SOLUTION: Set up the system of equations needed to solve the following problem. A company produces two types of widgets in two different factories. Factory A produces 25 standard and 50

Algebra ->  Systems-of-equations -> SOLUTION: Set up the system of equations needed to solve the following problem. A company produces two types of widgets in two different factories. Factory A produces 25 standard and 50       Log On


   

Question 834512: Set up the system of equations needed to solve the following problem.
A company produces two types of widgets in two different factories. Factory A produces 25 standard and 50 deluxe widgets per day. Factory B produces 150 standard and 75 deluxe widgets per day. How many days should the company operate each factory in order to exactly fill an order for 500 standard and 325 deluxe widgets?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
a= number of days factory A needs to work to exactly fill that order.
b= number of days factory B needs to work to exactly fill that order.

During a days, factory A can make 25a standard widgets.
During b days, factory B can make 150b standard widgets.
To make the 500 standard widgets a and b have to satisfy the equation
25a%2B150b=500 .
We can simplify that equation a bit by dividing both sides of the equal sign by 25.
%2825a%2B150b%29%2F25=500%2F25
25a%2F25%2B150b%2F25=500%2F25
highlight%28a%2B6b=20%29

During a days, factory A can make 50a deluxe widgets.
During b days, factory B can make 75b deluxe widgets.
To make the 325 deluxe widgets a and b have to satisfy the equation
50a%2B75b=325 .
We can simplify that equation a bit by dividing both sides of the equal sign by 25.
50a%2B75b=500
50a%2F25%2B75b%2F25=325%2F25
highlight%282a%2B3b=13%29

We can set up the simpler system
highlight%28system%28a%2B6b=20%2C2a%2B3b=13%29%29
or we could have stopped before simplifying the equations and end up with
highlight%28system%28a%2B6b=20%2C50a%2B75b=500%29%29

IF WE HAD TO SOLVE THE SYSTEM:
There are several ways to solve the system.
We could solve a%2B6b=20 for a to get an expression that is equal to a ,
a%2B6b=20 ---> a=20-6b
and the substitute the expression 20-6b for a in the other equation:
system%28a=20-6b%2C2a%2B3b=13%29 --> 2%2820-6b%29%2B3b=13
Then we solve the resulting equation for b
2%2820-6b%29%2B3b=13
2%2A20-2%2A6b%2B3b=13
40-12b%2B3b=13
40-9b=13
-9b=13-40
-9b=-27
b=%28-27%29%2F%28-9%29
highlight%28b=3%29
Then we would substitute the value found for b into a=20-6b and get
a=20-6%2A3
a=20-18
highlight%28a=2%29