Question 1031088: An automobile factory produces two models. The first model requires 7 widgets and 10 shims. The second model requires 15 widgets and 21 shims. The factory can obtain 800 widgets and 1130 shims per hour. How many cars of each model can it produce per hour if all parts available are used?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let x equal the number of first model cars.
let y equal the number of second model cars.
the first model requires 7 widgets and 10 shims.
the second model requires 15 widgets and 21 shims.
the total number of widgets used must be equal to 800.
the total number of shims used must be equal to 1130.
since there are 7 widgets required for the first model and 15 widgets requires for the second model, you get:
7x + 15y = 800
this means that 7 widgets times the number of first model cars plus 15 widgets times the number of second model cars must be equal to 800.
similarly, you get:
10x + 21y = 1130.
this means that 10 shims times the numb3r of first model cars plus 21 shims times the number of second model cars must be equal to 1130.
these are two equations that need to be solved simultaneously.
the two equations are:
7x + 15y = 800
10x + 21y = 1130
if you multiply both sides of the first equaiton by 10 and you multiply both sides of the second equaiton by 7, you will get:
70x + 150y = 8000
70x + 147y = 7910
if you subtract the second equation from the first, you will get:
3y = 90
solve for y to get y = 30.
replace y with 30 in the first equation and solve for x and you will get x = 50.
that's your solution.
x = 50
y = 30
go back to both your original equations and replace x with 50 and y with 30 and you will see that the equations are true.
for example:
10x + 21y = 1130 becomes 10*50 + 21*30 = 1130 which becomes 500 + 630 = 1130 which becomes 1130 = 1130 which is true.
the solutions need to be true for both equations, not just one.
7x + 15y = 800 becomes 7*50 + 15*30 = 800 which becomes 350 + 450 = 800 which becomes 800 = 800 which is also true.
the solutions satisfy requirements of both original equations so it is good.
the solution is:
x = 50
y = 30
number of first model cars is 50.
number of second model cars is 30.
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