4.3 Solve Mixture Applications with Systems of Equations
Topics covered in this section are:
- Solve mixture applications
- Solve interest applications
- Solve applications of cost and revenue functions
4.3.1 Solve Mixture Applications
Mixture application involve combining two or more quantities. When we solved mixture applications with coins and tickets earlier, we started by creating a table so we could organize the information. For a coin example with nickels and dimes, the table looked like this:
Using one variable meant that we had to relate the number of nickels and the number of dimes. We had to decide if we were going to let $n$ be the number of nickels and then write the number of dimes in terms of $n$, or if we would let $d$ be the number of dimes and write the number of nickels in terms of $d$.
Now that we know how to solve systems of equations with two variables, we’ll just let $n$ be the number of nickels and $d$ be the number of dimes. We’ll write one equation based on the total value column, like we did before, and the other equation will come from the number column.
For the first example, we’ll do a ticket problem where the ticket prices are in whole dollars, so we won’t need to use decimals just yet.
Example 1
Translate to a system of equations and solve:
A science center sold $1,363$ tickets on a busy weekend. The receipts totaled $\$12,126$. How many $\$12$ adult tickets and how many $\$7$ child tickets were sold?
Solution
| Step 1. Read the problem. | We will create a table to organize the information. |
| Step 2. Identify what we are looking for. | We are looking for the number of adult tickets and the number of child tickets sold. |
| Step 3. Name what we are looking for. | Let $a=$ the number of adult tickets and $c=$ the number of child tickets. |
| A table will help us organize the data. We have two types of tickets, adult and child. | Write in $a$ and $c$ for the number of tickets. |
| Write the total number of tickets sold at the bottom of the Number column. | Altogether $1,363 were sold. |
| Write the value of each type of ticket in the Value column. | The value of each adult ticket is $\$12$. The value of each child ticket is $\$7$. |
| The number times the value gives the total value, so the total value of adult tickets is $a \cdot 12 = 12a$, and the total value of child tickets is $c \cdot 7 =7c$. | Fill in the Total Value column. |
| Altogether the total value of the tickets was $\$12,146$. |  |
| Step 4. Translate into a system of equations. | |
| The Number column and Total value column give us the system of equations. | $\Bigg\{ \begin{align*} a+c&=1,363 \\12a+7c&=12,146 \end{align*}$ |
| Step 5. Solve this system using the elimination method. Multiply the first equation by $-7$. | $\Bigg\{ \begin{align*} -7(a+c)&=-7(1,363) \\12a+7c&=12,146 \end{align*}$ |
| Simplify and add, then solve for $a$. | $\Bigg\{ \begin{align*} -7a-c&=-9,541) \\12a+7c&=12,146 \end{align*}$ $\overline{ \ \ 5a \ \ \ \ \ \ \ =2,605}$ $a=521$ |
| Substitute $a=521$ into the first equation, then solve for $c$. | $\begin{align*} a+c&=1363 \\ \textcolor{red}{521}+c&=1363 \\ c&=842 \end{align*}$ |
| Step 6. Check the answer in the problem. | $521$ adult tickets at $\$12$ per ticket makes $\$6,252$. $842$ child tickets at $\$7$ per ticket makes $\$5,894$. The total receipts are $\$12,146 \ \checkmark$ |
| Step 7. Answer the question. | The science center sold $521$ adult tickets and $842$ child tickets. |
In the next example, we’ll solve a coin problem. Now that we know how to work with systems of two variables, naming the variables in the ‘number’ column will be easy.
Example 2
Translate into a system of equations and solve:
Juan has a pocketful of nickels and dimes. The total value of the coins is $\$8.10$. The number of dimes is $9$ less than twice the number of tickets. How many tickets and how many dimes does Juan have?
