Algebra II word problem: help?
A carpenter charges a set fee for each job in addition to his hourly charge. If a carpenter bills a customer $150 for a job which takes him 3 hours and bills a second customer $180 for a 6 hour job, how much would he charge for a job that takes him 9 hours?
Comments
Since the three extra hours cost an extra $30, you know each extra hours costs $10. So since 9 is 3 more than 6, add another $30 to the $180.
Okay, the set fee remains the same every time he charges for a job, so let's call that a.
He also separately charges an hourly rate, we can call that b.
For the first customer, the amount he charges in total would be the set charge plus the total of his hourly charge for 3 hours.
That would be a+ 3(hours)xb = 150.
The second one would then be the set charge plus the total hourly charge for 6 hours, which is a + 6b = 180.
The question is asking how much he would charge for a job that includes the set fee plus nine hours on the job, which is a + 9b.
Your known equations are:
a+ 3b = 150
a + 6b = 180
I assume you know about simultaneous equations, so just subtract the top equation from the bottom equation to eliminate a, so you only have one variable, b.
You get: 6b-3b = 180-150 which is 3b = 30, and b = 10.
Plug b into a + 3b = 150, and you get a: a + 30 = 150, a = 120.
Now you have the set fee, 120 dollars and the hourly charge 10 dollars/hour.
a + 9b = $120 + $9(10) = $210.
Tada!