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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

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

Apply the seven-step problem-solving strategy and a table to solve a coin mixture application using a system of linear equations.

**Problem:** 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?

1. **Read** the problem. A table will help organize the information.
2. **Identify** what to find: the number of nickels and the number of dimes.
3. **Name** the unknowns. Let $$n =$$ the number of nickels and $$d =$$ the number of dimes. Organize the data into a table by listing the coin types, their quantities, their monetary values, and their total values:

| Type | Number | Value (\$) | Total Value (\$) |
|---   |---     |---         |---               |
| Nickels | $$n$$ | $$0.05$$ | $$0.05n$$ |
| Dimes | $$d$$ | $$0.10$$ | $$0.10d$$ |
| Total | | | $$8.10$$ |

4. **Translate** into a system of equations.
   - The "Total Value" column gives the first equation:
     $$0.05n + 0.10d = 8.10$$
   - The relationship between the quantities (the number of dimes is $$9$$ less than twice the number of nickels) gives the second equation:
     $$d = 2n - 9$$

   The system of equations is:
   $$\left\{\begin{array}{l} 0.05n + 0.10d = 8.10 \\ d = 2n - 9 \end{array}\right.$$

5. **Solve** the system using the substitution method. Substitute $$2n - 9$$ for $$d$$ in the first equation:
   $$0.05n + 0.10(2n - 9) = 8.10$$
   
   Distribute the $$0.10$$:
   $$0.05n + 0.20n - 0.90 = 8.10$$
   
   Combine like terms:
   $$0.25n - 0.90 = 8.10$$
   
   Add $$0.90$$ to both sides:
   $$0.25n = 9.00$$
   
   Divide by $$0.25$$:
   $$n = 36$$
   
   Substitute $$n = 36$$ into the second equation to find $$d$$:
   $$d = 2(36) - 9 = 72 - 9 = 63$$

6. **Check** the result.
   $$36$$ nickels at \$0.05 each is \$1.80.
   $$63$$ dimes at \$0.10 each is \$6.30.
   $$1.80 + 6.30 = 8.10$$ $$\checkmark$$

7. **Answer** the question. Juan has $$36$$ nickels and $$63$$ dimes.

Solving a Coin Mixture Problem Using a System of Equations

Apply the seven-step problem-solving strategy and a table to solve a coin mixture application involving quarters and dimes using a system of linear equations.

**Problem:** Matilda has a handful of quarters and dimes, with a total value of \$8.55. The number of quarters is $$3$$ more than twice the number of dimes. How many dimes and how many quarters does she have?

1. **Read** the problem. A table will help organize the information.
2. **Identify** what to find: the number of dimes and the number of quarters.
3. **Name** the unknowns. Let $$q =$$ the number of quarters and $$d =$$ the number of dimes. Organize the data into a table:

| Type | Number | Value (\$) | Total Value (\$) |
|---   |---     |---         |---               |
| Quarters | $$q$$ | $$0.25$$ | $$0.25q$$ |
| Dimes | $$d$$ | $$0.10$$ | $$0.10d$$ |
| Total | | | $$8.55$$ |

4. **Translate** into a system of equations.
   - The "Total Value" column gives the first equation:
     $$0.25q + 0.10d = 8.55$$
   - The relationship between the quantities (quarters is $$3$$ more than twice dimes) gives the second equation:
     $$q = 2d + 3$$

   The system of equations is:
   $$\left\{\begin{array}{l} 0.25q + 0.10d = 8.55 \\ q = 2d + 3 \end{array}\right.$$

5. **Solve** the system using the substitution method. Substitute $$2d + 3$$ for $$q$$ in the first equation:
   $$0.25(2d + 3) + 0.10d = 8.55$$
   