Solution
| Step 1. Read the problem. | We will create a table to organize the information. |
| Step 2. Identify what we are looking for. | We are looking for the number of nickels and the number of dimes. |
| Step 3. Name what we are looking for. | Let $n=$ the number of nickels and $d=$ the number of dimes. |
| A table will help us organize the data. We have two types of tickets, nickels and dimes. | Write in $n$ and $d$ for the number of each type of coin. |
| Fill in the value column with the value of each type of coin. | The value of each nickel is $\$0.05$. The value of each child ticket is $\$0.10$. |
| The number times the value gives the total value, so, the total value of the nickels is $n(0.05) = 0.05n$, and the total value of dimes is $d(0.10)=0.10d$. Altogether the total value of the coins is $\$8.10$. |  |
| Step 4. Translate into a system of equations. | |
| The Total value column gives one equation. | $0.05n+0.10d=8.10$ |
| We also know the number of dimes is $9$ less than twice the number of nickels. Translate to get the second equation. | $d=2n-9$ |
| Now we have the system to solve. | $\Bigg\{ \begin{align*} 0.05n+0.10d&=8.10 \\d&=2n-9 \end{align*}$ |
| Step 5. Solve the system of equations. We will use the substitution method. | |
| Substitute $d=2n-9$ into the first equation. | $\begin{align*} 0.05n+0.10d&=8.10 \\0.05n+0.10(\textcolor{red}{2n-9})=8.10 \end{align*}$ |
| Simplify and solve for $n$. | $\begin{align*} 0.05n+0.2n-0.90&=8.10\\ 0.25n-0.90&=8.10\\0.25n&=9.00\\ n&=36 \end{align*}$ |
| To find the number of dimes, substitute $n=36$ into the second equation. | $\begin{align*} d&=2n-9 \\ d&=2 \cdot \textcolor{red}{36}-9\\ d&=63 \end{align*}$ |
| Step 6. Check the answer in the problem. | $63$ dimes at $\$0.10$ makes $\$6.30$. $36$ nickels at $\$0.05$ makes $\$1.80$. The total is $\$8.10 \ \checkmark$ |
| Step 7. Answer the question. | Juan has $36$ nickels and $63$ dimes. |
Some mixture applications involve combining foods or drinks. Example situations might include combining raisins and nuts to make a trail mix or using two types of coffee beans to make a blend.
Example 3
Translate to a system of equations and solve:
Carson wants to make $20$ pounds of trail mix using nuts and chocolate chips. His budget requires that the trail mix costs him $\$7.60$. per pound. Nuts cost $\$9.00$ per pound and chocolate chips cost $\$2.00$ per pound. How many pounds of nuts and how many pounds of chocolate chips should he use?
Solution
| Step 1. Read the problem. | We will create a table to organize the information. |
| Step 2. Identify what we are looking for. | We are looking for the number of pounds of nuts and the number of pounds of chocolate chips. |
| Step 3. Name what we are looking for. | Let $n=$ the number of pounds of nuts and $c=$ the number of pounds of chocolate chips. |
| Carson will mix nuts and chocolate chips to get trail mix. Write in $n$ and $c$ for the number of pounds of nuts and chocolate chips. |  |
| There will be 20 pounds of trail mix. Put the price per pound of each item in the Value column. Fill in the last column using Number $\cdot$ Value $=$ Total Value | |
| Step 4. Translate into a system of equations. We get the equations from the Number and Total Value columns. | $\Bigg\{ \begin{align*} n+c&=20\\9n+2c&=152 \end{align*}$ |
| Step 5. Solve the system of equations. We will use elimination to solve the system. Multiply the first equation by $-2$ to eliminate $c$. | $\Bigg\{ \begin{align*} -2(n+c)&=-2(20)\\9n+2c&=152 \end{align*}$ |
| Simplify and add. Solve for $n$. | $\Bigg\{ \begin{align*} -2n-2c&=-40\\9n+2c&=152 \end{align*}$ $\overline{ \ \ 7n \ \ \ \ \ =112}$ $n=16$ |
| To find the number of pounds of chocolate chips, substitute $n=16$ into the first equation, then solve for $c$. | $\begin{align*} n+c&=20 \\ \textcolor{red}{16}+c&=20\\ c&=4 \end{align*}$ |
| Step 6. Check the answer in the problem. | $16+4=20 \ \checkmark$ $9 \cdot 16 + 2 \cdot 4 =152 \ \checkmark$ |
| Step 7. Answer the question. | Carson should mix $16$ pounds of nuts and $4$ pounds of chocolate chips to create the trail mix. |
Another application of mixture problems relates to concentrated cleaning supplies, other chemicals, and mixed drinks. The concentration is given as a percent. For example, a $20\%$ concentrated household cleanser means that $20\%$ of the total amount is cleanser, and the rest is water. To make $35$ ounces of a $20\%$ concentration, you mix $7$ ounces ($20\%$ of $35$) of the cleanser with $28$ ounces of water.
