What is (13/5-1/2)+1/6 ?

The solution to (13/5-1/2)+1/6 is 2 4/15

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(13/5-1/2)+1/6

Solution

(\frac{13}{5} - \frac{1}{2}) + \frac{1}{6}

Solution

2 \frac{4}{15}

Solution steps

(\frac{13}{5} - \frac{1}{2}) + \frac{1}{6}

Follow the PEMDAS order of operations

Calculate within parentheses

(\frac{13}{5} - \frac{1}{2}) : \frac{21}{10}

\frac{13}{5} - \frac{1}{2}

\frac{13}{5} - \frac{1}{2} = \frac{21}{10}

\frac{13}{5} - \frac{1}{2}

Least Common Multiplier of

5, 2 : 10

5, 2

Least Common Multiplier (LCM)

Prime factorization of

5 : 5

5

5

is a prime number, therefore no factorization is possible

= 5

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Multiply each factor the greatest number of times it occurs in either

5

2

= 5 \cdot 2

Multiply the numbers:

5 \cdot 2 = 10

= 10

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

10

For

\frac{13}{5} :

multiply the denominator and numerator by

2

\frac{13}{5} = \frac{13 \cdot 2}{5 \cdot 2} = \frac{26}{10}

For

\frac{1}{2} :

multiply the denominator and numerator by

5

\frac{1}{2} = \frac{1 \cdot 5}{2 \cdot 5} = \frac{5}{10}

= \frac{26}{10} - \frac{5}{10}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{26 - 5}{10}

Subtract the numbers:

26 - 5 = 21

= \frac{21}{10}

= \frac{21}{10}

= \frac{21}{10} + \frac{1}{6}

Add and subtract (left to right)

\frac{21}{10} + \frac{1}{6} : \frac{34}{15}

\frac{21}{10} + \frac{1}{6}

\frac{21}{10} + \frac{1}{6} = \frac{34}{15}

\frac{21}{10} + \frac{1}{6}

Least Common Multiplier of

10, 6 : 30

10, 6

Least Common Multiplier (LCM)

Prime factorization of

10 : 2 \cdot 5

10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 5

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Multiply each factor the greatest number of times it occurs in either

10

6

= 2 \cdot 5 \cdot 3

Multiply the numbers:

2 \cdot 5 \cdot 3 = 30

= 30

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

30

For

\frac{21}{10} :

multiply the denominator and numerator by

3

\frac{21}{10} = \frac{21 \cdot 3}{10 \cdot 3} = \frac{63}{30}

For

\frac{1}{6} :

multiply the denominator and numerator by

5

\frac{1}{6} = \frac{1 \cdot 5}{6 \cdot 5} = \frac{5}{30}

= \frac{63}{30} + \frac{5}{30}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{63 + 5}{30}

Add the numbers:

63 + 5 = 68

= \frac{68}{30}

Cancel the common factor:

2

= \frac{34}{15}

= \frac{34}{15}

= \frac{34}{15}

Convert improper fractions to mixed numbers:

\frac{34}{15} = 2 \frac{4}{15}

\frac{34}{15} = 2

Remainder

4

\frac{34}{15}

Write the problem in long division format

15 \overline{∣ 34}

Divide

34

15

to get

2

Divide

34

15

to get

2

2 15 \overline{∣ 34}

Multiply the quotient digit

(2)

by the divisor

15

2 15 \overline{∣ 34} \underline{30}

Subtract

30

from

34

2 15 \overline{∣ 34} \underline{30} 4

2 15 \overline{∣ 34} \underline{30} 4

The solution for Long Division of

\frac{34}{15}

2

with remainder of

4

2 Remainder 4

Convert to mixed number: Quotient

\frac{Remainder}{Divisor}

\frac{34}{15} = 2 \frac{4}{15}

= 2 \frac{4}{15}

= 2 \frac{4}{15}

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Frequently Asked Questions (FAQ)

What is (13/5-1/2)+1/6 ?
The solution to (13/5-1/2)+1/6 is 2 4/15