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Chapter 7: Problem 11

Solve. If no solution exists, state this. \(\frac{3}{5}-\frac{2}{3}=\frac{x}{6}\)

Short Answer

Expert verified
x = -\frac{2}{5}

Step by step solution

01

Find a common denominator for the fractions on the left side

The fractions \(\frac{3}{5}\) and \(\frac{2}{3}\) have different denominators. To subtract them, convert them to have a common denominator. The least common multiple (LCM) of 5 and 3 is 15.
02

Convert the fractions to have a common denominator

Convert \(\frac{3}{5}\) to \(\frac{3 \times 3}{5 \times 3} = \frac{9}{15}\). Convert \(\frac{2}{3}\) to \(\frac{2 \times 5}{3 \times 5} = \frac{10}{15}\).
03

Subtract the fractions

Now subtract \(\frac{9}{15} - \frac{10}{15} = \frac{9-10}{15} = \frac{-1}{15}\). The equation now is \(\frac{-1}{15} = \frac{x}{6}\).
04

Solve for x

To solve \(\frac{-1}{15} = \frac{x}{6}\), cross-multiply to get \(-1 \times 6 = 15 \times x\), hence \(-6 = 15x\).
05

Isolate x

Divide both sides by 15 to isolate x: \(-6 \div 15 = x\), hence \(-\frac{6}{15} = x\). Simplify the fraction: \(-\frac{6}{15} = -\frac{2}{5}\). So, \ x = -\frac{2}{5} \.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Common Denominator
When dealing with fractions, it's often necessary to find a common denominator. This makes operations like addition and subtraction much easier.

A common denominator is a common multiple of the denominators of two or more fractions. For instance, in the problem \(\frac{3}{5}-\frac{2}{3} = \frac{x}{6}\), the denominators are 5 and 3.

To subtract these fractions, we need to find a common denominator. The least common multiple (LCM) of 5 and 3 is 15.

Here's how to convert the fractions:
  • \(\frac{3}{5}\) becomes \(\frac{9}{15}\) because \(3 \times 3 = 9\) and \(5 \times 3 = 15\).
  • \(\frac{2}{3}\) converts to \(\frac{10}{15}\) because \(2 \times 5 = 10\) and \(3 \times 5 = 15\).
Now both fractions can be easily subtracted because they have the same denominator.
Cross-Multiplication
Cross-multiplication is a powerful method for solving equations involving fractions. It usually comes into play when you have an equation where a fraction is equal to another fraction, like in the simplified step: \(\frac{-1}{15} = \frac{x}{6}\).

Here's the basic idea: you multiply the numerator of one fraction by the denominator of the other fraction, and set the two products equal to each other.

For instance:
  • In the equation \(\frac{-1}{15} = \frac{x}{6}\), cross-multiplying gives you \(-1 \times 6 = 15 \times x\).
  • This simplifies down to \(-6 = 15x\).
After cross-multiplying, you're left with a simple algebraic equation that you can solve for \(x\). This makes it much easier to isolate the variable and find the solution.
Subtracting Fractions
Subtracting fractions requires a few specific steps to ensure accuracy. It's crucial to make sure the fractions have a common denominator before you subtract them.

In our example, \(\frac{3}{5}-\frac{2}{3} = \frac{x}{6}\), we first converted the fractions to have a common denominator:
  • \(\frac{3}{5}\) became \(\frac{9}{15}\).
  • \(\frac{2}{3}\) became \(\frac{10}{15}\).

Once they have a common denominator, you can simply subtract the numerators:
  • \(\frac{9}{15} - \frac{10}{15} = \frac{9-10}{15} = \frac{-1}{15}\).
Now you can see that the fraction on the left side of the equation is \(\frac{-1}{15}\). This simplifies the subtraction process and prepares the equation for further steps, like cross-multiplication, to solve for \(x\).

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