For these kinds of mixture problems, we’ll use “percent” instead of “value” for one of the columns in our table.
Example 4
Translate to a system of equations and solve:
Sasheena is lab assistant at her community college. She needs to make $200$ milliliters of a $40\%$ solution of sulfuric acid for a lab experiment. The lab has only $25\%$ and $50\%$ solutions in the storeroom. How much should she mix of the $25\%$ and the $50\%$ solutions to make the $40\%$ solution?
Solution
| Step 1. Read the problem. A figure may help us visualize the situation, then we will create a table to organize the information. | Sasheena must mix some of the $25\%$ solution and some of the $50\%$ solution together to get $200$ ml of the $40\%$ solution. |
 | |
| Step 2. Identify what we are looking for. | We are looking for how much of each solution she needs. |
| Step 3. Name what we are looking for. | Let $x=$ the number of ml of $25\%$ solution and $y=$ number of ml of $50\%$ solution. |
| A table will help us organize the data. She will mix $x$ ml of $25\%$ with y ml of $50\%$ to get $200$ ml of $40\%$ solution. We write the percents as decimals in the chart. We multiply the number of units times the concentration to get the total amount of sulfuric acid in each solution. |  |
| Step 4. Translate into a system of equations. We get the equations from the Number column and the Amount column. Now we have the system. | $\Bigg\{ \begin{align*} x+y&=200 \\ 0.25x+0.50y&=0.40(200) \end{align*}$ |
| Step 5. Solve the system of equations. We will solve the system by elimination. Multiply the first equation by $−0.5$ to eliminate $y$. | $\Bigg\{ \begin{align*} -0.5(x+y)&=-0.5(200) \\ 0.25x+0.50y&=0.40(200) \end{align*}$ |
| Simplify and add to solve for $x$. | $\Bigg\{ \begin{align*} -0.5x-0.5y&=-100 \\ 0.25x+0.50y&=80\end{align*}$ $\overline{-0.25x \ \ \ \ \ \ \ \ =-20}$ $x=80$ |
| To solve for $y$, substitute $x=80$ into the first equation. | $\begin{align*} x+y&=200\\ \textcolor{red}{80}+y&=200\\ y&=120 \end{align*}$ |
| Step 6. Check the answer in the problem. | $\begin{align*} 80+120&=200 \ \checkmark \\ 0.25(80)+0.50(120)&=200 \ \checkmark \end{align*}$ |
| Step 7. Answer the question. | Sasheena should mix $80$ ml of the $25\%$ solution with $120$ ml of the $50\%$ solution to get $200$ ml of the $40\%$ solution. |
4.3.2 Solve Interest Applications
The formula to model simple interest applications is $I=Prt$. Interest, $I$, is the product of the principal, $P$, the rate, $r$, and the time, $t$. In our work here, we will calculate the interest earned in one year, so $t$ will be $1$.
We modify the column titles in the mixture table to show the formula for interest, as you’ll see in the next example.
