Solve Mixture Applications with Systems of Equations

Solve mixture applications

When we solved mixture applications earlier with coins and tickets, we started by creating a table with columns for number, value, and total value, following the model $\text{number} \cdot \text{value} = \text{total value}$. Using only one variable meant we had to relate the number of one type to the number of the other — deciding whether to let $n$ be the number of nickels and write the number of dimes in terms of $n$, or the other way around.

Now that we know how to solve systems of equations with two variables, we’ll just let one variable stand for the number of each type. One equation will come from the number column, and the other equation will come from the total value column.

We’ll start with a ticket problem where the ticket prices are in whole dollars, so we won’t need decimals just yet.

Example. A science center sold 1,363 tickets on a busy weekend. The receipts totaled $12,146. How many $12 adult and how many $7 child tickets were sold?

Let $a$ be the number of adult tickets and $c$ the number of child tickets.

The number column and the total value column give the system:

We’ll solve by elimination. Multiply the first equation by $-7$ so the $c$-terms cancel when we add:

So $a = 521$. Substituting into $a + c = 1{,}363$ gives $521 + c = 1{,}363$, so $c = 842$.

Check: $521$ adult tickets at $12 each make $521 \cdot 12 = 6{,}252$ dollars, and $842$ child tickets at $7 each make $842 \cdot 7 = 5{,}894$ dollars; together that’s $6{,}252 + 5{,}894 = 12{,}146$ dollars. ✓ The science center sold 521 adult tickets and 842 child tickets.

In the next example we solve a coin problem. Now that we can work with systems of two variables, naming the variables in the “number” column is easy — no more writing one count in terms of the other before we translate.

Example. 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 nickels. How many nickels and how many dimes does Juan have?

Let $n$ be the number of nickels and $d$ the number of dimes.

The total value column gives one equation. Translating “the number of dimes is 9 less than twice the number of nickels” gives the second:

We’ll solve by substitution, since the second equation is already solved for $d$. Substitute $d = 2n - 9$ into the first equation:

To find $d$, substitute $n = 36$ into $d = 2n - 9$: $d = 2(36) - 9 = 63$.

Check: $36$ nickels at $0.05 each make $36 \cdot 0.05 = 1.80$ dollars, and $63$ dimes at $0.10 each make $63 \cdot 0.10 = 6.30$ dollars; together that’s $1.80 + 6.30 = 8.10$ dollars. Also $2(36) - 9 = 63$. ✓ Juan has 36 nickels and 63 dimes.

Some mixture applications involve combining foods rather than coins or tickets — for example, mixing nuts and chocolate chips to make a trail mix.

Example. Carson wants to make 20 pounds of trail mix using nuts and chocolate chips. His budget requires that the trail mix cost him $7.60 a pound. Nuts cost $9.00 a pound and chocolate chips cost $2.00 a pound. How many pounds of nuts and how many pounds of chocolate chips should he use?

Let $n$ be the number of pounds of nuts and $c$ the number of pounds of chocolate chips.

The number column and the total value column give the system:

Solving by elimination — multiply the first equation by $-2$:

So $n = 16$. Substituting into $n + c = 20$ gives $c = 4$.

Check: $9(16) + 2(4) = 144 + 8 = 152$ dollars, and $16 + 4 = 20$ pounds. ✓ Carson should mix 16 pounds of nuts with 4 pounds of chocolate chips.

Another mixture application relates to concentrated cleaning supplies and other chemicals, where 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 of the cleanser with 28 ounces of water. For these problems we’ll use percent, instead of value, for one of the columns in our table.

Example. Sasheena is a 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?

Let $x$ be the number of milliliters of the $25\%$ solution and $y$ the number of milliliters of the $50\%$ solution.

We multiply the number of units by the concentration to get the total amount of sulfuric acid in each solution. The number column and the amount column give the system:

Solving by elimination — multiply the first equation by $-0.5$:

So $x = 80$. Substituting into $x + y = 200$ gives $y = 120$.

Check: $0.25(80) + 0.50(120) = 20 + 60 = 80$, which matches $0.40(200) = 80$, and $80 + 120 = 200$ ml. ✓ Sasheena should mix 80 ml of the $25\%$ solution with 120 ml of the $50\%$ solution.

