Example 4: The Justice family has a plan to pay off their debt. They have outstanding loans on two cars. Mr. Justice's car note balance is currently at $5,000 and they plan to pay $400 per month. Mrs. Justice's car note balance is currently at $6,000 and they plan to pay $500 per month. The family concluded that the outstanding balance on both cars would be the same after only 6 months. Was their conclusion reasonable?

System of equations:
(the outstanding balance of each car loan as a function of x months)

B(x) = 5,000 - 400x → Mr. Justice's car
B(x) = 6,000 - 500x → Mrs. Justice's car

To determine the outstanding balance of Mr. Justice's car loan after 6 months, use B(x) = 5,000 – 400x, or 5,000 – 400(6) = $2,600

To determine the outstanding balance of Mrs. Justice's car loan after 6 months, use B(x) = 6,000 - 500x, or 6,000 - 500(6) = $3,000

Therefore, the family's conclusion was not reasonable since the outstanding balance of each car loan was not the same after 6 months.

To determine the number of months needed for the outstanding balances to reach the same amount, use any method of solving the system of equations (substitution, elimination, or graphing).

Solve by substitution:
5,000 – 400x = 6,000 – 500x
           +500x            +500x
5,000 + 100x = 6,000
-5,000         -5,000
100x = 1,000
x = 10

So, the outstanding loans would be the same amount after 10 months.
B(x) = 5,000 - 400x, or 5,000 - 400(10 )= $1,000
B(x) = 6,000 - 500x, or 6,000 - 500(10) = $1,000

Example 5: A local department store sells a bag of 30 mini candy bars and 20 snack size candy bars for $7. They will also sell a bag of 10 mini candy bars and 50 snack size candy bars for $11. Marla estimated that in both bags, the mini candy bars cost $0.10 each and the snack size candy bars cost $0.20 each. Was Marla's estimate reasonable?

System of equations:

(m = price of the mini candy bars, s = price of the snack size candy bars)

30m + 20s = 7
10m + 50s = 11

To determine the cost of each size candy bars, solve the system using any method of solving the system of equations (substitution, elimination, or graphing).

Solve by elimination:

30m + 20s = 7
-3(10m + 50s = 11)

30m + 20s = 7
-30m + -150s = -33

-130s = -26
s = 0.2

30m + 20(0.2) = 7
30m + 4 = 7
30m = 3
m = 0.1

Close this window/tab to return to the lesson.