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Mixture Problems

Use linear systems to solve story problems

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Alcohol Solutions in Nursing
Teacher Contributed

Real World Applications – Algebra I


How do nurses use alcohol solutions?

Student Exploration

In the medical world, nurses, doctors and surgeons all use various types of materials in their practice. Rubbing alcohol, for example, is used quite often. Nurses and physicians, for example, use 70% alcohol solution as an antiseptic (like before giving injections), but refrain from using a 90% alcohol solution. Why do you think 70% is more beneficial to use than 90%?

Although 90% is stronger, it can be TOO strong when it comes to helping our bodies. 90% alcohol solution can potentially coagulate the protein in contact, meaning that the alcohol can change the state of the protein.

Sometimes, 40% alcohol solutions are used. In this application, we will learn how systems of equations have been used to help nurses and doctors in creating 40% alcohol solutions.

If a nurse has 20% alcohol solution and 70% alcohol solution, but needs 15 liters of a 40% alcohol solution, how do you think the nurse should proceed? Let’s use a system of linear equations and two different ways of solving this system of equations.

First, we need to identify our variables, since we need to find out how much 20% alcohol solution (let’s call it Mixture \begin{align*}A\end{align*}A) and how much 70% alcohol solution (let’s call it Mixture \begin{align*}B\end{align*}B) we need. Since our final mixture is a total of 15 liters, our first equation is \begin{align*}A + B = 15\end{align*}A+B=15. Since our final mixture should have 40% alcohol solution, our second equation is \begin{align*}0.20A + 0.70B = 6\end{align*}0.20A+0.70B=6.

Your job: Solve this system of equations two different ways.

Similar to the Mixture concept, let’s isolate one of the variables and then use the Substitution Method.

\begin{align*} A &= 15 - B\\ 0.20A + 0.70B &= 6\\ 0.20(15-B) + 0.70B &= 6\\ 3 - 0.20B + 0.70B &= 6\\ 0.50B &= 3\\ B &= 6\end{align*}A0.20A+0.70B0.20(15B)+0.70B30.20B+0.70B0.50BB=15B=6=6=6=3=6

Now, we find \begin{align*}A\end{align*}A. \begin{align*}A = 15-6 = 9\end{align*}A=156=9

The nurse will then need 9 liters of the 20% alcohol solution, and 6 liters of the 70% alcohol solution.

Now, let’s solve this system of linear equations using multiplication and then solve using the Elimination Method. Do you think we will get the same answers? Why or why not?

\begin{align*} A + B &= 15\\ 0.20A + 0.70B &= 6\end{align*}A+B0.20A+0.70B=15=6

Let’s choose to eliminate the “\begin{align*}A\end{align*}A” variable first, so we have to multiply the top equation by \begin{align*}-0.20\end{align*}0.20.

\begin{align*}-0.20A - 0.20B &= - 3\\ 0.20A + 0.70B &= 6\\ 0.50B &= 3\\ B &= 6\\ A + 6 &= 15\\ A &= 9\end{align*}0.20A0.20B0.20A+0.70B0.50BBA+6A=3=6=3=6=15=9

Again, we’ll need 9 liters of the 20% alcohol solution, and 6 liters of the 70% alcohol solution. Why do we get the same answers? Which strategy do you think was easier to solve this system of linear equations?

Resources Cited


Connections to other CK-12 Subject Areas


  • Homogeneous Mixture
  • Solute & Solvent
  • Preparing Solutions, Dilution

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