# MATH

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Furnace repairs

A repairman purchased several furnace -blower motors for a total cost of \$210. If is cost per motor had been \$5 less, he could have purchased one additional motor. How many motors did he buy at the regular rate?

• MATH -

regular rate = r
discount rate = (r-5)
number bought = n
number that would have been bought = (n+1)
so
210 = r n
and
210 = (r-5)(n+1) = r n + r - 5 n -5
from the first equation r = 210/n
then the second
215 = (210/n)n +210/n -5 n
5 = 210/n -5 n
5 n = 210 -5 n^2
n^2 + n - 42 = 0
(n-8)(n+7)=0
n = 8

• sorry factored wrong -

sorry
(n+7)(n-6) = 0
n = 6

• MATH -

check
r = 210/n = 35
r-5 = 30
n+1 = 7
30 * 7 = 210 check

• MATH -

thank you very much

• MATH -

HELP!!!!!I need a better explination.

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