### Learning Outcomes

- Solve mixture problems

## Write a system of linear equations representing a mixture problem, solve the system and interpret the results

One application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. A solution is a mixture of two or more different substances like water and salt or vinegar and oil. Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand. There are many other disciplines that use solutions as well.

The concentration or strength of a liquid solution is often described as a percentage. This number comes from the ratio of how much mass is in a specific volume of liquid. For example if you have [latex]50[/latex] grams of salt in a [latex]100mL[/latex] of water you have a [latex]50\%[/latex] salt solution based on the following ratio:

[latex]\frac{50\text{ grams }}{100\text{ mL }}=0.50\frac{\text{ grams }}{\text{ mL }}=50\text{ % }[/latex]

Solutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths. In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.

We will use the following table to help us solve mixture problems:

Amount | Concentration (%) | Total | |
---|---|---|---|

Solution 1 | |||

Solution 2 | |||

Final Solution |

To demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is [latex]120mL[/latex] of a [latex]9\%[/latex] solution, and the other is 75mL of a 23% solution. If we mix both of these solutions together we will have a new volume and a new mass of solute and with those we can find a new concentration.

First, find the total mass of solids for each solution by multiplying the volume by the concentration.

Amount | Concentration (%) | Total Mass | |
---|---|---|---|

Solution 1 | [latex]120 mL[/latex] | 0.09 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex] | [latex]\left(120\cancel{\text{ mL}}\right)\left(0.09\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=10.8\text{ grams }[/latex] |

Solution 2 | [latex]75 mL[/latex] | 0.23 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex] | [latex]\left(75\cancel{\text{ mL}}\right)\left(0.23\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=17.25\text{ grams }[/latex] |

Final Solution |

Next we add the new volumes and new masses.

Amount | Concentration (%) | Total Mass | |
---|---|---|---|

Solution 1 | [latex]120 mL[/latex] | 0.09 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex] | [latex]\left(120\cancel{\text{ mL}}\right)\left(0.09\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=10.8\text{ grams }[/latex] |

Solution 2 | [latex]75 mL[/latex] | 0.23 [latex]\frac{\text{ grams }}{\text{ mL }}[/latex] | [latex]\left(75\cancel{\text{ mL}}\right)\left(0.23\frac{\text{ grams }}{\cancel{\text{ mL }}}\right)=17.25\text{ grams }[/latex] |

Final Solution | [latex]195 mL[/latex] | [latex]\frac{28.05\text{ grams }}{ 195 \text{ mL }}=0.14=14\text{ % }[/latex] | [latex]10.8\text{ grams }+17.25\text{ grams }=28.05\text{ grams }[/latex] |

Now we have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution. In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.

### Example

A chemist has [latex]70 mL[/latex] of a [latex]50\%[/latex] methane solution. How much of an [latex]80\%[/latex] solution must she add so the final solution is [latex]60\%[/latex] methane?

The above problem illustrates how we can use the mixture table to define an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.

### Example

A farmer has two types of milk, one that is [latex]24\%[/latex] butterfat and another which is [latex]18\%[/latex] butterfat. How much of each should he use to end up with [latex]42[/latex] gallons of [latex]20\%[/latex] butterfat?

In the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.