To model simple interest applications with systems of equations, the mixture table can be adapted to reflect the simple interest formula, $$I = Prt$$. Since the interest earned is typically calculated for one year, the time variable $$t$$ is $$1$$, making the formula $$I = Pr$$. The table organizes the principal ($$P$$), the rate ($$r$$), and the interest ($$I$$) for each fund or loan. This structure allows the problem to be translated into a system of two equations: the first equation comes from the Principal column, representing the total amount of money invested or borrowed, and the second equation comes from the Interest column, representing the total interest earned or paid.

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When translating a mixture application to a system of equations, two separate variables are assigned to represent the unknown quantities, rather than expressing one in terms of the other. For instance, instead of letting one coin's count be $$c$$ and the other $$50 - c$$, a system approach allows letting $$n$$ be the number of nickels and $$d$$ be the number of dimes. This method naturally generates two equations: one drawn from the 'number' column of an organizing table (representing the total physical count) and a second drawn from the 'total value' column (representing the combined monetary or quantitative worth). Together, they form a system that can be solved using standard linear methods.

Setting Up a Mixture Application Using a System of Equations

https://openstax.org/details/books/intermediate-algebra-2e

OpenStax Intermediate Algebra

Apply the system of equations method to solve a ticket mixture application.

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

1. Name the unknowns. Let $$a =$$ the number of adult tickets and $$c =$$ the number of child tickets.
2. Translate into a system of equations.
 - The total number of tickets sold is $$1{,}363$$: $$a + c = 1{,}363$$
 - The total value of adult tickets is $$12a$$ and the total value of child tickets is $$7c$$. The total receipts are $$\$12{,}146$$: $$12a + 7c = 12{,}146$$
 The system of equations is:
 $$a + c = 1{,}363$$
 $$12a + 7c = 12{,}146$$
3. Solve the system using the elimination method. Multiply the first equation by $$-7$$:
 $$-7(a + c) = -7(1{,}363)$$
 $$-7a - 7c = -9{,}541$$
 Add this to the second equation:
 $$12a + 7c = 12{,}146$$
 $$5a = 2{,}605$$
 $$a = 521$$
 Substitute $$a = 521$$ into the first equation:
 $$521 + c = 1{,}363$$
 $$c = 842$$
4. Check the result.
 $$521$$ adult tickets at $$\$12$$ each is $$\$6{,}252$$.
 $$842$$ child tickets at $$\$7$$ each is $$\$5{,}894$$.
 $$6{,}252 + 5{,}894 = 12{,}146$$ $$\checkmark$$
5. Answer the question. The science center sold $$521$$ adult tickets and $$842$$ child tickets.

Solving a Science Center Ticket Mixture Application

To solve a coin mixture application where the total value and a relationship between the quantity of coins are known, set up a system of two equations. For example, if a pocketful of nickels and dimes has a total value of \$8.10, and the number of dimes is $$9$$ less than twice the number of nickels, we can represent this scenario by letting $$n$$ be the number of nickels and $$d$$ be the number of dimes. We multiply each quantity by its monetary value to write the total value equation: $$0.05n + 0.10d = 8.10$$. The relationship between the quantities provides the second equation: $$d = 2n - 9$$. We can then solve the resulting system using substitution. Substituting $$2n - 9$$ for $$d$$ in the first equation yields $$0.05n + 0.10(2n - 9) = 8.10$$, which simplifies to $$0.05n + 0.20n - 0.90 = 8.10$$ and gives $$0.25n = 9.00$$, so $$n = 36$$. Substituting this value back into the second equation yields $$d = 2(36) - 9 = 63$$. The pocket contains $$36$$ nickels and $$63$$ dimes.

