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Chapter 5: Problem 84

Perform the indicated operations. If possible, reduce the answer to its lowest terms. \(\frac{13}{15}-\frac{2}{45}\)

Short Answer

Expert verified
The result of the operation \(\frac{13}{15}-\frac{2}{45} = \frac{37}{45}\)

Step by step solution

01

Find the Least Common Denominator (LCD)

To subtract fractions, they must have the same denominator. The least common denominator for 15 and 45 is 45 since 45 is a multiple of 15. Hence, adjust the fractions to have the denominator as 45.
02

Adjust the fractions according to the LCD

Adjust the first fraction \(\frac{13}{15}\) to have the denominator as 45. This is done by multiplying the numerator and denominator by 3: \(\frac{13}{15}*\frac{3}{3}=\frac{39}{45}\). The second fraction is already \(\frac{2}{45}\). Hence, the fractions to subtract become \(\frac{39}{45}-\frac{2}{45}\)
03

Subtract fractions

Subtract the numerators of the fractions, keeping the common denominator the same: \(\frac{39}{45}-\frac{2}{45}=\frac{37}{45}\)
04

Reduce the result to the lowest terms

The resulting fraction \(\frac{37}{45}\) is already in its lowest terms since no common number other than 1 divides both 37 and 45 evenly.

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Most popular questions from this chapter

Find the indicated term for the geometric sequence with first term, \(a_{1}\), and common ratio, \(r\). Find \(a_{8}\), when \(a_{1}=12, r=\frac{1}{2}\).

Write the first six terms of the geometric sequence with the first term, \(a_{1}\), and common ratio, \(r\). \(a_{1}=-\frac{1}{16}, r=-4\)

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Write a formula for the general term (the nth term) of each geometric sequence. Then use the formula for \(a_{n}\) to find \(a_{7}\), the seventh term of the sequence. \(18,6,2, \frac{2}{3}, \ldots\)
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