SOLUTION: I am in College algebra and working on "Solving systems of equations by the Addition Method". I have the answers in front of me but I don't know how to work the problem.
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Question 700121: I am in College algebra and working on "Solving systems of equations by the Addition Method". I have the answers in front of me but I don't know how to work the problem.
The Word problems is as follow's...
A chemist needs 90 milliliters of a 72% solution but has only 68% and 77% solutions available. Find how many milliliter of each that should be mixed to get the desired solution.
I have tried using X+Y=72% over 68%X+77%Y=90 and adding the two together to single out the x and y values,but this just brings up some wierd number that are way to high. The answer in the book is 50 ml of 68% and 40ml of 77%. I cant see how they get this answer. Please help Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A chemist needs 90 milliliters of a 72% solution but has only 68% and 77% solutions available.
Find how many milliliter of each that should be mixed to get the desired solution.
:
Let x = amt of 77% solution required
Let y = amt of 68% solution required
:
"A chemist needs 90 milliliters of a 72% solution"
x + y = 90
y = (90-x); use this form for substitution
:
" but has only 68% and 77% solutions available."
Use the decimal equiv here
.77x + .68y = .72(90)
.77x + .68y = 64.8
Replace y with (90-x)
.77x + .68(90-x) = 64.8
.77x + 61.2 - .68x = 64.8
.77x - .68x = 64.8 - 61.2
.09x = 3.6
x = 3.6/.09
x = 40 ml of 77% solution
:
You can find the amt of 68% solution, check it in the original mixture equation
