What is 17/39-(4/39+1/9) ?

The solution to 17/39-(4/39+1/9) is 2/9

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Popular >

17/39-(4/39+1/9)

Solution

\frac{17}{39} - (\frac{4}{39} + \frac{1}{9})

Solution

\frac{2}{9}

Solution steps

\frac{17}{39} - (\frac{4}{39} + \frac{1}{9})

Follow the PEMDAS order of operations

Calculate within parentheses

(\frac{4}{39} + \frac{1}{9}) : \frac{25}{117}

\frac{4}{39} + \frac{1}{9}

\frac{4}{39} + \frac{1}{9} = \frac{25}{117}

\frac{4}{39} + \frac{1}{9}

Least Common Multiplier of

39, 9 : 117

39, 9

Least Common Multiplier (LCM)

Prime factorization of

39 : 3 \cdot 13

39

39

divides by

339 = 13 \cdot 3

= 3 \cdot 13

3, 13

are all prime numbers, therefore no further factorization is possible

= 3 \cdot 13

Prime factorization of

9 : 3 \cdot 3

9

9

divides by

39 = 3 \cdot 3

= 3 \cdot 3

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

39

9

= 3 \cdot 3 \cdot 13

Multiply the numbers:

3 \cdot 3 \cdot 13 = 117

= 117

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

117

For

\frac{4}{39} :

multiply the denominator and numerator by

3

\frac{4}{39} = \frac{4 \cdot 3}{39 \cdot 3} = \frac{12}{117}

For

\frac{1}{9} :

multiply the denominator and numerator by

13

\frac{1}{9} = \frac{1 \cdot 13}{9 \cdot 13} = \frac{13}{117}

= \frac{12}{117} + \frac{13}{117}

Since the denominators are equal, combine the fractions:

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

= \frac{12 + 13}{117}

Add the numbers:

12 + 13 = 25

= \frac{25}{117}

= \frac{25}{117}

= \frac{17}{39} - \frac{25}{117}

Add and subtract (left to right)

\frac{17}{39} - \frac{25}{117} : \frac{2}{9}

\frac{17}{39} - \frac{25}{117}

\frac{17}{39} - \frac{25}{117} = \frac{2}{9}

\frac{17}{39} - \frac{25}{117}

Least Common Multiplier of

39, 117 : 117

39, 117

Least Common Multiplier (LCM)

Prime factorization of

39 : 3 \cdot 13

39

39

divides by

339 = 13 \cdot 3

= 3 \cdot 13

3, 13

are all prime numbers, therefore no further factorization is possible

= 3 \cdot 13

Prime factorization of

117 : 3 \cdot 3 \cdot 13

117

117

divides by

3117 = 39 \cdot 3

= 3 \cdot 39

39

divides by

339 = 13 \cdot 3

= 3 \cdot 3 \cdot 13

3, 13

are all prime numbers, therefore no further factorization is possible

= 3 \cdot 3 \cdot 13

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

39

117

= 3 \cdot 3 \cdot 13

Multiply the numbers:

3 \cdot 3 \cdot 13 = 117

= 117

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

117

For

\frac{17}{39} :

multiply the denominator and numerator by

3

\frac{17}{39} = \frac{17 \cdot 3}{39 \cdot 3} = \frac{51}{117}

= \frac{51}{117} - \frac{25}{117}

Since the denominators are equal, combine the fractions:

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

= \frac{51 - 25}{117}

Subtract the numbers:

51 - 25 = 26

= \frac{26}{117}

Cancel the common factor:

13

= \frac{2}{9}

= \frac{2}{9}

= \frac{2}{9}

Popular Examples

33+(-27)-(-35)+(-47)

33 + (- 27) - (- 35) + (- 47)

1540\div 14+3960\div 12+240\div 6

1540 \div 14 + 3960 \div 12 + 240 \div 6

180/6-3*(4*(8-6)-30/10*2)

180/6 - 3 \cdot (4 \cdot (8 - 6) - 30/10 \cdot 2)

7-2^2

7 - 2^{2}

5^2-3\times 2^2-1

5^{2} - 3 \times 2^{2} - 1

Frequently Asked Questions (FAQ)

What is 17/39-(4/39+1/9) ?
The solution to 17/39-(4/39+1/9) is 2/9