What is 7/3+4-8/3+2 1/3 ?

The solution to 7/3+4-8/3+2 1/3 is 6

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7/3+4-8/3+2 1/3

\frac{7}{3} + 4 - \frac{8}{3} + 2 \frac{1}{3}

6

Solution steps

\frac{7}{3} + 4 - \frac{8}{3} + 2 \frac{1}{3}

Convert mixed numbers to improper fractions:

2 \frac{1}{3} = \frac{7}{3}

2 \frac{1}{3}

Convert mixed numbers to improper fraction:

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

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

= \frac{7}{3}

= \frac{7}{3} + 4 - \frac{8}{3} + \frac{7}{3}

Follow the PEMDAS order of operations

Add and subtract (left to right)

\frac{7}{3} + 4 - \frac{8}{3} + \frac{7}{3} : 6

\frac{7}{3} + 4 - \frac{8}{3} + \frac{7}{3}

\frac{7}{3} + 4 = \frac{19}{3}

\frac{7}{3} + 4

Convert element to fraction:

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

= \frac{4 \cdot 3}{3} + \frac{7}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{4 \cdot 3 + 7}{3}

4 \cdot 3 + 7 = 19

4 \cdot 3 + 7

Multiply the numbers:

4 \cdot 3 = 12

= 12 + 7

Add the numbers:

12 + 7 = 19

= 19

= \frac{19}{3}

= \frac{19}{3} - \frac{8}{3} + \frac{7}{3}

\frac{19}{3} - \frac{8}{3} = \frac{11}{3}

\frac{19}{3} - \frac{8}{3}

Apply rule

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

= \frac{19 - 8}{3}

Subtract the numbers:

19 - 8 = 11

= \frac{11}{3}

= \frac{11}{3} + \frac{7}{3}

\frac{11}{3} + \frac{7}{3} = 6

\frac{11}{3} + \frac{7}{3}

Apply rule

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

= \frac{11 + 7}{3}

Add the numbers:

11 + 7 = 18

= \frac{18}{3}

Divide the numbers:

\frac{18}{3} = 6

= 6

= 6

= 6

- (3 - 3)^{2} - 7

3^{3} \div 3^{2} + 5 \times (6^{2} - 2^{4}) - 3

- 3 \cdot (72 - 138)

108 - 180 + 63

4 \cdot 3^{4}