Question 1205707: When London spent 30 minutes on the elliptical trainer and then did circuit training for 10 minutes, her fitness app said she burned 470 calories. When she spent 10 minutes on the elliptical trainer and 30 minutes circuit training she burned 370 calories. Solve the system:
{ (30 e + 10 c = 470), (10 e + 30 c = 370) :}
for e , the number of calories she burned for each minute on the elliptical trainer, and c , the number of calories she burned for each minute of circuit training.
e =
c =
Found 3 solutions by Theo, greenestamps, math_tutor2020:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! your 2 equations are:
30e + 10c = 470
10e + 30c = 370
multiply the second equation by 3 and leave the first equation as is to get:
30e + 10c = 470
30e + 90c = 1110
subtract the first equation from the second to get:
80c = 640
solve for c to get:
c = 640 / 80 = 8
replace c in the first equation by 8 to get:
30e + 10 * 8= 470
subtract 10 * 8 from both sides of the equation to get:
30e = 470 - 80
simplify to get:
30e - 390
solve for e to get:
e = 390 / 30 = 13
you have c = 8 and e = 13
from the first equation, 30e + 10c becomes 30 * 13 + 10 * 8 = 390 + 80 = 470
from the second equation, 10e + 30c becomes 10 * 13 + 30 * 8 = 130 + 240 = 370
solution of c = 8 and e = 13 is confirmed to be correct.
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
The problem clearly involves solving a system of two linear equations; there are always a large number of choices of how to solve them.
e = calories burned per minute on elliptical trainer
c = calories burned per minute on circuit training
30e + 10c = 470
10e + 30c = 370
With those two equations, my clear choice for a solution path is the following.
Add the two starting equations and simplify:
40e + 40c = 840
10e + 10c = 210
Compare that equation to each of the starting equations:
20e = 260
20c = 160
Solve:
e = 13
c = 8
ANSWER: 13 calories per minute on the elliptical trainer; 8 per minute on the circuit training
Obviously this solution method is only useful because of the exact numbers used... but it is a worthwhile thing to look out for in a problem like this.
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
There are many approaches. I'll apply substitution.
Solve the first equation for the variable c.
30e+10c = 470
10c = 470-30e
c = (470-30e)/10
c = (470)/10-(30e)/10
c = 47 - 3e
Plug that into the other equation.
10e + 30c = 370
10e + 30(47 - 3e) = 370
10e + 1410 - 90e = 370
10e - 90e = 370 - 1410
-80e = -1040
e = -1040/(-80)
e = 13
Now we can determine c.
c = 47 - 3e
c = 47 - 3*13
c = 47 - 39
c = 8
Summary:
e = 13
c = 8
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