 |
 |
 |
 |
|
|
|
|
|
When evaluating expressions, we first must consider the order of operations, PE[MD][AS].
(1) 3 + 2 • 8 – 4 = 15
| For this expression, we work through the order of operations: P-Parentheses, E-Exponents, [M-Multiplication and D-Division], [A-Addition and S-Subtraction]. This is worked out in detail on the right. |
3 + 16 – 4 = 19 – 4 = 15
|
(2) {4 [2 (2 + 2)3 ÷ 2]} ÷ 2 = 128
This expression has a number of sets of parentheses and brackets that group parts of the expression. For each set of grouping symbols, the order of operations holds.
| Working from the inside out, we perform the order of operations on the inner set of parentheses, including the exponent operation associated with that set of parentheses. |
{4 [2 (2 + 2)3 ÷ 2]} ÷ 2 =
|
| Now we look at the inner set of brackets. Here we follow the order of operations to evaluate the expression contained in the brackets. |
{4 [128 ÷ 2]} ÷ 2=
|
| Now we have the final set of brackets to work with. In this case, there is only a multiplication to perform inside these braces. |
{4 [64]} ÷ 2 =
|
| Now that there are no grouping symbols left we perform the final operation, division. |
256 ÷ 2= 128
|
(3) (5rs + 7r) • r(4
÷ 2)3 = 40r2s + 56r2
In this expression, we have parentheses.
| First we evaluate the expressions inside the parentheses. The first set of parentheses contains two terms to be added. Since these are not like terms we cannot add them. But we can perform the indicated operations associated with the second set of parentheses. First divide 4 by 2, then perform the operation indicated by the exponent. Then multiply the result by the variable r. |
(5rs + 7r) • r(4 ÷
2)3 =
|
| By the distributive property we multiply 8r through the parentheses. |
(5rs + 7r) • 8r =
|
(4) 3b2 + 5b2 ÷
2b • 8b – 4b = 23b2
+ 4b
For this expression we work through the order of operations:
P-Parentheses, E-Exponents, M-Multiplication, D-Division, A-Addition,
D-division.
| There are no parentheses or exponents to perform so we move on to the multiplication and division in this expression. |
3b2 + 5b2 ÷ 2b
• 8b – 4b = |
| Now we are left with addition and subtraction. We add the two like terms, 3b2 and 20b2, but we cannot perform the subtraction since 23b2 and 4b are not like terms. | 3b2 + 20b2 – 4b = 23b2 – 4b |
NOTE: After reading through these detailed solutions and reviewing the parts of this unit where you are having problems, try the additional practice for this unit.
|
|
|
|
|
|
|
|
|
|