Question 1036606
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Hi !
I am having difficulties with this problem, we do not get along.

We have a 20% alcohol solution and a 50% solution. How many pints must be used from each to obtain 8 pints of a 30% solution ?
Set up and solve using a system of algebraic equation.

Thank you so much
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Let n be the number of pints of the 20% alcohol solution needed, and
let m be the number of pints of the 50% alcohol solution needed.

Then the total volume equation is 

n + m = 8.   (1)

n pints of the 20% alcohol solution contain 0.2n pints of the pure alcohol.
m pints of the 50% alcohol solution contain 0.5m pints of the pure alcohol.

In total, the mixt contains 0.2n + 0.5m pints of the pure alcohol.
So, the second equation for the alcohol percentage/contents in the mixt is

{{{(0.2n + 0.5m)/8}}} = 0.3.

You can rewrite the last equation in the form

0.2n + 0.5m = 0.3*8,   or

0.2n + 0.5m = 2.4.     (2)

These two equations, (1) and (2), form the system of two linear algebraic equations in two unknowns, n and m.


There are many ways to solve it. You can use the Substitution method, for example.
For it, express x from (1) as  n = 8-m  and substitute it into (2). You will get

0.2*(8-m) + 0.5m = 2.4,

1.6 -0.2m + 0.5m = 2.4,

0.3m = 2.4 - 1.6 = 0.8,

m = {{{0.8/0.3}}} = {{{2}}}{{{2/3}}} pint.

Then n = 8 - m = {{{5}}}{{{1/3}}} pint.

<U>Answer</U>.  {{{5}}}{{{1/3}}} pint of the 20% alcohol solution and  {{{2}}}{{{2/3}}} pint of the 50% alcohol solution.
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