Problem 55 - Intermediate Algebra

Question

Solve equation. If a solution is extraneous, so indicate. \(\frac{3}{y}+\frac{7}{2 y}=13\)

Step-by-Step Solution

Answer

The solution is \(y = \frac{1}{2}\), which is not extraneous.

Step 1: Identify and Simplify the Equation

The given equation is \( \frac{3}{y} + \frac{7}{2y} = 13 \). Begin by finding a common denominator for the fractions on the left-hand side. The common denominator is \(2y\), so rewrite the equation as \( \frac{6}{2y} + \frac{7}{2y} = 13 \).

Step 2: Combine the Fractions

Since both terms on the left-hand side have the same denominator, combine them: \( \frac{6 + 7}{2y} = \frac{13}{2y} \). Thus, you have \( \frac{13}{2y} = 13 \).

Step 3: Solve for y

To eliminate the fraction, multiply both sides of the equation by \(2y\): \(13 = 13 \times 2y \rightarrow 13 = 26y\). Now divide both sides by 26 to solve for \(y\): \(y = \frac{13}{26} = \frac{1}{2}.\)

Step 4: Check for Extraneous Solutions

Substitute \(y = \frac{1}{2}\) back into the original equation to check for validity: \(\frac{3}{\frac{1}{2}} + \frac{7}{2 \times \frac{1}{2}} = 13 \rightarrow 6 + 7 = 13,\) which simplifies correctly to 13. Therefore, \(y = \frac{1}{2}\) is the correct solution and is not extraneous.

Key Concepts

Equation Solving

In order to solve equations, we follow a series of steps to find the values of variables that satisfy the given equation. Consider our exercise: \( \frac{3}{y} + \frac{7}{2y} = 13 \). The key to solving such equations is to organize them in a way that allows easy manipulation and simplification. To begin solving, it's essential to bring all the variable terms on one side and constants on the other, if possible.

Here's a straightforward approach:

• Eliminate fractions: Find a common denominator to combine terms, making calculations simpler.
• Isolate the variable: This involves collecting all terms involving the variable on one side and constants on the other.
• Perform inverse operations: Use addition, subtraction, multiplication, or division to solve for the variable.
• Validate your solution: Substitute it back into the original equation to verify if it's correct, ensuring there are no errors.

By breaking down each step of equation solving, we help ensure clarity and accuracy throughout the process, leading to solutions such as \( y = \frac{1}{2} \).

Common Denominator

When dealing with equations that involve fractions, finding a common denominator can simplify the expression. For \( \frac{3}{y} + \frac{7}{2y} = 13 \), note that both fractions have denominators involving \( y \).

Here's how to find and use a common denominator effectively:

• Identify denominators: Look at all the denominators involved. Here, they are \( y \) and \( 2y \).
• Determine the least common denominator (LCD): The LCD for these terms is \( 2y \), as it accommodates both denominators.
• Rewrite each fraction: Express each term with the LCD to simplify addition or subtraction between them, like so: \( \frac{6}{2y} + \frac{7}{2y} \).
• Simplify the equation: With a common denominator, you can neatly combine terms effortlessly.

Utilizing a common denominator eliminates the fractions, making it easier to manipulate and solve equations.

Extraneous Solution

Extraneous solutions are solutions that arise during the solving process but do not satisfy the original equation. They can occur due to manipulations such as multiplying or squaring both sides of an equation. In our exercise, we ensured our solution wasn’t extraneous by substituting \( y = \frac{1}{2} \) back into the original equation.

Here’s how to manage and check for extraneous solutions:

• Substitute back: Always plug your solution back into the original equation.
• Verify each term: Ensure that each term in the substituted equation equals the original value.
• Logical check: Assess if the solution leads to any contradictions or undefined expressions, such as division by zero.

In this exercise, substituting back showed correctness, as \( 6 + 7 = 13 \) checks out completely. Hence, \( y = \frac{1}{2} \) was correct and not extraneous. Checking always helps prevent including false roots in your final answer.