SOLUTION: Internet Company A charges a $25 fee plus an additional $0.35 per hour. Internet Company B charges a $10 fee plus an additional $0.50 per hour. For what number of hours are the cha

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Internet Company A charges a $25 fee plus an additional $0.35 per hour. Internet Company B charges a $10 fee plus an additional $0.50 per hour. For what number of hours are the cha      Log On


   

Question 750136: Internet Company A charges a $25 fee plus an additional $0.35 per hour. Internet Company B charges a $10 fee plus an additional $0.50 per hour. For what number of hours are the charges for both companies the same?
Found 3 solutions by dkppathak, FrankM, tommyt3rd:
Answer by dkppathak(439) About Me  (Show Source):
You can put this solution on YOUR website!
let extra hours are y hrs
for a company 25+.35y
for B company 10+.50y
as per conditions both value are equal
therefor
25+.35y=10+.50y
15=.15y
y=15/.15=100
100hours
answer 100 hours

Answer by FrankM(1040) About Me  (Show Source):
You can put this solution on YOUR website!
2500+35X=1000+50X
1500=15X (I simplified by subtracting 1000 and 35X from both sides)
X=100
ANS 100 Hours
Note: setting the problem up as 'cents' helped avoid using decimals or fractions.

Answer by tommyt3rd(5050) About Me  (Show Source):
You can put this solution on YOUR website!
First we write their total charges as equations. Then we solve for t - the number of hours when both charges are equal
A=0.35t%2B25

B=0.50t%2B10

When they charge the same A=B and we can write

0.35t%2B25=0.50t%2B10

25-10=0.5t-0.35t

15=0.15t



so t=100 hours


:)