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

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1/4 (-1)+1

Solution

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

\frac{3}{4}

Solution steps

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

Follow the PEMDAS order of operations

Multiply and divide (left to right)

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

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

Apply rule

a \cdot (- b) = - a \cdot b

\frac{1}{4} (- 1) = - \frac{1}{4} \cdot 1 = - \frac{1}{4}

= - \frac{1}{4}

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

Add and subtract (left to right)

- \frac{1}{4} + 1 : \frac{3}{4}

- \frac{1}{4} + 1

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

- \frac{1}{4} + 1

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 \cdot 4 - 1 = 3

1 \cdot 4 - 1

Multiply the numbers:

1 \cdot 4 = 4

= 4 - 1

Subtract the numbers:

4 - 1 = 3

= 3

= \frac{3}{4}

= \frac{3}{4}

= \frac{3}{4}

\frac{5}{6} (\frac{2}{3} - \frac{1}{6})

3 \times + 2 \times 3

2 \cdot \frac{16}{41} + 5 \cdot \frac{25}{41}

4 - 6 - 9 - 7 - 2 + 8

(8 - \frac{3}{5}) \div 3 \frac{1}{2}