Applications of Linear Equations

1. Applicationsof Linear Equations

2. Applications of Linear Equations Six Steps to Solving Applied Problems Step 1Read the problem, several times if necessary, until you understand what is given and what is to be found. Step 2 Assign a variable to represent the unknown value, using diagrams or tables as needed. Write down what the variable represents. Express the other unknown values in terms of the variable. Step 3Write an equation using the variable expression(s). Step 4 Solve the equation. Step 5 State your answer. Does it seem reasonable? Step 6 Check your answer in the words of the original problem.

3. Example Applications of Linear Equations A box contains a combined total of 68 pens and pencils. If the box contains 16 more pens than pencils, how many of each are in the box? Read the problem carefully. We are given information about the total number of pens and pencils and asked to find the number of each in the box. Step 1 Assign a variable. Let x = the number pencils in the box. Then x + 16 = the number pens in the box. Step 2

4. Example (continued) Applications of Linear Equations A box contains a combined total of 68 pens and pencils. If the box contains 16 more pens than pencils, how many of each are in the box? Recall: x = # of pencils, x + 16 = # of pens Write an equation. Step 3 The total is the number of pens plus the number of pencils 68 ( x + 16 ) = + x

5. Example (continued) 52 2x = 2 2 Applications of Linear Equations A box contains a combined total of 68 pens and pencils. If the box contains 16 more pens than pencils, how many of each are in the box? Solve the equation. Step 4 68 = ( x + 16 ) + x 68 = 2x + 16 Combine terms. 68 – 16 = 2x + 16 – 16 Subtract 16. 52 = 2x Combine terms. Divide by 2. 26 = x or x= 26

6. Example (continued) Applications of Linear Equations A box contains a combined total of 68 pens and pencils. If the box contains 16 more pens than pencils, how many of each are in the box? Recall: x = # of pencils, x + 16 = # of pens State the answer. The variable x represents the number of pencils, so there are 26 pencils. Then the number of pens is x + 16 = 26 + 16 = 42. Step 5 Check. Since there are 26 pencils and 42 pens, the combined total number of pencils and pens is 26 + 42 = 68. Because 42 – 26 = 16, there are 16 more pens than pencils. This information agrees with what is given in the problem, so the answer checks. Step 6

7. Example Applications of Linear Equations A mixture of acid and water must be combined to fill a 286 ml beaker. If the mixture must contain 12 ml of water for each milliliter of acid, how many milliliters of water and how many milliliters of acid does it require to fill the beaker? Read the problem carefully. We must find how many milliliters of water and how many milliliters of acid are needed to fill the beaker. Step 1 Assign a variable. Let x = the number of milliliters of acid required. Then 12x = the number of milliliters of water required. Step 2

8. Example (continued) Applications of Linear Equations A mixture of acid and water must be combined to fill a 286 ml beaker. If the mixture must contain 12 ml of water for each milliliter of acid, how many milliliters of water and how many milliliters of acid does it require to fill the beaker? Recall: x = ml. of acid, 12x = ml. of water. Write an equation. A diagram is sometimes helpful. Step 3 Beaker Acidx Water12x = 286 x + 12x = 286

9. Example (continued) 13x 286 = 13 13 Applications of Linear Equations A mixture of acid and water must be combined to fill a 286 ml beaker. If the mixture must contain 12 ml of water for each milliliter of acid, how many milliliters of water and how many milliliters of acid does it require to fill the beaker? Solve. Step 4 x + 12x = 286 13x = 286 Combine terms. Divide by 13. x= 22

10. Example (continued) Applications of Linear Equations A mixture of acid and water must be combined to fill a 286 ml beaker. If the mixture must contain 12 ml of water for each milliliter of acid, how many milliliters of water and how many milliliters of acid does it require to fill the beaker? Recall: x = ml of acid, 12x = ml of water State the answer. The beaker requires 22 mlof acid and 12(22) = 264 ml of water. Step 5 Check. Since 22 + 264 = 286, and 264 is 12 times 22, the answer checks. Step 6

