SOLUTION: Your company produces two models of bicycles: Model A and Model B. Model A takes 2 hours to assemble, where Model B takes 3 hours to assemble. Model A costs $25 to make per bike
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Question 206054This question is from textbook Finite Mathematics
: Your company produces two models of bicycles: Model A and Model B. Model A takes 2 hours to assemble, where Model B takes 3 hours to assemble. Model A costs $25 to make per bike where Model B costs $30 to make per bike. If your company has a total of 34 hours and $350 available per day for these two models, how many of each model can be made in a day?
Solve using Elimination Method:
Equation:
120x+180y = 21,600 hours assembling per day
$25x+$30y = $750 costs
first equation:
120x + 180y = 21,600
second equation
(-5) 25 + (-5) 30y = (-5) 750
elimination 245x + 330y = -17,850
substitute back in to equation
120x + 180y ( ) = 21,600
x = bikes per day
I need help solving this equation with the steps
This question is from textbook Finite Mathematics
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Your company produces two models of bicycles: Model A and Model B.
Model A takes 2 hours to assemble, where Model B takes 3 hours to assemble.
Model A costs $25 to make per bike where Model B costs $30 to make per bike.
If your company has a total of 34 hours and $350 available per day for these
two models, how many of each model can be made in a day?
Solve using Elimination Method:
:
I am not sure what you are doing here, but you are making it way more complicated than it is:
:
Let a = no. of A bikes
Let b = no. of B bikes
:
2a + 3b = 34; total hrs available
:
25a + 30b = 350; total cost available
;
Multiply the 1st equation by 10 and subtract from the above equation
25a + 30b = 350
20a + 30b = 340
------------------subtraction eliminates b
5a = 10
a = 2 ea A models
:
use the time equation to find b
2(2) + 3b = 34
4 + 3b = 34
3b = 34 - 4
3b = 30
b =
b = 10 ea B models per day
;
:
Check solution in the cost equation
25(2) + 30(10) = $350 confirms our solution
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