to solve a wide variety of real-world problems. These applications often involve work with rates, proportions, mixtures, and other scenarios where quantities are related by ratios. In this section, we will explore several types of problems that can be solved using rational expressions and equations.

Key Applications of Rational Expressions

  • Work Problems: These problems involve rates of work and can often be solved by setting up rational equations.
  • Mixture Problems: These involve mixing two or more substances in a certain ratio, and can be modeled with rational expressions.
  • Rate Problems: Problems involving speed, distance, and time can often be expressed with rational equations.
  • Proportions: Solving proportions is one of the simplest applications of rational expressions, where we use cross-multiplication to solve for unknowns.

Example 1: Work Problem

Suppose two workers are hired to complete a job. The first worker can complete the job in 5 hours, and the second worker can complete it in 7 hours. How long will it take for them to complete the job working together?

Let t be the time it takes for both workers to complete the job together. The rate of work for each worker is the reciprocal of their time to complete the job. Therefore, we can write the equation:

\( \frac{1}{5} + \frac{1}{7} = \frac{1}{t} \)

Step 1: Find a common denominator for the left side:

\( \frac{7}{35} + \frac{5}{35} = \frac{1}{t} \)

Step 2: Simplify the left side:

\( \frac{12}{35} = \frac{1}{t} \)

Step 3: Solve for t by cross-multiplying:

\( t = \frac{35}{12} \)

Therefore, the time it takes for both workers to complete the job together is \( \frac{35}{12} \) hours, or approximately 2.92 hours.

Example 2: Mixture Problem

A chemist has 10 liters of a 40% alcohol solution. How many liters of a 60% alcohol solution should be added to obtain 20 liters of a 50% alcohol solution?

Let x represent the number of liters of the 60% alcohol solution to add. The amount of alcohol in the 40% solution is \( 0.4(10) \), and the amount of alcohol in the 60% solution is \( 0.6x \). The total amount of alcohol in the final solution is \( 0.5(20) \).

We can set up the following equation:

\( 0.4(10) + 0.6x = 0.5(20) \)

Step 1: Simplify the terms:

\( 4 + 0.6x = 10 \)

Step 2: Subtract 4 from both sides:

\( 0.6x = 6 \)

Step 3: Solve for x:

\( x = \frac{6}{0.6} = 10 \)

Therefore, 10 liters of the 60% alcohol solution should be added.

Practice Questions

  1. Question 1: A tank can be filled by two pipes. The first pipe can fill the tank in 4 hours, and the second pipe can fill it in 6 hours. How long will it take for both pipes to fill the tank together?
    Solution

    Let t be the time it takes for both pipes to fill the tank together. The rates of the pipes are \( \frac{1}{4} \) and \( \frac{1}{6} \), respectively. The equation is:

    \( \frac{1}{4} + \frac{1}{6} = \frac{1}{t} \)

    Step 1: Find a common denominator:

    \( \frac{3}{12} + \frac{2}{12} = \frac{1}{t} \)

    Step 2: Simplify the left side:

    \( \frac{5}{12} = \frac{1}{t} \)

    Step 3: Solve for t:

    \( t = \frac{12}{5} = 2.4 \, \text{hours} \)

    Therefore, it will take 2.4 hours for both pipes to fill the tank together.

  2. Question 2: A 50-liter solution contains 20% salt. How much of a 30% salt solution should be added to make the new solution 25% salt?
    Solution

    Let x be the amount of 30% salt solution to add. The amount of salt in the 50-liter solution is \( 0.2(50) = 10 \), and the amount of salt in the \( x \) liters of 30% solution is \( 0.3x \). The total amount of salt in the final solution is \( 0.25(50 + x) \).

    The equation is:

    \( 10 + 0.3x = 0.25(50 + x) \)

    Step 1: Simplify the equation:

    \( 10 + 0.3x = 12.5 + 0.25x \)

    Step 2: Subtract 0.25x from both sides:

    \( 10 + 0.05x = 12.5 \)

    Step 3: Subtract 10 from both sides:

    \( 0.05x = 2.5 \)

    Step 4: Solve for x:

    \( x = \frac{2.5}{0.05} = 50 \)

    Therefore, 50 liters of the 30% salt solution should be added.

  3. Question 3: Solve the proportion:

    \( \frac{x}{4} = \frac{3}{6} \)

    Solution

    To solve the proportion, cross-multiply:

    \( x \cdot 6 = 4 \cdot 3 \)

    Step 1: Simplify:

    \( 6x = 12 \)

    Step 2: Solve for x:

    \( x = \frac{12}{6} = 2 \)

    Therefore, \( x = 2 \).

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