# How do you solve mixture problems using system of equations?

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 $y$.

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

Sum of values of two acids = Value of mixture

Therefore, $0.20 x + 0.70 y = 20$

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

$2 x + 7 y = 200$

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,

$2 x + 2 y = 80$

This is equation (2)

Subtracting equation (2) from (1),

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

Thus,
$5 y = 120$

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

Substituting $y = 24$ in (2),

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

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.