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 type, number, value, and total value — filling in the value of each type of coin — so we could organize the information, following the model $\text{number} \cdot \text{value} = \text{total value}$. Using only one variable meant we had to relate the number of nickels and the number of dimes — 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 $n$ be the number of nickels and $d$ be the number of dimes. One equation will come from the Total Value column, like before, and the other equation will come from the Number 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. The box office at a movie theater sold 147 tickets for the evening show, and receipts totaled $1,302. How many $11 adult and how many $8 child tickets were sold?

We organize the information in a table. Let $a$ be the number of adult tickets and $c$ the number of child tickets sold.

The Number column gives one equation, and the Total Value column gives the other, so we have a system of two equations in two variables:

We’ll solve by elimination. Multiply the first equation by $-8$ so the $c$-terms cancel when we add:

So $a = 42$. Substituting into $a + c = 147$ gives $42 + c = 147$, so $c = 105$.

Check: $42$ adult tickets at $11 each make $42 \cdot 11 = 462$ dollars, and $105$ child tickets at $8 each make $105 \cdot 8 = 840$ dollars; together that’s $462 + 840 = 1{,}302$ dollars. ✓ The theater sold 42 adult tickets and 105 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.

Example. Priam has a collection of nickels and quarters, with a total value of $7.30. The number of nickels is six less than three times the number of quarters. How many nickels and how many quarters does he have?

Let $n$ be the number of nickels and $q$ the number of quarters.

The Total Value column gives one equation. Translating “the number of nickels is six less than three times the number of quarters” gives the second:

We’ll solve by substitution, since the second equation is already solved for $n$. Substitute $n = 3q - 6$ into the first equation:

To find $n$, substitute $q = 19$ into $n = 3q - 6$: $n = 3(19) - 6 = 51$.

Check: $19$ quarters at $0.25 each make $19 \cdot 0.25 = 4.75$ dollars, and $51$ nickels at $0.05 each make $51 \cdot 0.05 = 2.55$ dollars; together that’s $4.75 + 2.55 = 7.30$ dollars. Also $3 \cdot 19 - 6 = 51$. ✓ Priam has 51 nickels and 19 quarters.

Some mixture applications involve combining foods or drinks rather than coins — for example, mixing nuts and dried fruit to make a trail mix, or blending two kinds of coffee beans.

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: $80 + 120 = 200$ ml, and $0.25(80) + 0.50(120) = 20 + 60 = 80$, which matches $0.40(200) = 80$. ✓ 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 $t$. Since we’ll calculate the interest earned in one year, $t = 1$, 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: $0.08(32{,}800) + 0.03(7{,}200) = 2{,}624 + 216 = 2{,}840$ dollars, which matches $0.071(40{,}000) = 2{,}840$. ✓ Adnan should invest $32,800 in the stock fund and $7,200 in bonds.

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

Key terms

total value model — $\text{number} \cdot \text{value} = \text{total value}$: the number of coins, tickets, or units of an ingredient times the value of each gives the total value of that type; 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).

This section is adapted from Elementary Algebra 2e, Section 5.5: 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 coin/ticket/mixture/investment/loan tables as markdown tables; wrote each system of equations and its elimination or substitution steps as display-math blocks; omitted the “Be Prepared” readiness quiz, the Media links block, the Section 5.5 Exercises (“Practice Makes Perfect”) block, and the Self Check checklist; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.