Houghton Mifflin Harcourt Algebra 1, 2015
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Houghton Mifflin Harcourt Algebra 1, 2015 View details
2. Creating and Solving Equations
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Exercise 20 Page 53

Use a table to organize the information and see the relationships.

200 words

Practice makes perfect

To find the number of words that were initially on Bill's page, we can use tables to help us organize the information and see the relationships. Let w be the number of words that Charlie's page had in the beginning.

Person Initial number of words
Charlie w
Bill had 50words fewer than Charlie w -50
George had twice as many words as Bill 2(w - 50)

Notice that we only need to look at the number of words for Bill and Charlie since the exercise does not ask about the number of words on George's page. We are told that each person types 60 words per minute. We can write expressions for the number of words Charlie and Bill have after one minute.

Person Number of words after one minute
Charlie w + 60
Bill w - 50 + 60

The exercise then states that, after one minute, the difference between twice the number of words on Bill's page and the number of words on Charlie's page is 210. 2( w-50+60) - ( w +60) = 210 To solve the above equation, we need to isolate w using inverse operations and the Properties of Equality. Notice that we will need to distribute 2 throughout the first set parentheses and -1 throughout the second set.

2(w-50+60) - (w +60) = 210
â–¼
Solve for w
2(w+10) - (w +60) = 210
2w + 20 - w -60 = 210
w -40 =210
w -40 +40 =210 +40
w = 250

Since w is the number of words Charlie's page initially had, his page had 250 words. Now we know that w=250, we can use the expression we initially wrote for Bill to find the number of words that were on his page.

w-50
250-50
200

Bill's page originally contained 200 words.