Question

Simplify. Leave your answers as improper fractions. $$\frac{\frac{2}{3}+\frac{3}{4}}{\frac{1}{5}}$$

Short answer

\frac{85}{12}

Step-by-step solution

Simplify the Numerator

Add the fractions in the numerator. Since the denominators are not the same, find a common denominator, which for 3 and 4 is 12. Convert each fraction to have the common denominator:\[\frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}\] and\[\frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12}.\]Now add the fractions:\[\frac{8}{12} + \frac{9}{12} = \frac{8 + 9}{12} = \frac{17}{12}.\]

Recall Division of Fractions

Remember that dividing by a fraction is the same as multiplying by its reciprocal. The problem \[\frac{\frac{17}{12}}{\frac{1}{5}}\] becomes \[\frac{17}{12} \times \frac{5}{1}.\]

Multiply the Fractions

Multiply the numerators together and the denominators together:\[\frac{17}{12} \times \frac{5}{1} = \frac{17 \times 5}{12 \times 1} = \frac{85}{12}.\]This is the final simplified result as an improper fraction.

Key concepts

Improper Fractions

Improper fractions are fractions where the numerator (the top number) is larger than or equal to the denominator (the bottom number). This indicates that the fraction is greater than or equal to one whole unit. For example, in the simplified solution \(\frac{85}{12}\), the numerator (85) is larger than the denominator (12), making it an improper fraction.

Improper fractions are often encountered when adding or subtracting fractions, as well as when working with mixed numbers. They are just as valid as proper fractions (where the numerator is less than the denominator), and can be converted into mixed numbers if desired. However, in certain cases such as algebraic calculations, working with improper fractions can be preferable, as it keeps the format consistent and avoids additional steps of converting back and forth between mixed numbers and fractions.

Common Denominators

Finding common denominators is crucial when adding or subtracting fractions. It refers to making the denominators of two or more fractions the same. To add \(\frac{2}{3}\) and \(\frac{3}{4}\), we need to find a number that both 3 and 4 divide into evenly. In our example, 12 serves as the common denominator.

By converting \(\frac{2}{3}\) to \(\frac{8}{12}\) and \(\frac{3}{4}\) to \(\frac{9}{12}\), we could add them directly. This process involves multiplying the numerator and the denominator of each fraction by a factor that makes the denominators equal.

• For \(\frac{2}{3}\), we multiply both top and bottom by 4, yielding \(\frac{8}{12}\).
• For \(\frac{3}{4}\), we multiply by 3 to get \(\frac{9}{12}\).

This equal footing is a cornerstone in solving fraction problems as it allows for straightforward addition or subtraction of the numerators.

Dividing Fractions

Dividing fractions might initially seem challenging, but it becomes straightforward once you know the trick: multiply by the reciprocal of the divisor. The reciprocal of a fraction is simply switching the numerator and the denominator.

So, when we divide by a fraction, we flip the second fraction and change the division into multiplication. For our exercise, dividing \(\frac{17}{12}\) by \(\frac{1}{5}\) transforms into \(\frac{17}{12}\) multiplied by \(\frac{5}{1}\), which is the reciprocal of \(\frac{1}{5}\). This method of division removes the need for additional, complex steps and simplifies it to multiplication, a process we're more familiar with.

Multiplying Fractions

Multiplying fractions is a straightforward process; multiply the numerators together and the denominators together. Following our previous step, the division problem \(\frac{17}{12} \div\ \frac{1}{5}\) became a multiplication one: \(\frac{17}{12} \times\ \frac{5}{1}\).

In this step, you multiply the numerators 17 and 5 to get 85, and the denominators 12 and 1 to get 12, resulting in the final answer \(\frac{85}{12}\). Multiplying fractions does not require a common denominator, but when the multiplication results in an improper fraction, you might need to simplify or convert it, depending on the requirements of the problem.