What is -3\times 1/9+1 ?

The solution to -3\times 1/9+1 is 2/3

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-3\times 1/9+1

- 3 \cdot \frac{1}{9} + 1

\frac{2}{3}

Solution steps

- 3 \cdot \frac{1}{9} + 1

Follow the PEMDAS order of operations

Multiply and divide (left to right)

- 3 \cdot \frac{1}{9} : - \frac{1}{3}

- 3 \cdot \frac{1}{9}

Convert element to fraction:

3 = \frac{3}{1}

= - \frac{3}{1} \cdot \frac{1}{9}

Apply the fraction rule:

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

\frac{3}{1} \cdot \frac{1}{9} = \frac{3 \cdot 1}{1 \cdot 9}

= - \frac{3 \cdot 1}{1 \cdot 9}

\frac{3 \cdot 1}{1 \cdot 9} = \frac{1}{3}

\frac{3 \cdot 1}{1 \cdot 9}

\frac{3 \cdot 1}{1 \cdot 9} = \frac{3}{9}

\frac{3 \cdot 1}{1 \cdot 9}

Multiply the numbers:

3 \cdot 1 = 3

= \frac{3}{1 \cdot 9}

Multiply the numbers:

1 \cdot 9 = 9

= \frac{3}{9}

= \frac{3}{9}

Cancel the numbers:

\frac{3}{9} = \frac{1}{3}

= \frac{1}{3}

= - \frac{1}{3}

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

Add and subtract (left to right)

- \frac{1}{3} + 1 : \frac{2}{3}

- \frac{1}{3} + 1

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

- \frac{1}{3} + 1

Convert element to fraction:

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

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

1 \cdot 3 - 1 = 2

1 \cdot 3 - 1

Multiply the numbers:

1 \cdot 3 = 3

= 3 - 1

Subtract the numbers:

3 - 1 = 2

= 2

= \frac{2}{3}

= \frac{2}{3}

= \frac{2}{3}

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

(\frac{1}{2} + \frac{1}{4}) + \frac{2}{7}

3 \times + 4 \times + 5

2 (- 1)^{2} + 4 (- 1) - 2

2 (- 1)^{2} + 4 (- 1) - 6