   Distribute the $$0.25$$:
   $$0.50d + 0.75 + 0.10d = 8.55$$
   
   Combine like terms:
   $$0.60d + 0.75 = 8.55$$
   
   Subtract $$0.75$$ from both sides:
   $$0.60d = 7.80$$
   
   Divide by $$0.60$$:
   $$d = 13$$
   
   Substitute $$d = 13$$ into the second equation to find $$q$$:
   $$q = 2(13) + 3 = 26 + 3 = 29$$

6. **Check** the result.
   $$29$$ quarters at \$0.25 each is \$7.25.
   $$13$$ dimes at \$0.10 each is \$1.30.
   $$7.25 + 1.30 = 8.55$$ $$\checkmark$$

7. **Answer** the question. Matilda has $$13$$ dimes and $$29$$ quarters.

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

Apply the seven-step problem-solving strategy and a table to solve a coin mixture application involving nickels and quarters using a system of linear equations.

**Problem:** Priam has a collection of nickels and quarters, with a total value of \$7.30. The number of nickels is $$6$$ less than three times the number of quarters. How many nickels and how many quarters does he have?

1. **Read** the problem. A table will help organize the information.
2. **Identify** what to find: the number of nickels and the number of quarters.
3. **Name** the unknowns. Let $$n =$$ the number of nickels and $$q =$$ the number of quarters. Organize the data into a table:

| Type | Number | Value (\$) | Total Value (\$) |
|---   |---     |---         |---               |
| Nickels | $$n$$ | $$0.05$$ | $$0.05n$$ |
| Quarters | $$q$$ | $$0.25$$ | $$0.25q$$ |
| Total | | | $$7.30$$ |

4. **Translate** into a system of equations.
   - The "Total Value" column gives the first equation:
     $$0.05n + 0.25q = 7.30$$
   - The relationship between the quantities (nickels is $$6$$ less than three times quarters) gives the second equation:
     $$n = 3q - 6$$

   The system of equations is:
   $$\left\{\begin{array}{l} 0.05n + 0.25q = 7.30 \\ n = 3q - 6 \end{array}\right.$$

5. **Solve** the system using the substitution method. Substitute $$3q - 6$$ for $$n$$ in the first equation:
   $$0.05(3q - 6) + 0.25q = 7.30$$
   
   Distribute the $$0.05$$:
   $$0.15q - 0.30 + 0.25q = 7.30$$
   
   Combine like terms:
   $$0.40q - 0.30 = 7.30$$
   
   Add $$0.30$$ to both sides:
   $$0.40q = 7.60$$
   
   Divide by $$0.40$$:
   $$q = 19$$
   
   Substitute $$q = 19$$ into the second equation to find $$n$$:
   $$n = 3(19) - 6 = 57 - 6 = 51$$

6. **Check** the result.
   $$51$$ nickels at \$0.05 each is \$2.55.
   $$19$$ quarters at \$0.25 each is \$4.75.
   $$2.55 + 4.75 = 7.30$$ $$\checkmark$$

7. **Answer** the question. Priam has $$51$$ 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 solve sulfuric acid mixture applications. For example, if $$150$$ milliliters of a $$30\%$$ sulfuric acid solution is needed, but only $$25\%$$ and $$50\%$$ solutions are available, we can determine the required amounts of each. Let $$x$$ be the volume of the $$25\%$$ solution and $$y$$ be the volume of the $$50\%$$ solution. The system of equations is set up as follows:
$$x + y = 150$$
$$0.25x + 0.5y = 0.30(150)$$
Solving this system provides the specific volumes of each starting solution needed to achieve the target mixture.

Solving a Sulfuric Acid Mixture Application Using a System of Equations

A system of equations can be used to solve hydrochloric acid mixture applications. For instance, if $$250$$ milliliters of a $$25\%$$ hydrochloric acid solution is required, and only $$10\%$$ and $$40\%$$ solutions are available, the necessary volumes of each can be found. Let $$x$$ be the volume of the $$10\%$$ solution and $$y$$ be the volume of the $$40\%$$ solution. The system of equations translates to:
$$x + y = 250$$
$$0.10x + 0.40y = 0.25(250)$$
Solving this system yields the exact amounts of the $$10\%$$ and $$40\%$$ solutions that must be mixed.

