Solve {l}{-|-33|+(-12)+18+|45-40|-57}{|40-37|-|13-52|}

$$\left. \begin{array} { l } { - | - 33 | + ( - 12 ) + 18 + | 45 - 40 | - 57 } \\ { | 40 - 37 | - | 13 - 52 | } \end{array} \right.$$

Sort

$$-79,\ -36$$

Solution Steps

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -33 is 33.

$$sort(-33-12+18+|45-40|-57,|40-37|-|13-52|)$$

Subtract 12 from -33 to get -45.

$$sort(-45+18+|45-40|-57,|40-37|-|13-52|)$$

Add -45 and 18 to get -27.

$$sort(-27+|45-40|-57,|40-37|-|13-52|)$$

Subtract 40 from 45 to get 5.

$$sort(-27+|5|-57,|40-37|-|13-52|)$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of 5 is 5.

$$sort(-27+5-57,|40-37|-|13-52|)$$

Add -27 and 5 to get -22.

$$sort(-22-57,|40-37|-|13-52|)$$

Subtract 57 from -22 to get -79.

$$sort(-79,|40-37|-|13-52|)$$

Subtract 37 from 40 to get 3.

$$sort(-79,|3|-|13-52|)$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of 3 is 3.

$$sort(-79,3-|13-52|)$$

Subtract 52 from 13 to get -39.

$$sort(-79,3-|-39|)$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -39 is 39.

$$sort(-79,3-39)$$

Subtract 39 from 3 to get -36.

$$sort(-79,-36)$$

The list values are already in order.

$$-79,-36$$

Evaluate

$$-79,\ -36$$