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Ch 6 Sec 4: Slide #1 Columbus State Community College Chapter 6 Section 4 An Introduction to Applications of Linear Equations

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Ch 6 Sec 4: Slide #2 An Introduction to Applications of Linear Equations 1.Learn the six steps for solving applied problems. 2.Solve problems involving unknown numbers. 3.Solve problems involving sums of quantities. 4.Solve problems involving supplementary and complementary angles. 5.Solve problems involving consecutive integers.

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

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Ch 6 Sec 4: Slide #4 Problem-Solving Hint The third step in solving an applied problem is often the hardest. To translate the problem into an equation, write the given phrases as mathematical expressions. Replace any words that mean equal or same with an = sign. Other forms of the verb to be, such as is, are, was, and were also translate as an = sign. The = sign leads to an equation to be solved.

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Ch 6 Sec 4: Slide #5 Finding the Value of an Unknown Number EXAMPLE 1 Finding the Value of an Unknown Number The product of 3, and a number decreased by 5, is 87. Find the number. Step 1Read the problem carefully. We are asked to find a number. Step 2Assign a variable to represent the unknown quantity. Let n = the number. There are no other unknown quantities to find.

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Ch 6 Sec 4: Slide #6 Finding the Value of an Unknown Number EXAMPLE 1 Finding the Value of an Unknown Number The product of 3, and a number decreased by 5, is 87. Find the number. Step 3Write an equation. The product of 3, a number and is87 3 n decreased by 5, –5=87()

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Ch 6 Sec 4: Slide #7 Finding the Value of an Unknown Number EXAMPLE 1 Finding the Value of an Unknown Number The product of 3, and a number decreased by 5, is 87. Find the number. Step 4Solve the equation. 3 ( n – 5 ) = 87 3n – 15 = 87 3n = 102 n = 34 Distributive property Add 15. Divide by 3.

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Ch 6 Sec 4: Slide #8 Finding the Value of an Unknown Number EXAMPLE 1 Finding the Value of an Unknown Number The product of 3, and a number decreased by 5, is 87. Find the number. Step 5State the answer. The number is 34. Step 6Check. When 34 is decreased by 5, we get 34 – 5 = 29. If 3 is multiplied by 29, we get 87, as required. The answer, 34, is correct.

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Ch 6 Sec 4: Slide #9 Finding the Value of Two Unknowns During Jarrods senior year of high school, he won 15 more wrestling matches than he lost. If he had a total of 27 matches, what was Jarrods record for wins and losses? Step 1Read the problem. We are given the total number of matches and asked to find Jarrods record in terms of wins and losses. Step 2Assign a variable. Let x = the number of matches lost. Thenx + 15 = the number of matches won. EXAMPLE 2 Finding the Number of Wrestling Matches

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Ch 6 Sec 4: Slide #10 Finding the Value of Two Unknowns During Jarrods senior year of high school, he won 15 more wrestling matches than he lost. If he had a total of 27 matches, what was Jarrods record for wins and losses? Step 3Write an equation. EXAMPLE 2 Finding the Number of Wrestling Matches The number of wins += total matches x + 15 the number of losses x=27+

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Ch 6 Sec 4: Slide #11 Finding the Value of Two Unknowns During Jarrods senior year of high school, he won 15 more wrestling matches than he lost. If he had a total of 27 matches, what was Jarrods record for wins and losses? Step 4Solve the equation. EXAMPLE 2 Finding the Number of Wrestling Matches x + x + 15 = 27 2x = 12 x = 6 Subtract 15. Divide by 2. 2x + 15 = 27Combine terms.

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Ch 6 Sec 4: Slide #12 Finding the Value of Two Unknowns During Jarrods senior year of high school, he won 15 more wrestling matches than he lost. If he had a total of 27 matches, what was Jarrods record for wins and losses? Step 5State the answer. Jarrod lost 6 matches and won 6 + 15 = 21 matches. EXAMPLE 2 Finding the Number of Wrestling Matches Step 6Check. Since Jarrod lost 6 matches and won 21 matches, the total number of matches was 6 + 21 = 27. Because 21 – 6 = 15, Jarrod won 15 more matches than he lost. The answers are correct.

