Solve -8+|-6+4|*5-(3-|-10-32|:7)

$$- 8 + | - 6 + 4 | \cdot 5 - ( 3 - | - 10 - 32 | : 7 )$$

Evaluate

$$5$$

Solution Steps

Add -6 and 4 to get -2.

$$-8+|-2|\times 5-\left(3-\frac{|-10-32|}{7}\right)$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -2 is 2.

$$-8+2\times 5-\left(3-\frac{|-10-32|}{7}\right)$$

Multiply 2 and 5 to get 10.

$$-8+10-\left(3-\frac{|-10-32|}{7}\right)$$

Add -8 and 10 to get 2.

$$2-\left(3-\frac{|-10-32|}{7}\right)$$

Subtract 32 from -10 to get -42.

$$2-\left(3-\frac{|-42|}{7}\right)$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -42 is 42.

$$2-\left(3-\frac{42}{7}\right)$$

Divide 42 by 7 to get 6.

$$2-\left(3-6\right)$$

Subtract 6 from 3 to get -3.

$$2-\left(-3\right)$$

The opposite of -3 is 3.

$$2+3$$

Add 2 and 3 to get 5.

$$5$$

Factor

$$5$$