Solving a Hydrochloric Acid Mixture Problem Using a System of Equations

When applying the standard problem-solving strategy to a food mixture application, a system of linear equations is used to determine the exact amounts of two ingredients. For example, if a person wants to prepare $$20$$ pounds of trail mix using nuts (priced at $$\$9.00$$ per pound) and chocolate chips (priced at $$\$2.00$$ per pound) with a target budget of $$\$7.60$$ per pound, two variables can be defined: let $$n$$ represent the pounds of nuts and $$c$$ represent the pounds of chocolate chips. The first equation represents the total weight constraint: $$n + c = 20$$. The second equation represents the total cost constraint (the value of the nuts plus the value of the chips equals the value of the entire mixture): $$9n + 2c = 20(7.60)$$, which simplifies to $$9n + 2c = 152$$. This forms a system of two equations. To solve by elimination, the first equation can be multiplied by $$-2$$ to give $$-2n - 2c = -40$$. Adding this to the second equation yields $$7n = 112$$, meaning $$n = 16$$. Substituting $$16$$ back into the first equation ($$16 + c = 20$$) reveals that $$c = 4$$. After verifying the total cost ($$16(9) + 4(2) = 144 + 8 = 152$$), it is concluded that $$16$$ pounds of nuts and $$4$$ pounds of chocolate chips are needed.

Solving a Trail Mix Mixture Problem Using a System of Equations

A system of linear equations is an effective tool for solving food mixture applications that involve combining two ingredients to meet specific weight and cost requirements. Consider a scenario where $$5$$ pounds of a nut mix is desired, using peanuts (priced at $$\$4$$ per pound) and cashews (priced at $$\$9$$ per pound), with an overall budget constraint of $$\$6$$ per pound. By letting $$p$$ be the pounds of peanuts and $$c$$ be the pounds of cashews, the total weight constraint is modeled as $$p + c = 5$$. The total cost constraint is modeled by adding the value of each component and setting it equal to the total value of the mixture: $$4p + 9c = 5(6)$$, which gives $$4p + 9c = 30$$. Solving this system using the elimination method involves multiplying the weight equation by $$-4$$ (resulting in $$-4p - 4c = -20$$) and adding it to the cost equation. This eliminates $$p$$ and yields $$5c = 10$$, so $$c = 2$$. Substituting $$c = 2$$ back into the first equation ($$p + 2 = 5$$) gives $$p = 3$$. Verifying the cost confirms that $$3$$ pounds of peanuts and $$2$$ pounds of cashews accurately produce the required $$5$$-pound mixture at the target price.

Solving a Nut Mix Application Using a System of Equations

Systems of linear equations can determine the exact quantities of different ingredients required in a recipe to satisfy both a weight requirement and a budget constraint. For instance, if a large batch of chili requires a combined total of $$20$$ pounds of beans and ground beef, with an average target cost of $$\$3$$ per pound, this scenario can be modeled algebraically. If beans cost $$\$1$$ per pound and ground beef costs $$\$5$$ per pound, one can assign variables such as $$b$$ for the pounds of beans and $$g$$ for the pounds of ground beef. The total weight constraint translates to the equation $$b + g = 20$$. The cost constraint is expressed by equating the individual costs to the total mixture value: $$1b + 5g = 20(3)$$, which simplifies to $$b + 5g = 60$$. To solve this system by elimination, the weight equation can be multiplied by $$-1$$ (resulting in $$-b - g = -20$$) and added to the cost equation. This eliminates $$b$$ and yields $$4g = 40$$, meaning $$g = 10$$. Substituting $$10$$ back into the first equation ($$b + 10 = 20$$) reveals that $$b = 10$$. Thus, exactly $$10$$ pounds of beans and $$10$$ pounds of ground beef must be purchased to perfectly meet the recipe and budget requirements.

