Algebra 1 college

How much of a 12% saline solution should be mixed with a 4% saline solution to obtain 40 milliliters of a 6% saline solution?. Use a system of equations and elimination method to solve. can someone please see if I am on the right track.
12T + 4F=6x40
-4 -4F 4
8T=2x40
8T=80
8T/8=80/8
T=10 ml

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asked by Sophia
  1. First of all, you always define your variables.
    I deduced from your work that you:
    let the amount of 12% solution used be T ml
    let the amount of 4% solution used be F ml

    clearly T + F = 40 is our first equation
    ( don't see that anywhere)

    then .12T + .04F = .06(40)
    times 100
    12T + 4F = 240
    or, dividing by 4 ....
    3T + F = 60 , again, where is that equation?

    (you had 12T + 4T = 6(40)
    which is correct but I don't know how you are using it.
    the line -4 -4f 4 below it means nothing to me)

    so we have 2 equations:
    3T + F = 60
    T + F = 40
    since we are to use elimination, subtract them
    2T = 20
    T = 10

    back in T+F=40
    10+F =40
    F = 30

    They should use 10 ml of the 12%, and 30 ml of the 4% solution.

    You did get T = 10, but did not finish the question.
    Your instructions were to use elimination, thus you needed two equations.

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    posted by Reiny

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