Solve interest applications

We can also use the mixture model to solve investment and loan applications using simple interest. The formula to model interest applications is $I = Prt$: the interest $I$ is the product of the principal $P$, the rate $r$, and the time $t$. Here we’ll calculate the interest earned in one year, so $t = 1$ and the formula becomes $I = Pr$. We modify the column headings in the mixture table to show this formula.

Example. 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 earn $3\%$ per year. How much money should he put into each account?

Let $s$ be the amount invested in the stock fund and $b$ the amount invested in bonds.

Notice that the principal column represents the total amount of money invested, while the interest column represents only the interest earned. The principal column and the interest column give the system:

Solving by elimination — multiply the first equation by $-0.03$:

So $s = 32{,}800$. Substituting into $s + b = 40{,}000$ gives $b = 7{,}200$.

Check: $0.08(32{,}800) + 0.03(7{,}200) = 2{,}624 + 216 = 2{,}840$, which matches $0.071(40{,}000) = 2{,}840$, and $32{,}800 + 7{,}200 = 40{,}000$. ✓ Adnan should invest $32,800 in the stock fund and $7,200 in bonds.

The next example requires that we find the principal, given the amount of interest earned.

Example. Rosie owes $21,540 on two student loans. The interest rate on her bank loan is $10.5\%$ and the interest rate on her 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?

Let $b$ be the principal for the bank loan and $f$ the principal for the federal loan.

The principal column and the interest column give the system:

We’ll solve by substitution. Solving the first equation for $b$ gives $b = -f + 21{,}540$; substituting into the second equation:

To find $b$, substitute $f = 12{,}870$ into $b + f = 21{,}540$: $b = 8{,}670$.

Check: $0.105(8{,}670) + 0.059(12{,}870) = 910.35 + 759.33 = 1{,}669.68$, and $8{,}670 + 12{,}870 = 21{,}540$. ✓ The principal for Rosie’s bank loan is $8,670 and the principal for her federal loan is $12,870.

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 cost to produce $x$ units. This is the cost to manufacture 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 — 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.

Example. 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.

(a) Find the cost function $C$ when $x$ benches are manufactured.

(b) Find the revenue function $R$ when $x$ benches are sold.

(c) Show the break-even point by graphing both the revenue and cost functions on the same grid.

(d) Find the break-even point. Interpret what the break-even point means.

(a) 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:

(b) The manufacturer sells each weight training bench for $245. We get the total revenue by multiplying the revenue per unit by the number of units sold:

(c) Together, the cost and revenue functions form a system of linear equations:

Graphing $y = 105x + 7{,}000$ and $y = 245x$ on the same grid (with benches on the horizontal axis and dollars on the vertical axis), the cost line starts at $(0, 7{,}000)$ and climbs gently, while the revenue line starts at the origin $(0,0)$ and climbs more steeply, since $245 > 105$. Because the revenue line starts lower but climbs faster, the two lines cross exactly once — at the break-even point:

(d) To find the actual value, remember the break-even point occurs when costs equal revenue:

When 50 benches are sold, the costs equal the revenue: $C(50) = 105(50) + 7{,}000 = 12{,}250$ and $R(50) = 245(50) = 12{,}250$. This corresponds to the ordered pair $(50, 12{,}250)$ — when 50 benches are sold, both the cost and the revenue are $12,250.

Key terms

total value model — $\text{number} \cdot \text{value} = \text{total value}$: in a mixture problem with two unknown types, the number column and the total value column each give an equation, and together they form a system of two equations in two variables. simple interest formula — $I = Prt$, or $I = Pr$ when $t = 1$ year; in an investment or loan mixture problem, the principal column gives one equation (the total amount invested or owed) and the interest column gives the other (the total interest earned or paid). cost function — $C(x) = (\text{cost per unit}) \cdot x + \text{fixed costs}$, the total cost of manufacturing $x$ units. revenue function — $R(x) = (\text{selling price per unit}) \cdot x$, the total money brought in from selling $x$ units. break-even point — the point where the cost and revenue functions are equal, $C(x) = R(x)$.

This section is adapted from Intermediate Algebra 2e, Section 4.3: Solve Mixture Applications with Systems of Equations by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the ticket/coin/mixture/investment/loan tables as markdown tables; wrote each system of equations and its elimination or substitution steps as display-math blocks; recreated the cost-and-revenue break-even graph with the site’s coordinate-graph component; omitted the “Be Prepared” readiness quiz, the Media link, the Section 4.3 Exercises (“Practice Makes Perfect”) block, and the Self Check checklist; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.