Problem 55 - Prealgebra

Question

Solve each equation. Check your solution. $$y+\frac{3}{4}=\frac{2}{3}$$

Step-by-Step Solution

Answer

The solution is \( y = -\frac{1}{12} \).

Step 1: Isolate the Variable

To solve the equation, we need to isolate the variable \( y \). Start by subtracting \( \frac{3}{4} \) from both sides of the equation:\[y + \frac{3}{4} - \frac{3}{4} = \frac{2}{3} - \frac{3}{4}\]This simplifies to:\[y = \frac{2}{3} - \frac{3}{4}\]

Step 2: Find a Common Denominator

To subtract the fractions, we must first find a common denominator. The least common multiple of 3 and 4 is 12. Convert each fraction to have a denominator of 12:\[\frac{2}{3} = \frac{8}{12} \quad \text{and} \quad \frac{3}{4} = \frac{9}{12}\]

Step 3: Subtract the Fractions

Now that both fractions have a common denominator, subtract them:\[y = \frac{8}{12} - \frac{9}{12} \]Which simplifies to:\[y = -\frac{1}{12}\]

Step 4: Check the Solution

Substitute \( y = -\frac{1}{12} \) back into the original equation to verify:\[y + \frac{3}{4} = \frac{2}{3}\]Replace \( y \) with \( -\frac{1}{12} \):\[-\frac{1}{12} + \frac{3}{4} = \frac{2}{3}\]Convert \( \frac{3}{4} \) to \( \frac{9}{12} \) and find the sum:\[-\frac{1}{12} + \frac{9}{12} = \frac{8}{12} = \frac{2}{3}\]The original equation holds true.

Key Concepts

Subtracting Fractions

When it comes to subtracting fractions, the key is to ensure that both fractions share the same denominator. Unlike addition where you simply sum up the fractions, subtraction requires closely paying attention to the numerators while keeping the common denominator intact.
Imagine you want to subtract \( \frac{3}{4} \) from \( \frac{2}{3} \). You can't do this directly because the denominators are different. So, first, you'll need to find equivalent fractions with a shared denominator.
When you achieve the same denominator for both fractions, you subtract only the numerators while the common denominator remains unchanged.
In the exercise we're looking at, once both fractions \( \frac{2}{3} \) and \( \frac{3}{4} \) are expressed with a common denominator of 12, we subtract \(8 - 9\) to get \(-1\). Thus, the resulting fraction is \(-\frac{1}{12}\).
It's essential to keep practice subtracting fractions as it sharpens the ability to handle similar tasks in algebra.

Common Denominator

Finding a common denominator is a crucial step in making different fractions compatible for operations like addition or subtraction. This concept involves altering the fractions into equivalent forms that have the same denominator.
Think of the denominators as units that must be the same before you can carry out any operation on the numerators. To do this, find a number that is a multiple of both denominators. This number is called the least common denominator (LCD).

• For the problem \(\frac{2}{3}\) and \(\frac{3}{4}\), we find that 12 is the least common multiple of 3 and 4, hence 12 is the LCD.
• Convert each fraction: \(\frac{2}{3}\) becomes \(\frac{8}{12}\) because \(2 \times 4 = 8\), and \(\frac{3}{4}\) becomes \(\frac{9}{12}\) because \(3 \times 3 = 9\).

This common denominator approach allows seamless operations on the fractions, making solving equations involving fractions much simpler and clear.

Checking Solutions

After solving an equation, particularly with fractions, it's prudent to double-check your solution. This assures you that your computations are correct.
To check the solution, substitute the found value back into the original equation to see if it satisfies the equation. For instance, if you find that \(y = -\frac{1}{12}\), substitute it back into the equation \( y + \frac{3}{4} = \frac{2}{3} \).

• Replace \(y\):
  \(-\frac{1}{12} + \frac{3}{4} = \frac{2}{3}\).
• Now, convert \(\frac{3}{4}\) to \(\frac{9}{12}\), so the operation is \(-\frac{1}{12} + \frac{9}{12} = \frac{8}{12}\), which simplifies back to \(\frac{2}{3}\).

If both sides of the equation match, your solution is verified, confirming the correctness of your work. This step not only ensures clarity but also builds confidence in your mathematical solving skills.