How do you solve mixture problems using system of equations?

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Answer 1 · 2014-11-26T17:41:43

Just to make it easier for me, I usually make a small table with the headings "Components", "Unit Value", "Amount" and "Value".

Consider the following question:

**How many pints of 20% acid solution and 70% acid solution must be mixed to obtain 40 pints of 50% acid solution? **

First, I'll set up my table. I'll fill in the unknowns with variables $x$ $x$ and $y$ $y$ .

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From this, we can easily set up the two equations.

Sum of values of two acids = Value of mixture

Therefore, $0.20 x + 0.70 y = 20$ $0.20x + 0.70y = 20$

For convenience, we'll multiply the entire equation by 10,

$2 x + 7 y = 200$ $2x + 7y=200$

This is equation (1)

Setting up the second equation,

Sum of amounts of each acid = Amount of mixture

$x + y = 40$ $x+y=40$

To make at least one term of this equation identical to a term of equation (1), we'll multiply the entire equation by 2,

$2 x + 2 y = 80$ $2x+2y=80$

This is equation (2)

Subtracting equation (2) from (1),

$(+) 2 x + 7 y = 200$ $(+) 2x + 7y=200$
$(-) 2 x + 2 y = 80$ $(-)2x+2y=80$
$- - - - - - - -$ $--------$
$(=) 0 + 5 y = 120$ $(=)0+5y=120$

Thus,
$5 y = 120$ $5y=120$

$y = \frac{120}{5} = 24$ $y=120/5=24$

Substituting $y = 24$ $y=24$ in (2),

$2 x + 2 (24) = 80$ $2x+2(24)=80$
$2 x + 48 = 80$ $2x+48=80$
$2 x = 80 - 48 = 32$ $2x=80-48=32$
$x = \frac{32}{2} = 16$ $x=32/2=16$

So, we have
$x = 16$ $x=16$ and $y = 24$ $y=24$

We can conclude that 16 pints of 20% acid solution must be mixed with 24 pints of 70% solution to obtain 40 pints of 50% solution.

I hope your question was answered.