Solving a Coin Mixture Problem Using a System of Equations

For a mixture problem involving quarters and dimes, a system of two equations can be translated from the given conditions to find the quantity of each coin. For instance, if a handful of quarters and dimes has a total value of \$8.55, and the number of quarters is $$3$$ more than twice the number of dimes, we let $$q$$ be the number of quarters and $$d$$ be the number of dimes. The total monetary value provides the first equation: $$0.10d + 0.25q = 8.55$$. The quantity relationship gives the second equation: $$q = 2d + 3$$. Substituting $$2d + 3$$ for $$q$$ into the first equation allows us to solve for $$d$$. Finding $$0.10d + 0.25(2d + 3) = 8.55$$ gives $$0.10d + 0.50d + 0.75 = 8.55$$, so $$0.60d = 7.80$$, which means $$d = 13$$. Substituting $$d = 13$$ into the second equation reveals $$q = 2(13) + 3 = 29$$. There are $$13$$ dimes and $$29$$ quarters.

Solving a Quarters and Dimes Mixture Problem Using a System of Equations

A mixture problem involving nickels and quarters can be modeled as a system of equations by translating the total value and the relationship between the quantities. For example, if a collection of nickels and quarters has a total value of \$7.30, and the number of nickels is $$6$$ less than three times the number of quarters, we can let $$n$$ be the number of nickels and $$q$$ be the number of quarters. The total value equation is $$0.05n + 0.25q = 7.30$$. The relationship between the coins translates to $$n = 3q - 6$$. Using substitution, replacing $$n$$ with $$3q - 6$$ in the first equation yields $$0.05(3q - 6) + 0.25q = 7.30$$. Distributing gives $$0.15q - 0.30 + 0.25q = 7.30$$, which simplifies to $$0.40q = 7.60$$, resulting in $$q = 19$$. Substituting $$q = 19$$ into the second equation determines $$n = 3(19) - 6 = 51$$. The collection consists of $$51$$ nickels and $$19$$ quarters.

Solving a Nickels and Quarters Mixture Problem Using a System of Equations

To solve an acid mixture application, use a system of linear equations by organizing the volumes and concentrations in a table. For instance, if a lab assistant needs $$200$$ milliliters of a $$40\%$$ sulfuric acid solution by combining a $$25\%$$ solution and a $$50\%$$ solution, let $$x$$ be the volume of the $$25\%$$ solution and $$y$$ be the volume of the $$50\%$$ solution. The first equation represents the total mixture volume: $$x + y = 200$$. The second equation represents the amount of pure acid (volume multiplied by concentration as a decimal): $$0.25x + 0.50y = 0.40(200)$$, which simplifies to $$0.25x + 0.50y = 80$$. To solve this system by elimination, multiply the first equation by $$-0.5$$ to eliminate $$y$$, resulting in $$-0.5x - 0.5y = -100$$. Adding this to the second equation yields $$-0.25x = -20$$, which gives $$x = 80$$. Substituting $$x = 80$$ into the first equation ($$80 + y = 200$$) gives $$y = 120$$. Therefore, combining $$80$$ milliliters of the $$25\%$$ solution with $$120$$ milliliters of the $$50\%$$ solution will produce the desired $$40\%$$ mixture.

Solving a Laboratory Acid Mixture Problem Using a System of Equations

A system of equations can be used to determine the necessary volumes of different sulfuric acid concentrations needed to create a specific mixture. For example, if a student needs $$150$$ milliliters of a $$30\%$$ sulfuric acid solution and only has access to $$25\%$$ and $$50\%$$ solutions, let $$x$$ represent the amount of the $$25\%$$ solution and $$y$$ represent the amount of the $$50\%$$ solution. The total volume constraint gives the first equation: $$x + y = 150$$. The pure acid content constraint gives the second equation: $$0.25x + 0.50y = 0.30(150)$$, which simplifies to $$0.25x + 0.50y = 45$$. Solving this system by substituting $$y = 150 - x$$ into the second equation yields $$0.25x + 0.50(150 - x) = 45$$. Distributing gives $$0.25x + 75 - 0.50x = 45$$, which simplifies to $$-0.25x = -30$$, resulting in $$x = 120$$. Substituting this back gives $$y = 150 - 120 = 30$$. The student should mix $$120$$ milliliters of the $$25\%$$ solution with $$30$$ milliliters of the $$50\%$$ solution.

