Problem 17 - Elementary Algebra

Question

For exercises 11-30, (a) solve. (b) check. $$ \frac{3}{10}+\frac{7}{m}=\frac{14}{m}+\frac{1}{15} $$

Step-by-Step Solution

Answer

m = 210

Step 1 - Combine like terms

First, isolate the terms that involve the variable on one side of the equation. Subtract \(\frac{7}{m}\) from both sides: \(\frac{3}{10} = \frac{14}{m} - \frac{7}{m} + \frac{1}{15}\)

Step 2 - Simplify the equation

Combine the fractions on the right side: \(\frac{3}{10} = \frac{7}{m} + \frac{1}{15}\)

Step 3 - Eliminate the denominators

Multiply both sides of the equation by 30m to eliminate the denominators: \(30m \times \frac{3}{10} = 30m \times \frac{7}{m} + 30m \times \frac{1}{15}\) \(3m = 210 + 2m\)

Step 4 - Solve for m

Rearrange the equation to isolate m: \(3m - 2m = 210\) \(m = 210\)

Step 5 - Check the solution

Substitute \(m = 210\) back into the original equation to verify: \(\frac{3}{10} + \frac{7}{210} = \frac{14}{210} + \frac{1}{15}\) \(\frac{3}{10} + \frac{1}{30} = \frac{1}{15} + \frac{1}{15}\) \Since both sides equal \(0.35\), the solution is correct.

Key Concepts

Combining Like Terms

Combining like terms is a foundational concept in solving equations. When we say 'like terms,' we are referring to terms that have the same variable raised to the same power. Consider the example equation: \( \frac{3}{10} + \frac{7}{m} = \frac{14}{m} + \frac{1}{15} \). To make the equation easier to handle, we need to gather all terms involving the variable (in this case, \( m \)) to one side of the equation. We start by subtracting \( \frac{7}{m} \) from both sides: \( \frac{3}{10} = \frac{14}{m} - \frac{7}{m} + \frac{1}{15} \). This step helps us simplify our work as it consolidates our variable terms for easier manipulation later on.

Eliminating Denominators

One of the most powerful strategies for solving rational equations is eliminating the denominators. This allows us to deal with a simpler linear equation. In our example, the equation after combining like terms becomes: \( \frac{3}{10} = \frac{7}{m} + \frac{1}{15} \). To clear the fractions, we multiply every term by the least common multiple (LCM) of the denominators. For denominators 10, \( m \), and 15, the LCM is 30\( m \). We multiply both sides by 30\( m \): \( 30m \times \frac{3}{10} = 30m \times \frac{7}{m} + 30m \times \frac{1}{15} \). This simplifies to: \( 3m = 210 + 2m \). This step is crucial as it converts the original rational equation into a much simpler form for solving.

Checking Solutions

After isolating and solving for the variable, it's absolutely essential to check if the solution satisfies the original equation. This prevents errors and ensures the solution is valid. For instance, after solving for \( m \), we get: \( m = 210 \). We substitute \( m = 210 \) back into the original expression: \( \frac{3}{10} + \frac{7}{210} \) and \( \frac{14}{210} + \frac{1}{15} \). Simplifying both sides, we find both equal 0.35, confirming our solution is correct. Ensuring the solution works in the original equation is a necessary step that verifies the accuracy of our work.

Rational Equations

Rational equations are equations involving fractions whose numerators and denominators contain variables. Handling these requires special steps. For example: \( \frac{3}{10} + \frac{7}{m} = \frac{14}{m} + \frac{1}{15} \). These types of equations often need extra steps, like finding a common denominator or multiplying through to clear the denominators. Solving rational equations typically involves:

• Combining like terms
• Eliminating denominators
• Isolating the variable
• Checking the solution

This structured approach helps in managing and solving the equations effectively.

Isolation of Variable

The ultimate goal in solving equations is to isolate the variable to determine its value. After simplifying the equation and eliminating the denominators, we had: \( 3m = 210 + 2m \). To isolate \( m \), we subtract \( 2m \) from both sides: \( 3m - 2m = 210 \). This simplifies to: \( m = 210 \). Isolating the variable is a step where we get down to the final solution by performing inverse operations, ensuring that the variable is on one side of the equation and its value is clearly seen on the other side. Getting comfortable with these steps is crucial for mastering equation solving.