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.