What is 12-[9+(7/6-5/4)-(1/2 × 3/4)] ?

The solution to 12-[9+(7/6-5/4)-(1/2 × 3/4)] is 3 11/24

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Popular Pre Algebra >

12-[9+(7/6-5/4)-(1/2 × 3/4)]

Solution

12 - [9 + (\frac{7}{6} - \frac{5}{4}) - (\frac{1}{2} \times \frac{3}{4})]

Solution

3 \frac{11}{24}

Decimal

3.45833 \dots

Solution steps

12 - [9 + (\frac{7}{6} - \frac{5}{4}) - (\frac{1}{2} \cdot \frac{3}{4})]

Follow the PEMDAS order of operations

Calculate within parentheses

[9 + (\frac{7}{6} - \frac{5}{4}) - (\frac{1}{2} \cdot \frac{3}{4})] : \frac{205}{24}

9 + (\frac{7}{6} - \frac{5}{4}) - (\frac{1}{2} \cdot \frac{3}{4})

Calculate within parentheses

(\frac{7}{6} - \frac{5}{4}) : - \frac{1}{12}

\frac{7}{6} - \frac{5}{4}

\frac{7}{6} - \frac{5}{4} = - \frac{1}{12}

\frac{7}{6} - \frac{5}{4}

Least Common Multiplier of

6, 4 : 12

6, 4

Least Common Multiplier (LCM)

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

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

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

6

4

= 2 \cdot 2 \cdot 3

Multiply the numbers:

2 \cdot 2 \cdot 3 = 12

= 12

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

12

For

\frac{7}{6} :

multiply the denominator and numerator by

2

\frac{7}{6} = \frac{7 \cdot 2}{6 \cdot 2} = \frac{14}{12}

For

\frac{5}{4} :

multiply the denominator and numerator by

3

\frac{5}{4} = \frac{5 \cdot 3}{4 \cdot 3} = \frac{15}{12}

= \frac{14}{12} - \frac{15}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{14 - 15}{12}

Subtract the numbers:

14 - 15 = - 1

= \frac{- 1}{12}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{12}

= - \frac{1}{12}

= 9 - \frac{1}{12} - (\frac{1}{2} \cdot \frac{3}{4})

Calculate within parentheses

(\frac{1}{2} \cdot \frac{3}{4}) : \frac{3}{8}

\frac{1}{2} \cdot \frac{3}{4}

\frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}

\frac{1}{2} \cdot \frac{3}{4}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot 3}{2 \cdot 4}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{3}{8}

= \frac{3}{8}

= 9 - \frac{1}{12} - (\frac{1}{2} \cdot \frac{3}{4})

Add and subtract (left to right)

9 - \frac{1}{12} - (\frac{1}{2} \cdot \frac{3}{4}) : \frac{205}{24}

9 - \frac{1}{12} - (\frac{1}{2} \cdot \frac{3}{4})

9 - \frac{1}{12} = \frac{107}{12}

9 - \frac{1}{12}

Convert element to fraction:

9 = \frac{9 \cdot 12}{12}

= \frac{9 \cdot 12}{12} - \frac{1}{12}

Since the denominators are equal, combine the fractions:

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

= \frac{9 \cdot 12 - 1}{12}

9 \cdot 12 - 1 = 107

9 \cdot 12 - 1

Multiply the numbers:

9 \cdot 12 = 108

= 108 - 1

Subtract the numbers:

108 - 1 = 107

= 107

= \frac{107}{12}

= \frac{107}{12} - (\frac{1}{2} \cdot \frac{3}{4})

\frac{107}{12} - (\frac{1}{2} \cdot \frac{3}{4}) = \frac{205}{24}

= \frac{205}{24}

= \frac{205}{24}

= 12 - \frac{205}{24}

Add and subtract (left to right)

12 - \frac{205}{24} : \frac{83}{24}

12 - \frac{205}{24}

12 - \frac{205}{24} = \frac{83}{24}

12 - \frac{205}{24}

Convert element to fraction:

12 = \frac{12 \cdot 24}{24}

= \frac{12 \cdot 24}{24} - \frac{205}{24}

Since the denominators are equal, combine the fractions:

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

= \frac{12 \cdot 24 - 205}{24}

12 \cdot 24 - 205 = 83

12 \cdot 24 - 205

Multiply the numbers:

12 \cdot 24 = 288

= 288 - 205

Subtract the numbers:

288 - 205 = 83

= 83

= \frac{83}{24}

= \frac{83}{24}

= \frac{83}{24}

Convert improper fractions to mixed numbers:

\frac{83}{24} = 3 \frac{11}{24}

\frac{83}{24} = 3

Remainder

11

\frac{83}{24}

Write the problem in long division format

24 \overline{∣ 83}

Divide

83

24

to get

3

Divide

83

24

to get

3

3 24 \overline{∣ 83}

Multiply the quotient digit

(3)

by the divisor

24

3 24 \overline{∣ 83} \underline{72}

Subtract

72

from

83

3 24 \overline{∣ 83} \underline{72} 11

3 24 \overline{∣ 83} \underline{72} 11

The solution for Long Division of

\frac{83}{24}

3

with remainder of

11

3 Remainder 11

Convert to mixed number: Quotient

\frac{Remainder}{Divisor}

\frac{83}{24} = 3 \frac{11}{24}

= 3 \frac{11}{24}

= 3 \frac{11}{24}

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Frequently Asked Questions (FAQ)

What is 12-[9+(7/6-5/4)-(1/2 × 3/4)] ?
The solution to 12-[9+(7/6-5/4)-(1/2 × 3/4)] is 3 11/24

12-[9+(7/6-5/4)-(1/2 × 3/4)]

Solution

Solution

Popular Examples

Frequently Asked Questions (FAQ)

What is 12-[9+(7/6-5/4)-(1/2 × 3/4)] ?

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