# 4.3 Solve Mixture Applications with Systems of Equations

Topics covered in this section are:

## 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 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

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 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 ## 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

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

## 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. • 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. • 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. When$50$benches are sold, the costs equal the revenue.$C(x)=105x+7000 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R(x)=245xC(\textcolor{red}{50})=105(\textcolor{red}{50})+7000 \ \ \ \ \ \ \ R(\textcolor{red}{50})=245 \cdot \textcolor{red}{50}C(50)=12,250 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R(50)=12,250$When 50 benches are sold, the revenue and costs are both$\$12,250$. Notice this corresponds to the ordered pair $(50, 12250)$.