Solving a Recipe Mixture Application Using a System of Equations

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.

Setting Up an Interest Application Using a System of Equations

Apply the seven-step problem-solving strategy and a mixture table to solve an interest application involving two investments. Problem: 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? 1. Read the problem. A chart will help organize the information. 2. Identify what to find: the amount to invest in each fund. 3. Name the unknowns. Let $$s$$ = the amount invested in stocks, and $$b$$ = the amount invested in bonds. Multiply Principal $$\cdot$$ Rate $$\cdot$$ Time ($$t=1$$) to get the Interest. Organize into a table: / Fund / Principal (\$) / Rate / Interest (\$) / /---/---/---/---/ / Stocks / $$s$$ / $$0.08$$ / $$0.08s$$ / / Bonds / $$b$$ / $$0.03$$ / $$0.03b$$ / / Total / $$40{,}000$$ / $$0.071$$ / $$0.071(40{,}000)$$ / 4. Translate into a system of equations from the Principal and Interest columns: $$\left\{\begin{array}{l} s + b = 40{,}000 \ 0.08s + 0.03b = 0.071(40{,}000) \end{array}ight.$$ 5. Solve the system by elimination. Multiply the top equation by $$-0.03$$: $$-0.03s - 0.03b = -1{,}200$$ $$0.08s + 0.03b = 2{,}840$$ Add the equations: $$0.05s = 1{,}640$$ $$s = 32{,}800$$ Substitute $$s = 32{,}800$$ into the first equation: $$32{,}800 + b = 40{,}000$$ $$b = 7{,}200$$ 6. Check the answer. $$32{,}800 + 7{,}200 = 40{,}000$$ $$\checkmark$$ $$0.08(32{,}800) + 0.03(7{,}200) = 2{,}624 + 216 = 2{,}840$$ $$0.071(40{,}000) = 2{,}840$$ $$\checkmark$$ 7. Answer the question. Adnan should invest $$\$32{,}800$$ in stock and $$\$7{,}200$$ in bonds.

Solving an Investment Fund Interest Application Using a System of Equations

Apply the seven-step problem-solving strategy and a mixture table to solve an interest application involving two investments. Problem: Leon had $$\$50{,}000$$ to invest and hopes to earn $$6.2\%$$ interest per year. He will put some of the money into a stock fund that earns $$7\%$$ per year and the rest in to a savings account that earns $$2\%$$ per year. How much money should he put into each fund? 1. Read the problem. 2. Identify what to find: the amount to invest in each fund. 3. Name the unknowns. Let $$s$$ = the amount invested in stocks, and $$a$$ = the amount invested in the savings account. Organize into a table: / Fund / Principal (\$) / Rate / Interest (\$) / /---/---/---/---/ / Stocks / $$s$$ / $$0.07$$ / $$0.07s$$ / / Savings / $$a$$ / $$0.02$$ / $$0.02a$$ / / Total / $$50{,}000$$ / $$0.062$$ / $$0.062(50{,}000)$$ / 4. Translate into a system of equations: $$\left\{\begin{array}{l} s + a = 50{,}000 \ 0.07s + 0.02a = 0.062(50{,}000) \end{array}ight.$$ 5. Solve the system by elimination. Multiply the top equation by $$-0.02$$: $$-0.02s - 0.02a = -1{,}000$$ $$0.07s + 0.02a = 3{,}100$$ Add the equations: $$0.05s = 2{,}100$$ $$s = 42{,}000$$ Substitute $$s = 42{,}000$$ into the first equation: $$42{,}000 + a = 50{,}000$$ $$a = 8{,}000$$ 6. Check the answer. $$42{,}000 + 8{,}000 = 50{,}000$$ $$\checkmark$$ $$0.07(42{,}000) + 0.02(8{,}000) = 2{,}940 + 160 = 3{,}100$$ $$0.062(50{,}000) = 3{,}100$$ $$\checkmark$$ 7. Answer the question. Leon 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

