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Math Help - More simultaneous equations

  1. #1
    Junior Member
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    More simultaneous equations

    A chemical manufacturer has an order for 500 litres of a 25% acid solution (i.e. 25% by volume is acid). Solutions of 30% and 18% are available in stock. The manufacturer wishes to make up the 500 litres from a mixture of 30% and 18% solutions.
    Let x denote the amount of 30% solution required.
    Let y denote the amount of 18% solution required.
    Use simultaneous equations in x and y to determine the amount of each solution required.

    I'm unsure how to set up the simultabeous equations therefore I am struggling with this question aby help would be appreciated!!
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  2. #2
    Super Member 11rdc11's Avatar
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    x + y = 500

    .30x + .18y =125

    You can also do this problem using just one variable.

    .18(500-x) + .30(x) = 500(.25)
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  3. #3
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    Thank you that was very helpful!!
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