What is (4/3-1)-(2/3+3) ?

The solution to (4/3-1)-(2/3+3) is -10/3

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(4/3-1)-(2/3+3)

(\frac{4}{3} - 1) - (\frac{2}{3} + 3)

- \frac{10}{3}

Solution steps

(\frac{4}{3} - 1) - (\frac{2}{3} + 3)

Follow the PEMDAS order of operations

Calculate within parentheses

(\frac{4}{3} - 1) : \frac{1}{3}

\frac{4}{3} - 1

\frac{4}{3} - 1 = \frac{1}{3}

\frac{4}{3} - 1

Convert element to fraction:

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

= - \frac{1 \cdot 3}{3} + \frac{4}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot 3 + 4}{3}

- 1 \cdot 3 + 4 = 1

- 1 \cdot 3 + 4

Multiply the numbers:

1 \cdot 3 = 3

= - 3 + 4

Add/Subtract the numbers:

- 3 + 4 = 1

= 1

= \frac{1}{3}

= \frac{1}{3}

= \frac{1}{3} - (\frac{2}{3} + 3)

Calculate within parentheses

(\frac{2}{3} + 3) : \frac{11}{3}

\frac{2}{3} + 3

\frac{2}{3} + 3 = \frac{11}{3}

\frac{2}{3} + 3

Convert element to fraction:

3 = \frac{3 \cdot 3}{3}

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

Since the denominators are equal, combine the fractions:

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

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

3 \cdot 3 + 2 = 11

3 \cdot 3 + 2

Multiply the numbers:

3 \cdot 3 = 9

= 9 + 2

Add the numbers:

9 + 2 = 11

= 11

= \frac{11}{3}

= \frac{11}{3}

= \frac{1}{3} - \frac{11}{3}

Add and subtract (left to right)

\frac{1}{3} - \frac{11}{3} : - \frac{10}{3}

\frac{1}{3} - \frac{11}{3}

\frac{1}{3} - \frac{11}{3} = - \frac{10}{3}

\frac{1}{3} - \frac{11}{3}

Apply rule

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

= \frac{1 - 11}{3}

Subtract the numbers:

1 - 11 = - 10

= \frac{- 10}{3}

Apply the fraction rule:

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

= - \frac{10}{3}

= - \frac{10}{3}

= - \frac{10}{3}

3 \frac{1}{15} \div 4 - \frac{7}{2}

4 (- 2)^{2} - 6

6 (9) + 5

(- 2) \cdot (- 7) + (- 3)^{2}

(3 - \frac{1}{4}) \div (2 - \frac{1}{6})