Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
4. Solving Equations With Variables on Both Sides
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Exercise 42 Page 107

Write an expression for each day's work and then equate them.

60 potatoes

Practice makes perfect

We have been asked to calculate the number of potatoes in a bucket. Let's begin by answering the questions posed.

Known Quantities

From the problem we know the amount of time it took the deli worker to complete the pies on each day, as well as the rate at which the potatoes were peeled. Let's write the relationship as an equation. Time=Pies+Rate* Number of potatoes

Unknown Quantities

The unknown value is the number of potatoes. The known values are the times to make pies and the rates to peel potatoes. First, we need to find the product of the number of potatoes p and the rates.

1.5p and 1p Then we add the respective time to make pies to each of these. Since it took the same total time on both days, these expressions can then be set equal to one another to form an equation.

Solve the Problem

We are assuming that a bucket roughly holds the same number of potatoes each day, p. Since we have two different time units, we need to choose which unit to work with. For simplicity, let's use minutes.

Phrase Expression
2 hours to make pies on Monday 2 hours = 120 minutes
1.5 minutes to peel each potato on Monday 1.5p
2.5 hours to make pies on Tuesday 2.5 hours = 150 minutes
1 minute to peel each potato on Tuesday 1p
Now we can write an expression describing how many minutes the deli worker took to complete the work on Monday. 120+1.5p Next we can write an expression for the number of minutes the deli worker took on Tuesday. 150+p Since the total amount of time for both days was the same, we can form an equation by setting the two expressions equal to each other. 120+1.5p= 150+p From there we solve for p to find the number of potatoes. To do this we will use the inverse operations.
120+1.5p=150+p
â–Ľ
LHS-120=RHS-120
120+1.5p-120=150+p-120
1.5p=30+p
â–Ľ
LHS-p=RHS-p
1.5p-p=30+p-p
0.5p=30
â–Ľ
.LHS /0.5.=.RHS /0.5.
0.5p/0.5=30/0.5
p=60
We have shown that there are 60 potatoes in a bucket.

Checking Our Answer

Checking our solution
Now we can check our work by substituting 60 for each instance of p in the original equation.
120+1.5p=150+p
120+1.5( 60)? = 150+ 60
120+90 ? = 150+60
210=210
Since we found a statement that is always true, we solved the equation correctly with a solution of p=60.