Problem 85 - Algebra 2

Question

Solve each equation. \(x-\frac{3}{5}=\frac{2}{3}\)

Step-by-Step Solution

Answer

The solution is \( x = \frac{19}{15} \).

Step 1: Understand the Equation

The given equation is \( x - \frac{3}{5} = \frac{2}{3} \). Our task is to solve for the variable \( x \).

Step 2: Isolate the Variable

To solve for \( x \), we need to isolate it on one side of the equation. We'll do this by adding \( \frac{3}{5} \) to both sides of the equation: \[ x - \frac{3}{5} + \frac{3}{5} = \frac{2}{3} + \frac{3}{5} \] This simplifies to: \[ x = \frac{2}{3} + \frac{3}{5} \]

Step 3: Find a Common Denominator

The fractions \( \frac{2}{3} \) and \( \frac{3}{5} \) have different denominators. We need a common denominator to add them. The least common multiple of 3 and 5 is 15. Convert each fraction: \[ \frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15} \] and \[ \frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15} \]

Step 4: Add the Fractions

With the common denominator, add the fractions: \[ \frac{10}{15} + \frac{9}{15} = \frac{19}{15} \]

Step 5: Write the Solution for x

We have: \[ x = \frac{19}{15} \] Thus, the solution to the equation is \( x = \frac{19}{15} \).

Key Concepts

Common Denominator

When dealing with fractions, a crucial step in operations like addition and subtraction is finding a common denominator. This allows us to easily combine the fractions. Denominators are numbers under the fraction bar that determine the size of each fraction part.

To add or subtract fractions, we need their denominators to be the same. This ensures we're comparing equal parts. If our denominators are different, each fraction represents different-sized portions, making direct addition impossible. So instead, we find the "least common multiple" (LCM) of these denominators to use as a common denominator.

In our original problem, we had fractions \( \frac{2}{3} \) and \( \frac{3}{5} \). To find a common denominator, we determined that the LCM of 3 and 5 is 15. This is because it's the smallest number that 3 and 5 both divide into without leaving a remainder.

Fractions Addition

Once we've found a common denominator, adding fractions becomes straightforward. Let's walk through this to ensure complete understanding.

Think of the common denominator as a new standardized basis upon which all fractions will be established, making their size comparable.

• First, convert each fraction to an equivalent one with the common denominator. For example, \( \frac{2}{3} \) turns into \( \frac{10}{15} \) because \( 2 \times 5 = 10 \) and \( 3 \times 5 = 15 \).
• Similarly, \( \frac{3}{5} \) becomes \( \frac{9}{15} \).

This conversion means we've expanded or adjusted the fractions to play nicely together on the same field size.

Now, adding them is as easy as summing their numerators: \( \frac{10}{15} + \frac{9}{15} = \frac{19}{15} \). Notice how the denominator remains consistent throughout.

Thus, through common denominators and equal-sized parts, fractions arithmetic becomes intuitive.

Isolating Variables

One of the fundamental skills in solving equations is isolating the variable. This means getting the variable by itself on one side of the equation to determine its value.

In our exercise, we started with the equation \( x - \frac{3}{5} = \frac{2}{3} \). The goal was to solve for \( x \), indicating we needed \( x \) alone on one side. Here's how we did it:

• We added \( \frac{3}{5} \) to both sides. Adding or subtracting the same value helps keep the equation balanced while simplifying it. So, we had: \( x - \frac{3}{5} + \frac{3}{5} = \frac{2}{3} + \frac{3}{5} \).
• This step removed \( \frac{3}{5} \) from the left side, leaving \( x \).

Thus, the equation simplified to \( x = \frac{2}{3} + \frac{3}{5} \).

Isolating variables is pivotal as it allows us to focus on the lone variable and solve for its exact value. Understanding and practicing this skill will greatly enhance your equation-solving abilities.