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Ch 6 Sec 4: Slide #13 Problem-Solving Hint The problem in Example 2 could also be solved by letting x represent the number of matches Jarrod won. Then x – 15 would represent the number of matches Jarrod lost. The equation would be ( x – 15 ) + x = 27. The solution of this equation is 21, which is the number of matches Jarrod won. The number of matches he lost would be 27 – 21 = 6. The answers are the same, whichever approach is used.

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Ch 6 Sec 4: Slide #14 Finding the Value of Two Unknowns Chunkys sells 12 chocolate chip cookies for every oatmeal cookie they sell. One day, the company sold 546 of these two types of cookies combined. How many of each were sold? Step 1Read the problem. We are given the total number of cookies and asked to find the number of oatmeal and chocolate chip cookies sold. Step 2Assign a variable. Let x = the number of oatmeal cookies sold. Then 12x = the number of chocolate chip cookies sold. EXAMPLE 3 Analyzing the Number of Cookies Sold

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Ch 6 Sec 4: Slide #15 Finding the Value of Two Unknowns Chunkys sells 12 chocolate chip cookies for every oatmeal cookie they sell. One day, the company sold 546 of these two types of cookies combined. How many of each were sold? Step 3Write an equation. EXAMPLE 3 Analyzing the Number of Cookies Sold The number of oatmeal cookies += total number of cookies x the number of chocolate chip cookies 12x=546+

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Ch 6 Sec 4: Slide #16 Finding the Value of Two Unknowns Chunkys sells 12 chocolate chip cookies for every oatmeal cookie they sell. One day, the company sold 546 of these two types of cookies combined. How many of each were sold? EXAMPLE 3 Analyzing the Number of Cookies Sold Step 4Solve the equation. x + 12x = 546 x = 42 Divide by 13. 13x = 546Combine terms.

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Ch 6 Sec 4: Slide #17 Finding the Value of Two Unknowns Chunkys sells 12 chocolate chip cookies for every oatmeal cookie they sell. One day, the company sold 546 of these two types of cookies combined. How many of each were sold? EXAMPLE 3 Analyzing the Number of Cookies Sold Step 5State the answer. There were 42 oatmeal cookies sold and 12 42 = 504 chocolate chip cookies sold. Step 6Check. Since 42 + 504 = 546, and 504 is 12 times 42, the answers are correct.

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Ch 6 Sec 4: Slide #18 Problem-Solving Hint Sometimes it is necessary to find three unknown quantities in an applied problem. Frequently the three unknowns are compared in pairs. When this happens, it is usually easiest to let the variable represent the unknown found in both pairs. The next example illustrates this.

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Ch 6 Sec 4: Slide #19 Finding the Value of Three Unknowns A 103-inch chain is cut into 3 pieces. The second piece of chain is twice as long as the first piece and the third piece is 5 inches shorter than the first piece. Find the lengths of each section of chain. Step 1Read the problem. We are given the length of the original chain. We are being asked to find the length of three pieces of chain. Step 2Assign a variable. Let x = the length of the first piece of chain, 2x = the length of the second piece of chain, andx – 5 = the length of the third piece of chain. EXAMPLE 4 Dividing a Chain into Pieces

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Ch 6 Sec 4: Slide #20 Finding the Value of Three Unknowns A 103-inch chain is cut into 3 pieces. The second piece of chain is twice as long as the first piece and the third piece is 5 inches shorter than the first piece. Find the lengths of each section of chain. Step 3Write an equation. EXAMPLE 4 Dividing a Chain into Pieces First piece += Total length x Second piece 2x2x=103+ + Third piece x – 5+

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Ch 6 Sec 4: Slide #21 Finding the Value of Three Unknowns A 103-inch chain is cut into 3 pieces. The second piece of chain is twice as long as the first piece and the third piece is 5 inches shorter than the first piece. Find the lengths of each section of chain. EXAMPLE 4 Dividing a Chain into Pieces Step 4Solve the equation. x + 2x + x – 5 = 103 4x = 108 x = 27 Add 5. Divide by 4. 4x – 5 = 103Combine terms.

