SOLUTION: Separate 800 into three parts, such that the sum of the first, 1/2 of the second, and 2/5 of the third is 400; and the sum of the second, 3/4 of the first, and 1/4 of the third is

Algebra ->  Square-cubic-other-roots -> SOLUTION: Separate 800 into three parts, such that the sum of the first, 1/2 of the second, and 2/5 of the third is 400; and the sum of the second, 3/4 of the first, and 1/4 of the third is       Log On


   



Question 1112145: Separate 800 into three parts, such that the sum of the first, 1/2 of the second, and 2/5 of the third is 400; and the sum of the second, 3/4 of the first, and 1/4 of the third is 400.
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The given information tells us...
x%2By%2Bz+=+800 (1)
x%2B%281%2F2%29y%2B%282%2F5%29z+=+400 (2)
%283%2F4%29x%2By%2B%281%2F4%29z+=+400 (3)

Solve the system by first eliminating one variable to reduce the problem to two equations in two variables. With the given coefficients, eliminating y first looks to be the easiest.

x%2By%2Bz+=+800; 2x%2By%2B%284%2F5%29z+=+800; --> -x%2B%281%2F5%29z+=+0

x%2By%2Bz+=+800; %283%2F4%29x%2By%2B%281%2F4%29z+=+400; --> %281%2F4%29x%2B%283%2F4%29z+=+400 --> x%2B3z+=+1200

Adding the last two equations,
%2816%2F5%29z+=+1600 --> z+=+500

Plugging back into earlier equations, x=100 and y=200.

CHECK:
100%2B200%2B500+=+800
100%2B%281%2F2%29200%2B%282%2F5%29500+=+100%2B100%2B200+=+400
{{(3/4)(100)+200+(1/4)(500) = 75+200+125 = 400}}}