Question

Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and

thumb drive. Troy bought 5 notebooks and 15 thumb drives for $395. Lisa bought 12 notebooks and 8 thumb
drives for $248. Find the cost of each notebook and each thumb drive.
The cost of each notebook =
The cost of each thumb drive =

Answer 1

• notebook: $4
• thumb drive: $25

Explanation:

You want the cost of a notebook and a thumb drive when 5 notebooks and 15 thumb drives costs $395, while 12 notebooks and 8 thumb drives cost $248.

Equations

The two purchases can be described by the equations ...

• 5n +15d = 395 ⇒ n +3d = 79
• 12n +8d = 248 ⇒ 3n +2d = 62

Solution

Subtracting the second equation from 3 times the first, we have ...

3(n +3d) -(3n +2d) = 3(79) -(62)

7d = 175

d = 25 . . . . . . . divide by 7

Substituting into the first equation gives ...

n +3(25) =79

n = 4 . . . . . . . subtract 75

The cost of each notebook is $4; the cost of each thumb drive is $25.

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Additional comment

It is often easier to solve a system of equations when they are written in standard form. Here, we made the coefficients mutually prime by dividing each equation by the greatest common factor. For the first equation, we divided by 5; for the second equation, we divided by 4. This is especially useful if one of the coefficients ends up being 1.