Solve (-2*9)div(-2-1)-3*(7-5)+2*(-5)*|-3-1|div2

$$( - 2 \times 9 ) \div ( - 2 - 1 ) - 3 \times ( 7 - 5 ) + 2 \times ( - 5 ) \times | - 3 - 1 | \div 2$$

Evaluate

$$-20$$

Solution Steps

Multiply -2 and 9 to get -18.

$$\frac{-18}{-2-1}-3\left(7-5\right)+\frac{2\left(-5\right)|-3-1|}{2}$$

Subtract 1 from -2 to get -3.

$$\frac{-18}{-3}-3\left(7-5\right)+\frac{2\left(-5\right)|-3-1|}{2}$$

Divide -18 by -3 to get 6.

$$6-3\left(7-5\right)+\frac{2\left(-5\right)|-3-1|}{2}$$

Subtract 5 from 7 to get 2.

$$6-3\times 2+\frac{2\left(-5\right)|-3-1|}{2}$$

Multiply 3 and 2 to get 6.

$$6-6+\frac{2\left(-5\right)|-3-1|}{2}$$

Subtract 6 from 6 to get 0.

$$0+\frac{2\left(-5\right)|-3-1|}{2}$$

Cancel out 2 and 2.

$$0-5|-3-1|$$

Anything plus zero gives itself.

$$-5|-3-1|$$

Subtract 1 from -3 to get -4.

$$-5|-4|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -4 is 4.

$$-5\times 4$$

Multiply -5 and 4 to get -20.

$$-20$$

Factor

$$-20$$