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Ch 6 Sec 4: Slide #22 Finding the Value of Three Unknowns A 103-inch chain is cut into 3 pieces. The second piece of chain is twice as long as the first piece and the third piece is 5 inches shorter than the first piece. Find the lengths of each section of chain. EXAMPLE 4 Dividing a Chain into Pieces Step 5State the answer. The first piece is 27 inches long. The second piece is 2 ( 27 ) = 54 inches long. The third piece is 27 – 5 = 22 inches long. Step 6Check. The sum of the lengths is 103 inches. All conditions of the problem are satisfied. The answers are correct.

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Ch 6 Sec 4: Slide #23 ComplementarySupplementary Angle 1Angle 2Angle 1Angle 2 1) 60 120 2) 15 40 3) 40.2 100.4 4) x m Angles A right angle is an angle that measures 90. Note: measures is a keyword for equal to. Two angles are said to be: complementary if their sum is 90, and supplementary if their sum is 180. 30° 75° 49.8° ( 90 – x )° 60° 140° 79.6° ( 180 – m )°

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Ch 6 Sec 4: Slide #24 a Supplementary ( 180º ) Use the variable in each diagram below to write an algebraic expression for each missing angle. ( 90 – n )°( 180 – a )° Angles ?? n Complementary ( 90º )

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Ch 6 Sec 4: Slide #25 Problem-Solving Hint If x represents the degree of an angle, then 90 – x represents the degree measure of its complement, and 180 – x represents the degree measure of its supplement.

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Ch 6 Sec 4: Slide #26 Finding the Measure of an Angle Find the measure of an angle whose supplement is 35º more than twice its complement. Step 1Read the problem. We are to find the measure of an angle, given information about its complement and supplement. Step 2Assign a variable. Let x = the degree measure of the angle. Then 90 – x = the degree measure of its complement; 180 – x = the degree measure of its supplement. EXAMPLE 5 Finding the Measure of an Angle

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Ch 6 Sec 4: Slide #27 Finding the Measure of an Angle Find the measure of an angle whose supplement is 35º more than twice its complement. Step 3Write an equation. EXAMPLE 5 Finding the Measure of an Angle Supplementisits complement 180 – x 35º 2 = more thantwice 35 +x–90()

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Ch 6 Sec 4: Slide #28 Finding the Measure of an Angle Find the measure of an angle whose supplement is 35º more than twice its complement. Step 4Solve. EXAMPLE 5 Finding the Measure of an Angle 180 – x = 35 + 2 ( 90 – x ) 180 – x = 215 – 2x x = 35 Combine terms. Add 2x. 180 – x = 35 + 180 – 2xDistributive property 180 + x = 215 Subtract 180.

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Ch 6 Sec 4: Slide #29 Finding the Measure of an Angle Find the measure of an angle whose supplement is 35º more than twice its complement. Step 5State the answer. The measure of the angle is 35º. EXAMPLE 5 Finding the Measure of an Angle Step 6Check. The supplement of 35º is 145º and the complement of 35º is 55º. Also, 145º is equal to 35º more than twice 55º. The answer is correct.

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Ch 6 Sec 4: Slide #30 Problem-Solving Hint When solving consecutive integer problems, if x = the first integer, then for any two consecutive integers, usex,x + 1; two consecutive even integers, usex,x + 2; two consecutive odd integers, usex,x + 2. Notice the setup for two consecutive odd integers is the same as the setup for two consecutive even integers.

