Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
7. The Distributive Property
Continue to next subchapter

Exercise 74 Page 51

Construct a numerical expression that indicates how many gallons of water per minute we save when using the new shower head.

36 gallons

Practice makes perfect

Notice that the old shower head uses 7 gallons of water per minute, while the new one uses only 2.5. Old:& 7 galmin New:& 2.5 galmin From this information we can construct a numerical expression that indicates how many gallons of water per minute we will save when using the new shower head.

Writing the Expression

The amount of water saved per minute is the difference between the water usage of the old and the new shower heads. 7 gal - 2.5 gal

Using the Distributive Property

Multiplying the difference of the water usage by the time it takes to take a shower will give us two things.

  1. The amount of water we use with the old shower head.
  2. The amount of water we use with the new shower head.
By calculating the difference, we will be able to find the gallons of water saved with the new shower head. Let's use the Distributive Property.
8 min (7 galmin - 2.5 galmin)
8 min * 7 galmin - 8 min * 2.5 galmin
8 min * 7 Galmin - 8 min * 2.5 Galmin
56 gal - 20 gal
Here, 56 gallons is the amount of water that we use with the old shower head. With the new one, 20 gallons of water are used. Let's calculate the difference.
56 gal - 20 gal
36 Gal
Therefore, the total amount of water we will save is 36 gallons.