Apply the seven-step problem-solving strategy and a mixture table to solve an interest application involving two investments. Problem: Julius invested $$\$7{,}000$$ into two stock investments. One stock paid $$11\%$$ interest and the other stock paid $$13\%$$ interest. He earned $$12.5\%$$ interest on the total investment. How much money did he put in each stock? 1. Read the problem. 2. Identify what to find: the amount invested in each stock. 3. Name the unknowns. Let $$x$$ = the amount invested at $$11\%$$, and $$y$$ = the amount invested at $$13\%$$. Organize into a table: / Fund / Principal (\$) / Rate / Interest (\$) / /---/---/---/---/ / 11% Stock / $$x$$ / $$0.11$$ / $$0.11x$$ / / 13% Stock / $$y$$ / $$0.13$$ / $$0.13y$$ / / Total / $$7{,}000$$ / $$0.125$$ / $$0.125(7{,}000)$$ / 4. Translate into a system of equations: $$\left\{\begin{array}{l} x + y = 7{,}000 \ 0.11x + 0.13y = 0.125(7{,}000) \end{array}ight.$$ 5. Solve the system by elimination. Multiply the top equation by $$-0.11$$: $$-0.11x - 0.11y = -770$$ $$0.11x + 0.13y = 875$$ Add the equations: $$0.02y = 105$$ $$y = 5{,}250$$ Substitute $$y = 5{,}250$$ into the first equation: $$x + 5{,}250 = 7{,}000$$ $$x = 1{,}750$$ 6. Check the answer. $$1{,}750 + 5{,}250 = 7{,}000$$ $$\checkmark$$ $$0.11(1{,}750) + 0.13(5{,}250) = 192.5 + 682.5 = 875$$ $$0.125(7{,}000) = 875$$ $$\checkmark$$ 7. Answer the question. Julius invested $$\$1{,}750$$ at $$11\%$$ and $$\$5{,}250$$ at $$13\%$$.

Solving a Dual Stock Interest Application Using a System of Equations

Apply the seven-step problem-solving strategy and a mixture table to find the principal given the amount of interest earned. Problem: 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? 1. Read the problem. A chart will help organize the information. 2. Identify what to find: the principal of each loan. 3. Name the unknowns. Let $$b$$ = the principal for the bank loan, and $$f$$ = the principal for the federal loan. Multiply Principal $$\cdot$$ Rate $$\cdot$$ Time ($$t=1$$) to get the Interest. Organize into a table: / Loan / Principal (\$) / Rate / Interest (\$) / /---/---/---/---/ / Bank / $$b$$ / $$0.105$$ / $$0.105b$$ / / Federal / $$f$$ / $$0.059$$ / $$0.059f$$ / / Total / $$21{,}540$$ / / $$1{,}669.68$$ / 4. Translate into a system of equations from the Principal and Interest columns: $$\left\{\begin{array}{l} b + f = 21{,}540 \ 0.105b + 0.059f = 1{,}669.68 \end{array}ight.$$ 5. Solve the system using substitution. Solve the first equation for $$b$$: $$b = -f + 21{,}540$$ Substitute into the second equation: $$0.105(-f + 21{,}540) + 0.059f = 1{,}669.68$$ $$-0.105f + 2{,}261.7 + 0.059f = 1{,}669.68$$ $$-0.046f + 2{,}261.7 = 1{,}669.68$$ $$-0.046f = -592.02$$ $$f = 12{,}870$$ To find $$b$$, substitute $$f = 12{,}870$$ into the first equation: $$b = -12{,}870 + 21{,}540$$ $$b = 8{,}670$$ 6. Check the answer. $$8{,}670 + 12{,}870 = 21{,}540$$ $$\checkmark$$ $$0.105(8{,}670) + 0.059(12{,}870) = 910.35 + 759.33 = 1{,}669.68$$ $$\checkmark$$ 7. Answer the question. The principal for the bank loan was $$\$8{,}670$$ and the principal for the federal loan was $$\$12{,}870$$.

