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Homework Help: Systems of equations

  1. Nov 18, 2005 #1
    Hello, I am stuck on the following problem.

    How many milliliters of a 30% acid solution and a 55% acid solution must be mixed to obtain 100 milliliters of a 50% acid solution?

    My solution:

    Let X = The unknown milliliters of 30% acid.
    Let Y = The unknown milliliters of 55% acid.

    (1) X + Y = 100
    (2) 0.3X + 0.55Y = 0.5

    Multiplying (2) by 100 in order to work with integers,

    (2) 30X + 55Y = 50

    Multiplying (1) by -30 in order to eliminate X

    (1) -30X - 30Y = -3000

    So our new equations are,

    (1) -30X - 30Y = -3000
    (2) 30X + 55Y = 50

    Adding the two equations in order to eliminate X,

    25Y = -2950 => Y = -118

    Back-substituting Y in to (1) in order to solve for X,

    -30X - 30(-118) = -3000

    -30X + 3540 = -3000

    -30X = -6540

    X = 218

    The answer in the book says:

    20 milliliters of 30% acid solution and 80 milliliters of 55% acid solution.


    Any hints as to where I am going wrong?
     
    Last edited: Nov 18, 2005
  2. jcsd
  3. Nov 18, 2005 #2
    The above part is wrong. the RHS is wrong. think?
     
  4. Nov 18, 2005 #3
    For starters, let's get the right first 2 equations. I hope you understand the following:

    50 = .3x + .55y
    100 = x + y

    y = 100 - x
    50 = .3x + .55 (100 - x)
    50 = .3x + 55 - .55x
    -5 = -.25x
    (-5/-.25) = x
    x = 20
    100 - 20 = y
    y = 80

    So it's 20 mills of 30%, and 80 mills of 55%
     
    Last edited: Nov 18, 2005
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