Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
5. Standard Form
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Exercise 5 Page 325

Our variables represent the number of each kind of gift card purchased.

Equation in Standard Form: 10x+25y=285
Example Solutions: See solution.

Practice makes perfect

We are told that gift cards can be purchased for $10 or for $25. We want to know how many of each kind we could have purchased when spending $285.

Write the Equation

First, we need to write an equation in the standard form to model this situation. Let x be the number of $10 gift cards purchased and y be the number of $25 gift cards purchased. We get the following equation. 10x+25y=285

Possible Gift Card Combinations

We need to find three combinations for the numbers of each card purchased that fit our equation. These numbers need to be whole numbers because we cannot purchase a portion of a gift card. The best method here is to guess and check. But, we can make educated guesses! We have some things to consider when making these educated guesses.

  • Think ahead about the steps we will take when solving the equation. For example, will we need to divide by the coefficient.
  • In order to have a whole number for x, we need our y value to be an odd number. An odd number value for y ensures that, when solving for x, our right-hand side is divisible by 10. Let's solve for x when y=1.
10x+25y=285
10x+25( 1)=285
10x+25=285
10x=260
See how the odd number for y gave us a number divisible by 10 on the right-hand side? Let's continue solving.
10x=260
x=26
This means that if we bought 1 $25 gift card, we also bought 26 $10 gift cards. Now let's try substituting y=3 and y=5.
y 10x+25y=285 x
3 10x+25* 3=285 21
5 10x+25* 5=285 16

This means that if we bought 5 $25 gift cards, we also bought 16 $10 gift cards. While there are more possibilities, three example combinations are as follows.

The Number of $25 Gift Cards The Number of $10 Gift Cards
1 26
3 21
5 16