Example 5
Translate to a system of equations and solve:
Adnan has $\$40,000$ to invest and hopes to earn $7.1\%$ interest per year. He will put some of the money into a stock fund that earns $8\%$ per year and the rest into bonds that earns $3\%$ per year. How much money should he put into each fund?
Solution
| Step 1. Read the problem. | A chart will help us organize the information. |
| Step 2. Identify what we are looking for. | We are looking for the amount to invest in each fund. |
| Step 3. Name what we are looking for. | Let $s=$ the amount invested in stocks and $b=$ the amount invested in bonds. |
| Write the interest rate as a decimal for each fund. Multiply: Principal • Rate • Time |  |
| Step 4. Translate into a system of equations. We get our system of equations from the Principal column and the Interest column. | $\Bigg\{ \begin{align*} s+b&=40,000 \\ 0.08s+0.03b&=0.071(40,000) \end{align*}$ |
| Step 5. Solve the system of equations by elimination. Multiply the top equation by $-0.03$. | $\Bigg\{ \begin{align*} \textcolor{red}{-0.03}(s+b)&=\textcolor{red}{-0.03}40,000 \\ 0.08s+0.03b&=0.071(40,000) \end{align*}$ |
| Simplify and add to solve for $s$. | $\Bigg\{ \begin{align*} -0.03s-0.03b&=1,200 \\ 0.08s+0.03b&=2,840 \end{align*}$ $\overline{\ \ \ \ \ \ 0.05s=1,640}$ $s=32,800$ |
| To find $b$, substitute $s=32,800$ into the first equation. | $\begin{align*} s+b&=40,000 \\ \textcolor{red}{32,800}+b&=40,000\\ b&=7,2000 \end{align*}$ |
| Step 6. Check the answer in the problem. | We leave the check to you. |
| Step 7. Answer the question. | Adnan should invest $\$32,800$ in stocks and $\$7,200$ in bonds. |
Did you notice that the Principal column represents the total amount of money invested while the Interest column represents only the interest earned? Likewise, the first equation in our system, $s+b=40,000$, represents the total amount of money invested and the second equation, $0.08s+0.03b=0.071(40,000)$, represents the interest earned.
The next example requires that we find the principal given the amount of interest earned.
Example 6
Translate to a system of equations and solve:
Rosie owes $\$21,540$ on her two student loans. The interest rate on her bank loan is $10.5\%$ and the interest rate on the federal loan is $5.9\%$. The total amount of interest she paid last year was $\$1,669.68$. What was the principal for each loan?
Solution
| Step 1. Read the problem. | A chart will help us organize the information. |
| Step 2. Identify what we are looking for. | We are looking for the principal of each loan. |
| Step 3. Name what we are looking for. | Let $b=$ the principal for the bank loan and $f=$ the principal on the federal loan. |
| The total loans are $\$21,540$. | |
| Record the interest rates as decimals in the chart. Multiply using the formula $I=Prt$ to get the Interest. |  |
| Step 4. Translate into a system of equations. The system of equations from the Principal column and the Interest column. | $\Bigg\{ \begin{align*} b+f&=21540 \\ 0.105b+0.059f&=1669.68 \end{align*}$ |
| Step 5. Solve the system of equations. We will use substitution to solve. Solve the first equation for $b$. | $\begin{align*} b+f&=21540 \\ b&=-f+21540 \end{align*}$ |
| Substitute $b=-f+21,540$ into the second equation. Then simplify and solve for $f$. | $\begin{align*} 0.105(\textcolor{red}{-f+21,540})+0.059f&=1669.68 \\ -0.105f+2261.70+0.059f&=1669.68 \\ -0.046f+2261.70&=1669.68 \\ -0.046f&=-592.02 \\ f&=12870 \end{align*}$ |
| To find $b$, substitute $f=12,870$ into the first equation. | $\begin{align*} b+f&=21,540 \\ b+\textcolor{red}{12,870}&=21540\\ b&=8670 \end{align*}$ |
| Step 6. Check the answer in the problem. | We leave the check to you. |
| Step 7. Answer the question. | The principal of the federal loan was $\$12,870$ and the principal for the bank loan was $\$8,670$. |
4.3.3 Solve applications of cost and revenue functions
Suppose a company makes and sells $x$ units of a product. The cost to the company is the total costs to produce $x$ units. This is the cost to manufacture for each unit times $x$, the number of units manufactured, plus the fixed costs.