Solving a Student Loan Interest Application Using a System of Equations

Apply the seven-step problem-solving strategy and a mixture table to find the principal given the amount of interest earned. Problem: Laura owes $$\$18{,}000$$ on her student loans. The interest rate on the bank loan is $$2.5\%$$ and the interest rate on the federal loan is $$6.9\%$$. The total amount of interest she paid last year was $$\$1{,}066$$. What was the principal for each loan? 1. Read the problem. 2. Identify what to find: the principal of each loan. 3. Name the unknowns. Let $$b$$ = the principal for the bank loan, and $$f$$ = the principal for the federal loan. Organize into a table: / Loan / Principal (\$) / Rate / Interest (\$) / /---/---/---/---/ / Bank / $$b$$ / $$0.025$$ / $$0.025b$$ / / Federal / $$f$$ / $$0.069$$ / $$0.069f$$ / / Total / $$18{,}000$$ / / $$1{,}066$$ / 4. Translate into a system of equations: $$\left\{\begin{array}{l} b + f = 18{,}000 \ 0.025b + 0.069f = 1{,}066 \end{array}ight.$$ 5. Solve the system using substitution. Solve the first equation for $$b$$: $$b = -f + 18{,}000$$ Substitute into the second equation: $$0.025(-f + 18{,}000) + 0.069f = 1{,}066$$ $$-0.025f + 450 + 0.069f = 1{,}066$$ $$0.044f + 450 = 1{,}066$$ $$0.044f = 616$$ $$f = 14{,}000$$ To find $$b$$, substitute $$f = 14{,}000$$ into the first equation: $$b + 14{,}000 = 18{,}000$$ $$b = 4{,}000$$ 6. Check the answer. $$4{,}000 + 14{,}000 = 18{,}000$$ $$\checkmark$$ $$0.025(4{,}000) + 0.069(14{,}000) = 100 + 966 = 1{,}066$$ $$\checkmark$$ 7. Answer the question. The principal for the bank loan was $$\$4{,}000$$ and the principal for the federal loan was $$\$14{,}000$$.

Solving a Dual Student Loan Interest Application Using a System of Equations

In algebra, the systems of equations model used for coin and ticket problems can be directly extended to food and drink mixture applications. These applications involve blending two or more consumable items—such as combining raisins and nuts to make a trail mix or using two types of coffee beans to make a custom blend. By defining separate variables for the unknown amount of each ingredient, you can construct a system of linear equations to determine the exact quantities needed to meet a specific total amount and a desired overall value or concentration.

Food and Drink Mixture Applications

An inventory specialist is mixing two types of bulk products for a shipment. Match each part of the 'System of Equations' model with the specific relationship or component it represents in a mixture application.

As a production supervisor, you are setting up a system of equations to determine the exact amounts of two different grades of raw materials needed for a manufacturing batch. When using the system of equations method for this mixture application, how do you initially represent the unknown amounts of the two materials?

A procurement officer is blending two different grades of metal alloys for a large-scale construction project. To model this mixture scenario using a system of equations, the officer must establish two separate equations: one representing the total physical quantity (such as the total weight) of the alloys and a second representing the total monetary or quantitative worth (the total value) of those alloys.

A procurement officer is using a system of equations to determine the required amounts of two different materials for a mixture. According to the standard setup for mixture problems using an organizing table, what are the two specific types of 'totals' that must be identified from the table to create the two equations in the system?

Components of a Mixture System Setup

A manufacturing supervisor is blending two different grades of metal alloys for a specific production run. Arrange the following steps in the correct order to set up a system of equations that models this mixture application.

An inventory specialist is blending two different grades of metal alloys to meet a manufacturing specification. To find the required amount of each alloy, the specialist sets up a system of equations with two separate variables, $x$ and $y$. The first equation is created from the total physical count or weight, and the second equation is created from the total ____ of the components in the mixture.

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