Problem 44 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve each equation. Using the addition property of equality. Be sure to check your proposed solutions. $$y+\frac{7}{11}=\frac{7}{11}$$

Step-by-Step Solution

Answer

The solution for the equation \(y+\frac{7}{11}=\frac{7}{11}\) is \(y = 0\).

Step 1: Identify the Problem

In this equation, \(y+\frac{7}{11}=\frac{7}{11}\), the variable we need to solve for is \(y\).

Step 2: Use Addition property of equality to isolate \(y\)

To do this, we will subtract \(\frac{7}{11}\) from both sides of the equation: \(y + \frac{7}{11} - \frac{7}{11} = \frac{7}{11} - \frac{7}{11}\).

Step 3: Simplify and find the value of \(y\)

When we simplify both sides, we get \(y = 0\).

Step 4: Check the proposed solution

We substitute \(y = 0\) into the original equation to verify our solution: \(0 + \frac{7}{11} = \frac{7}{11}\). Indeed, \(\frac{7}{11} = \frac{7}{11}\), confirming that \(y = 0\) is the correct solution.

Key Concepts

Solving Equations

Understanding how to solve equations is fundamental in algebra. When dealing with equations, our goal is to determine the value(s) of the variable that make the equation true. Once we have an equation, we start by identifying the operations applied to the variable and look for ways to reverse these operations to isolate the variable on one side of the equation.

In our example, we start with the equation \(y + \frac{7}{11} = \frac{7}{11}\). Notice the variable \(y\) is added to \(\frac{7}{11}\). According to the addition property of equality, we can add or subtract the same number to both sides of the equation without changing the equality. This means that the original and the new equations are equivalent. Here, we'll use subtraction to help in solving the equation, as subtracting is the inverse operation of adding.

By carefully applying these properties and operations, we simplify to find values for the unknown variable and solve the equation.

Variable Isolation

Variable isolation involves having the variable alone on one side of the equation. This process is crucial because it allows us to find the variable's value directly. In algebra, isolation often requires a series of steps that may involve adding, subtracting, multiplying, or dividing both sides of the equation by the same number.

Let's break down our example: \(y + \frac{7}{11} = \frac{7}{11}\). To isolate \(y\), we need to remove the \(\frac{7}{11}\) added to it. Since the opposite of adding \(\frac{7}{11}\) is subtracting \(\frac{7}{11}\), we subtract it from both sides. Thus, we get \(y + \frac{7}{11} - \frac{7}{11} = \frac{7}{11} - \frac{7}{11}\). This simplifies to \(y = 0\).

When you see variable isolation in action, remember it is about finding the simplest equivalent form, where you "undo" operations to free the variable, establishing the key to unlock the equation.

Checking Solutions

After solving an equation, it's essential to check if the solution is correct. Checking solutions verifies that you've applied algebraic operations correctly. Let's go back to our initial equation: \(y + \frac{7}{11} = \frac{7}{11}\).

We found \(y = 0\). To ensure this is correct, substitute \(y\) back into the original equation to see if it satisfies the equation. Plugging \(y = 0\) we have \(0 + \frac{7}{11} = \frac{7}{11}\), which balances the equation correctly.

This validation process confirms that our solution is accurate. Always double-check your answers by substituting your solution back into the original equation. This step acts as a safety net against possible calculation errors and assures full understanding of the solution process, enhancing your problem-solving confidence.