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Ch 6 Sec 4: Slide #31 Finding Consecutive Integers Abdul and Damon have lockers next to each other. The lockers are consecutively numbered and their sum is 395. What are the locker numbers? Step 1Read the problem. We are finding two consecutive integers. The sum of the two locker numbers is 395. EXAMPLE 6 Finding Consecutive Integers

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Ch 6 Sec 4: Slide #32 Finding Consecutive Integers Abdul and Damon have lockers next to each other. The lockers are consecutively numbered and their sum is 395. What are the locker numbers? EXAMPLE 6 Finding Consecutive Integers xx + 1 x Step 2Assign a variable. Let x = the first locker number. Then x + 1 = the second locker number.

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Ch 6 Sec 4: Slide #33 Finding Consecutive Integers Abdul and Damon have lockers next to each other. The lockers are consecutively numbered and their sum is 395. What are the locker numbers? EXAMPLE 6 Finding Consecutive Integers xx + 1 x Step 3Write an equation. The sum of the locker numbers is 395. x + x + 1 = 395

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Ch 6 Sec 4: Slide #34 Finding Consecutive Integers Abdul and Damon have lockers next to each other. The lockers are consecutively numbered and their sum is 395. What are the locker numbers? EXAMPLE 6 Finding Consecutive Integers xx + 1 x Step 4Solve. x + x + 1 = 395 2x + 1 = 395Combine terms. 2x = 394 Subtract 1. x = 197 Divide by 2.

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Ch 6 Sec 4: Slide #35 Finding Consecutive Integers Abdul and Damon have lockers next to each other. The lockers are consecutively numbered and their sum is 395. What are the locker numbers? EXAMPLE 6 Finding Consecutive Integers 197198 xx + 1 Step 5State the answer. The first locker number is 197 and the second locker number is 198. 197198

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Ch 6 Sec 4: Slide #36 Finding Consecutive Integers Abdul and Damon have lockers next to each other. The lockers are consecutively numbered and their sum is 395. What are the locker numbers? EXAMPLE 6 Finding Consecutive Integers 197198 Step 6Check. The sum of 197 and 198 is 395. The answers are correct. 197198

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Ch 6 Sec 4: Slide #37 Finding Consecutive Odd Integers If 13 is added to twice the larger of two consecutive odd integers, the result is 3 times the smaller integer. Find the two integers. Let x be the smaller integer. Since the two numbers are consecutive odd integers, then x + 2 is the larger. EXAMPLE 7 Finding Consecutive Odd Integers Now, write an equation. 13is added to 13 twice the larger =+ the result is three times the smaller 2 ( x + 2 )3x3x

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Ch 6 Sec 4: Slide #38 Finding Consecutive Odd Integers If 13 is added to twice the larger of two consecutive odd integers, the result is 3 times the smaller integer. Find the two integers. 13 + 2 ( x + 2 ) = 3x EXAMPLE 7 Finding Consecutive Odd Integers 13 + 2x + 4 = 3x Distributive property 17 + 2x = 3x Combine terms. 17 = x Subtract 2x. Now solve the equation.

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Ch 6 Sec 4: Slide #39 Finding Consecutive Odd Integers If 13 is added to twice the larger of two consecutive odd integers, the result is 3 times the smaller integer. Find the two integers. EXAMPLE 7 Finding Consecutive Odd Integers The first integer is 17 and the second is 17 + 2 = 19. To check, we see that when 13 is added to twice 19 (the larger integer), we get 51, which is 3 times 17 (the smaller integer). The answers are correct.

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Ch 6 Sec 4: Slide #40 An Introduction to Applications of Linear Equations Chapter 6 Section 4 – Completed Written by John T. Wallace

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Do Now Solve. 1. x – 17 = 32 2. y + 11 = 41 3. = 18 4. 12x = 108 5. x – 9 = 20 x = 49 y = 30 w 5 w = 90 x = 9 x = 29 Hwk: p29 odd, Test this Friday.

Do Now Solve. 1. x – 17 = 32 2. y + 11 = 41 3. = 18 4. 12x = 108 5. x – 9 = 20 x = 49 y = 30 w 5 w = 90 x = 9 x = 29 Hwk: p29 odd, Test this Friday.

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