How do you solve mixture problems using system of equations?

1 Answer
Nov 26, 2014

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 and 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, 0.20x + 0.70y = 20

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

2x + 7y=200

This is equation (1)

Setting up the second equation,

Sum of amounts of each acid = Amount of mixture

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,

2x+2y=80

This is equation (2)

Subtracting equation (2) from (1),

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

Thus,
5y=120

y=120/5=24

Substituting y=24 in (2),

2x+2(24)=80
2x+48=80
2x=80-48=32
x=32/2=16

So, we have
x=16 and 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.