11. Example Applications of Linear Equations Gerald has a wire 96 inches long that he must cut into three pieces. The longest piece must be three times the length of the middle-sized piece and the shortest piece must be 14 inches shorter than the middle-sized piece. How long must each piece be? Read the problem carefully. Three lengths must be found. Step 1 Assign a variable. x = the length of the middle-sized piece, 3x = the length of the longest piece, and x – 14 = the length of the shortest piece. Step 2

12. Example (continued) 96 inches 3x x x – 14 Applications of Linear Equations Gerald has a wire 96 inches long that he must cut into three pieces. The longest piece must be three times the length of the middle-sized piece and the shortest piece must be 14 inches shorter than the middle-sized piece. How long must each piece be? Write an equation. Step 3 Longest Middle-sized Shortest is Total length 3x + x + x – 14 = 96

13. Example (continued) 110 5x = 5 5 Applications of Linear Equations Gerald has a wire 96 inches long that he must cut into three pieces. The longest piece must be three times the length of the middle-sized piece and the shortest piece must be 14 inches shorter than the middle-sized piece. How long must each piece be? Solve. 96 = 3x + x + x – 14 Step 4 96 = 5x – 14 Combine terms. Add 14. 96 + 14 = 5x – 14 + 14 110 = 5x Combine terms. Divide by 5. 22 = x or x= 22

14. Example (continued) Applications of Linear Equations Gerald has a wire 96 inches long that he must cut into three pieces. The longest piece must be three times the length of the middle-sized piece and the shortest piece must be 14 inches shorter than the middle-sized piece. How long must each piece be? Recall: x = length of middle-sized piece, 3x = length of longest piece, and x – 14 = length of shortest piece. State the answer. The middle-sized piece is 22 in. long, the longest piece is 3(22) = 66 in. long, and the shortest piece is 22 – 14 = 8 in. long. Step 5 Check. The sum of the lengths is 96 in. All conditions of the problem are satisfied. Step 6

15. “Work problems” are problems that ask you to find long it takes to complete a job when 2 or more people (machines) are working together. In order to solve these problems use the formula:work = rate of work x time. Rate of work compares how much can be completed to a period of time.

16. Pipe 1 fills a tank in 6 minutes and pipe 2 fills the same tank in 8 minutes. How long does it take for both pipes together to fill the tank? • The rate for work for pipe 1 is • The rate of work for pipe 2 is • Let x = time together

17. work = rate of work x timework for pipe 1 = (x) work for pipe 2 = (x)

18. To finish, solve the equation

19. Bob has scored 85, 88, 91, and 86 on his four math tests. He has one more test left to take. What does he need to score on the last test in order get an A in the class? His average has to be at least 89.5 to get an A

20. Your grade in this class is a "weighted" average which means not all scores count equally. Homework is 7%, quiz average is 10%, tests average is 48% and the final exam is 35% of the final grade. What does a student need to make on the final exam in order to pass (69.5) if they have the following grades: HW =80 , QA =72 , and TA =60 ?

21. Solving a Mixture Problem A chemist must mix 12 L of a 30% acid solution with some 80% solution to get a 60% solution. How much of the 80% solution should be used? EXAMPLE 7 Step 1 Read the problem. The problem asks for the amount of 80% solution to be used. Step 2 Assign a variable. Let x = the number of liters of 80% solution to be used. 80% = + 30% 80% 30% 12 L Unknown number of liters, x (12 + x)L

22. A chemist must mix 12 L of a 30% acid solution with some 80% solution to get a 60% solution. How much of the 80% solution should be used? • Write an equation. = + 30% 80% 12 L x (12 + x)L

23. A chemist must mix 12 L of a 30% acid solution with some 80% solution to get a 60% solution. How much of the 80% solution should be used? • Solve the equation.