McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
2. Order of Operations
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Exercise 60 Page 14

See solution.

Practice makes perfect

To evaluate these expressions, we should use the order of operations to know which operation we should perform first. Let's explain how we would do it for the first expression and then for the second one!

a[(b-c) Ă· d]-f

According to the order of operations, we should begin with evaluating the expressions inside grouping symbols. Therefore, the first expression we should evaluate is inside the brackets. [(b-c) Ă· d]

But, notice that it involves another expression inside parentheses, (b-c). Thus, we will evaluate this one first! [ (b-c) Ă· d] Then, we can divide this difference by d. Our next steps will be any multiplication or division by terms outside of the grouping symbols. Therefore, we should multiply a by the number we found by reducing the expression inside the square brackets. Finally, we can subtract f.

Step Expression to evaluate
1 (b-c)
2 [(b-c) Ă· d]
3 a[(b-c) Ă· d]
4 a[(b-c) Ă· d] -f

a * b-c Ă· d-f

Now let's analyze solving this expression when it is written without the grouping symbols! Since there are no parentheses or powers, according to the order of operations, we will begin with multiplying or dividing from left to right. a * b- c Ă· d -f First we will multiply a and b and then divide c and d. After evaluating those, the expression will involve subtraction only, which is the last step in the order of operations. Therefore, we will subtract the quotient of c and d from the product of a and b first. Finally, we will subtract f from this difference!

Step Expression to evaluate
1 a * b-c Ă· d -f
2 a * b-c Ă· d -f
3 a * b - c Ă· d -f
4 a * b-c Ă· d -f