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Kiara's question at Yahoo! Answers regarding addition/subtraction of mixed numbers

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MarkFL

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Feb 24, 2012
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MarkFL

Administrator
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Feb 24, 2012
13,775
Hello Kiara,

I would convert the mixed numbers to improper fractions, get a common denominator, then add or subtract, and the convert back to a mixed number

A mixed number consists of an integer and a proper fraction. A proper fraction simply means the numerator is less than the denominator, otherwise it is an improper fraction.

To convert a mixed number to an improper fraction, take the denominator in the fractional part, multiply it by the integer, then add the numerator of the fractional part, and this is the numerator of the improper fraction. The denominator is the same as the denominator of the fractional part.

So, let's convert the 3 problems to improper fractions:

1.) \(\displaystyle 3\frac{1}{2}-2\frac{3}{4}=\frac{2\cdot3+1}{2}-\frac{4\cdot2+3}{4}=\frac{7}{2}-\frac{11}{4}\)

2.) \(\displaystyle 2\frac{1}{10}+2\frac{1}{6}=\frac{10\cdot2+1}{10}+ \frac{6\cdot2+1}{6}=\frac{21}{10}+\frac{13}{6}\)

3.) \(\displaystyle 8\frac{1}{3}-2\frac{1}{8}=\frac{3\cdot8+1}{3}-\frac{8\cdot2+1}{8}=\frac{25}{3}-\frac{17}{8}\)

Now we want to get common denominators, so we use the least common multiple LCM as the least common denominator LCD.

1.) The denominators here are 2 and 4, and since 4 is a multiple of 2, the LCD is 4, and in order to give the first term a denominator of 4 without changing its value we will multiply by \(\displaystyle 1=\frac{2}{2}\) and the difference becomes:

\(\displaystyle \frac{7}{2}\cdot\frac{2}{2}-\frac{11}{4}=\frac{14}{4}-\frac{11}{4}=\frac{14-11}{4}=\frac{3}{4}\)

2.) The denominators here are 6 and 10. If you are unsure what the LCM is, you may find the prime factorization of both numbers, and take the highest power of each factor found in either factorization:

$6=2\cdot3$

$10=2\cdot5$

So the LCM is then:

$2\cdot3\cdot5=30$

The first term need to be multiplied by \(\displaystyle 1=\frac{3}{3}\) and the second term needs to be multiplied by \(\displaystyle 1=\frac{5}{5}\). Notice these are formed from the factors in the LCM not present in the original denominators. So we have:

\(\displaystyle \frac{21}{10}\cdot\frac{3}{3}+\frac{13}{6} \cdot\frac{5}{5}=\frac{63}{30}+\frac{65}{30}=\frac{63+65}{30}=\frac{128}{30}=\frac{2\cdot64}{2\cdot15}=\frac{64}{15}\)

3.) The denominators here are 3 and 8, and if we make the observation that these numbers are co-prime, that is they share no common factors, we may simply take the LCM as the product of the two numbers $3\cdot8=24$. So we want to multiply the first term by \(\displaystyle 1=\frac{8}{8}\) and the second term by \(\displaystyle 1=\frac{3}{3}\) to get:

\(\displaystyle \frac{25}{3}\cdot\frac{8}{8}-\frac{17}{8}\cdot\frac{3}{3}=\frac{200}{24}-\frac{51}{24}=\frac{149}{24}\)

Now, the last step is to convert these results to mixed numbers, as this is how the numbers in the problem were originally given. To do this, we find the greatest multiple of the denominator that is less than or equal to the numerator, and express the numerator as the quotient and remainder:

1.) This result is a proper fraction so there is no need to convert. It is \(\displaystyle \frac{3}{4}\).

2.) \(\displaystyle \frac{64}{15}=\frac{4\cdot15+4}{15}=4+\frac{4}{15}=4\frac{4}{15}\)

3.) \(\displaystyle \frac{149}{24}=\frac{6\cdot24+5}{24}=6+\frac{5}{24}=6\frac{5}{24}\)

To Kiara and other visitors viewing this topic, I would encourage you to register and post similar questions in our Pre-Algebra and Algebra forum. (Happy)