Solving a Sulfuric Acid Mixture Application Using a System of Equations

To find the exact volumes of two hydrochloric acid solutions needed to form a target mixture, we can set up a system of equations. For example, to make $$250$$ milliliters of a $$25\%$$ hydrochloric acid solution using available $$10\%$$ and $$40\%$$ solutions, let $$x$$ be the volume of the $$10\%$$ solution and $$y$$ be the volume of the $$40\%$$ solution. The equation for the total volume is $$x + y = 250$$. The equation for the pure acid amount is $$0.10x + 0.40y = 0.25(250)$$, or $$0.10x + 0.40y = 62.5$$. By solving the system using substitution (where $$y = 250 - x$$), we get $$0.10x + 0.40(250 - x) = 62.5$$. This expands to $$0.10x + 100 - 0.40x = 62.5$$, simplifying to $$-0.30x = -37.5$$, meaning $$x = 125$$. Substituting back finds $$y = 250 - 125 = 125$$. Therefore, mixing $$125$$ milliliters of the $$10\%$$ solution and $$125$$ milliliters of the $$40\%$$ solution produces the required $$25\%$$ mixture.

Solving a Hydrochloric Acid Mixture Problem Using a System of Equations

To solve a mixture application involving solid ingredients, you can translate the problem's constraints into a system of equations by organizing the known and unknown amounts and values. For instance, suppose Carson wants to make $$20$$ pounds of trail mix using nuts and chocolate chips, with a target cost of $$\$7.60$$ per pound. If nuts cost $$\$9.00$$ per pound and chocolate chips cost $$\$2.00$$ per pound, we can determine the required amounts.

First, define the variables: Let $$n$$ represent the pounds of nuts and $$c$$ represent the pounds of chocolate chips.

Next, set up the system of equations. The first equation represents the total amount: $$n + c = 20$$. The second equation represents the total value. The value of the nuts is $$9n$$ and the value of the chocolate chips is $$2c$$. The total expected value of the mixture is $$20(7.60) = 152$$. Therefore, the second equation is $$9n + 2c = 152$$.

Finally, solve the system. Using the elimination method, multiply the first equation by $$-2$$ to get $$-2n - 2c = -40$$. Adding this to the second equation eliminates $$c$$, resulting in $$7n = 112$$, which simplifies to $$n = 16$$. Substituting $$n = 16$$ back into the first equation ($$16 + c = 20$$) yields $$c = 4$$. Thus, the mixture should consist of $$16$$ pounds of nuts and $$4$$ pounds of chocolate chips.

Solving a Trail Mix Mixture Problem Using a System of Equations

A system of equations can be applied to solve a solid mixture problem to determine how much of each ingredient is needed to achieve a specific total amount and cost. For example, if Greta wants to make $$5$$ pounds of a nut mix using peanuts and cashews tailored to cost $$\$6$$ per pound, she can use an algebraic system to find the exact amounts.

Given that peanuts cost $$\$4$$ per pound and cashews cost $$\$9$$ per pound, variables can be defined for each unknown, such as $$p$$ for pounds of peanuts and $$c$$ for pounds of cashews. The first equation represents the total desired weight ($$p + c = 5$$), while the second equation represents the total financial value ($$4p + 9c = 5(6)$$, or $$4p + 9c = 30$$). By solving this system of two equations, Greta can accurately calculate the required pounds of peanuts and cashews for her mixture.

Solving a Nut Mix Application Using a System of Equations

Systems of equations are highly useful for determining the necessary quantities of different ingredients in a recipe while adhering to a budget constraint. For example, suppose Sammy is making a large batch of chili and needs a combined total of $$20$$ pounds of beans and ground beef. If his budget is set at $$\$3$$ per pound, he can translate this scenario into an algebraic system.

