Question 1077881
The first acid solution contains <b>25%</b> acid, the second contains <b>40%</b> and the third contains <b>85%</b> . He wants to use all three solutions to obtain a mixture of 108 liters containing 45% acid, using 2 times as much of the 85% solution as the 40% solution. How many liters of each solution should be used?
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x of 25%
y of 40%
z of 85%
{{{z/y=2}}}


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Account for the pure acid
{{{25x+40y+85*2y=108*45}}}
{{{5x+8y+17*2y=108*9}}}
{{{5x+8y+34y=972}}}
{{{5x+42y=972}}}


Account for the volume
{{{x+y+2y=108}}}
{{{x+3y=108}}}


Simpler system in two variables:  {{{system(5x+42y=972,x+3y=108)}}}
Use five times the second equation and subtract from first equation.


{{{5x+42y-5x-15y=972-540}}}
{{{42y-15y=432}}}
{{{27y=432}}}
{{{highlight(y=16)}}}


{{{highlight(x=60)}}}


{{{highlight(z=32)}}}


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60 liters of 25%
16 liters of 40%
32 liters of 85%