Question 1145012: Two mechanics worked on a car. The first mechanic charged $65 per hour, and the second mechanic charged $115 per hour. The mechanics worked for a combined total of 35 hours, and together they charged a total of $3025. How long did each mechanic work?
Found 3 solutions by josgarithmetic, ikleyn, MathTherapy:
Answer by josgarithmetic(39623)   (Show Source):
Answer by ikleyn(52839)  (Show Source):
You can put this solution on YOUR website! .
For a TEMPLATE for this problem, look into the lesson
- Two mechanics work on a car
in this site.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this online textbook under the topic "Systems of two linear equations in two unknowns".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.
Answer by MathTherapy(10555)  (Show Source):
You can put this solution on YOUR website! Two mechanics worked on a car. The first mechanic charged $65 per hour, and the second mechanic charged $115 per hour. The mechanics worked for a combined total of 35 hours, and together they charged a total of $3025. How long did each mechanic work?
Let first mechanic's hours be F, and second's, S
Then we get: F + S = 35_____F = 35 - S ------- eq (i)
Also, 65F + 115S = 3,025_____5(13F + 23S) = 5(605)______13F + 23S = 605 ------ eq (ii)
13(35 - S) + 23S = 605 ------- Substituting 35 - S for F in eq (ii)
13(35) - 13S + 23S = 605
10S = 605 - 13(35)
S, or hours worked by the second mechanic = 
Substitute 15 for S in eq (i) to get F, the hours worked by the first mechanic
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