With beans priced at $$\$1$$ per pound and ground beef at $$\$5$$ per pound, Sammy can set up one equation to model the total weight of the two ingredients combined, and a second equation to model their total overall cost. Solving this system algebraically allows him to compute exactly how many pounds of beans and how many pounds of ground beef to purchase to meet both his recipe requirements and his overall budget.

Solving a Recipe Mixture Application Using a System of Equations

Setting Up an Interest Application Using a System of Equations

To solve an interest application where a principal is divided between two investments, apply a system of equations by adapting the mixture table method. For example, if Adnan has $$\$40{,}000$$ to invest and hopes to earn $$7.1\%$$ interest per year, splitting it between a stock fund earning $$8\%$$ and bonds earning $$3\%$$, let $$s$$ be the amount in stocks and $$b$$ be the amount in bonds. The total principal gives the first equation: $$s + b = 40{,}000$$. The total interest gives the second equation: $$0.08s + 0.03b = 0.071(40{,}000)$$. Solving this system, such as by multiplying the top equation by $$-0.03$$ to eliminate $$b$$, yields $$s = 32{,}800$$ and $$b = 7{,}200$$. Adnan should invest $$\$32{,}800$$ in stocks and $$\$7{,}200$$ in bonds.

Solving an Investment Fund Interest Application Using a System of Equations

A system of equations can solve an application with two investments seeking a target overall interest rate. For instance, if Leon has $$\$50{,}000$$ to invest to earn $$6.2\%$$ interest per year, allocating it between a stock fund earning $$7\%$$ and a savings account earning $$2\%$$, let $$s$$ and $$a$$ represent the amounts in the stock fund and savings account, respectively. The total principal equation is $$s + a = 50{,}000$$. The total interest equation is $$0.07s + 0.02a = 0.062(50{,}000)$$. Solving this system provides the optimal allocation: he should invest $$\$42{,}000$$ in the stock fund and $$\$8{,}000$$ in the savings account.

Solving a Savings and Stock Interest Application Using a System of Equations

A system of equations can model an application involving two stock investments. For example, if Julius invested $$\$7{,}000$$ into two stocks—one paying $$11\%$$ interest and the other paying $$13\%$$—and earned $$12.5\%$$ interest on the overall investment, let $$x$$ and $$y$$ be the amounts invested in the respective stocks. The total principal equation is $$x + y = 7{,}000$$. The interest earned equation is $$0.11x + 0.13y = 0.125(7{,}000)$$. Solving the system reveals that he placed $$\$1{,}750$$ into the $$11\%$$ stock and $$\$5{,}250$$ into the $$13\%$$ stock.

Solving a Dual Stock Interest Application Using a System of Equations

To determine the separate principal amounts for different loans when only the total principal and total interest are given, set up a system of equations by adapting mixture models. For example, if Rosie owes $$\$21{,}540$$ on two student loans—a bank loan at $$10.5\%$$ interest and a federal loan at $$5.9\%$$ interest—and her total interest paid last year was $$\$1{,}669.68$$, let $$b$$ be the bank loan principal and $$f$$ be the federal loan principal. The total loans equation is $$b + f = 21{,}540$$. The total interest equation is $$0.105b + 0.059f = 1{,}669.68$$. Using substitution, this system yields $$f = 12{,}870$$ and $$b = 8{,}670$$. The principal was $$\$8{,}670$$ for the bank loan and $$\$12{,}870$$ for the federal loan.

Solving a Student Loan Interest Application Using a System of Equations

A system of equations can solve for the principal amounts of two distinct student loans using mixture principles. For instance, if Laura owes $$\$18{,}000$$ on a bank loan at $$2.5\%$$ interest and a federal loan at $$6.9\%$$ interest, and she paid a total of $$\$1{,}066$$ in interest, let $$b$$ be the bank loan amount and $$f$$ be the federal loan amount. The principal equation is $$b + f = 18{,}000$$ and the interest equation is $$0.025b + 0.069f = 1{,}066$$. Solving this system establishes that her bank loan principal was $$\$4{,}000$$ and her federal loan principal was $$\$14{,}000$$.

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