The revenue is the money the company brings in as a result of selling x units. This is the selling price of each unit times the number of units sold.
When the costs equal the revenue we say the business has reached the break-even point.
COST AND REVENUE FUNCTIONS
The cost function is the cost to manufacture each unit times $x$, the number of units manufactured, plus the fixed costs.
$C(x)=(\text{cost per unit}) \cdot x+\text{fixed costs}$
The revenue function is the selling price of each unit times x, the number of units sold.
$R(x)=(\text{selling price per unit}) \cdot x$
The break-even point is when the revenue equals the costs.
$C(x)=R(x)$
Example 7
The manufacturer of a weight training bench spends $\$105$ to build each bench and sells them for $\$245$. The manufacturer also has fixed costs each month of $\$7,000$.
- Find the cost function $C$ when $x$ benches are manufactured.
- Find the revenue function $R$ when $x$ benches are sold.
- Show the break-even point by graphing both the Revenue and Cost functions on the same graph.
- Find the break-even point. Interpret what the break-even point means.
Solution
- The manufacturer has $\$7,000$ of fixed costs no matter how many weight training benches it produces. In addition to the fixed costs, the manufacturer also spends $\$105$ to produce each bench. Suppose $x$ benches are sold.
| Write the general Cost function formula. | $C(x)=(\text{cost per unit}) \cdot x + \text{fixed costs}$ |
| Substitute in the cost values. | $C(x)=105x+7000$ |
- The manufacturer has $\$7,000$ of fixed costs no matter how many weight training benches it produces. In addition to the fixed costs, the manufacturer also spends $\$105$ to produce each bench. Suppose $x$ benches are sold.
| Write the general Revenue function. | $R(x)=(\text{selling price per unit}) \cdot x$ |
| Substitute in the revenue per unit. | $R(x)=245x$ |
- Essentially we have a system of linear equations. We will show the graph of the system as this helps make the idea of a break-even point more visual.
| $\Bigg\{ \begin{align*} C(x)&=105x+7000 \\ R(x)&=245x \end{align*}$ | or | $\Bigg\{ \begin{align*} y&=105x+7000 \\ y&=245x \end{align*}$ |
- To find the actual value, we remember the break-even point occurs when costs equal revenue.
| Write the break-even formula. | $C(x)=R(x)$ $105x+7000=245x$ |
| Solve. | $\begin{align*} 7000&=140x\\ 50&=x \end{align*}$ |
When $50$ benches are sold, the costs equal the revenue.
$C(x)=105x+7000 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R(x)=245x$
$C(\textcolor{red}{50})=105(\textcolor{red}{50})+7000 \ \ \ \ \ \ \ R(\textcolor{red}{50})=245 \cdot \textcolor{red}{50}$
$C(50)=12,250 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R(50)=12,250$
When 50 benches are sold, the revenue and costs are both $\$12,250$. Notice this corresponds to the ordered pair $(50, 12250)$.
Licenses & Attributions
CC Licensed Content, Original
- Revision and Adaption. Provided by: Minute Math. License: CC BY 4.0
CC Licensed Content, Shared Previously
- Marecek, L., & Mathis, A. H. (2020). Solve Mixture Applications with Systems of Equations. In Intermediate Algebra 2e. OpenStax. https://openstax.org/books/intermediate-algebra-2e/pages/4-3-solve-mixture-applications-with-systems-of-equations. License: CC BY 4.0. Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
