Algebra Examples

$15 \div (- 3) + 5 | - 2 | \div (- 5)$

Step 1

Simplify each term.

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Step 1.1

Divide $15$ by $- 3$ .

$- 5 + 5 | - 2 | \div (- 5)$

Step 1.2

Rewrite the division as a fraction.

$- 5 + \frac{5 | - 2 |}{- 5}$

Step 1.3

Cancel the common factor of $5$ and $- 5$ .

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Step 1.3.1

Factor $5$ out of $5 | - 2 |$ .

$- 5 + \frac{5 (| - 2 |)}{- 5}$

Step 1.3.2

Move the negative one from the denominator of $\frac{| - 2 |}{- 1}$ .

$- 5 - 1 \cdot | - 2 |$

Step 1.4

Rewrite $- 1 \cdot | - 2 |$ as $- | - 2 |$ .

$- 5 - | - 2 |$

Step 1.5

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$- 5 - 1 \cdot 2$

Step 1.6

Multiply $- 1$ by $2$ .

$- 5 - 2$

Step 2

Subtract $2$ from $- 5$ .

$- 7$