Question 1184332: A hospital needed supplies of an expensive medicine company a has the most supplies available 1.4 mg which is twice the difference between the weight of company b supplies and c company and 0.8 mg more than company c supplies how many milligrams can the hospital get from these three companies
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! A = 1.4 mg.
A = 2 * abs(B - C)
A = C + .8
from A = C + .8, you get:
1.4 = C + .8
subtract .8 from both sides of that equation to get:
1.4 - .8 = C
solve for C to get:
C = .6
from A = 2 * (B - C), you get:
1.4 = 2 * (B - .6)
simplify to get:
1.4 = 2 * B - 1.2
add 1.2 to both sides of the equation to get:
1.4 + 1.2 = 2 * B
combine like terms to get:
2.6 = 2 * B
solve for B to get:
B = 1.3
your solution should be A + B + C = 1.4 + 1.3 + .6 = 3.3 mg.
to confirm the numbers are good:
A = 1.4, B = 1.3, C = .6
A = 2 * (B - C) becomes:
A = 2 * (1.3 - .6) which becomes:
A = 2 * (.7) which becomes:
A = 1.4 which is true.
A = 1.4, C = .6
A = C + .8 becomes:
A = .6 + .8 which becomes:
A = 1.4 which is true.
all original statements are true when A = 1.4 and B = 1.3 and C = .6
this confirms the numbers are correct.
the solution is the sum of those numbers = 3.3.
|
|
|
