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2.4 An Introduction to Applications of Linear Equations. Objective 1 . Learn the six steps for solving applied problems. Slide 2.4-3. Learn the six steps for solving applied problems.
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Objective 1 Learn the six steps for solving applied problems. Slide 2.4-3
Learn the six steps for solving applied problems. While there is no one specific method that enables you to solve all kinds of applied problems, the following six-step method is often applicable. Solving an Applied Problem Step 1:Readthe problem carefully. What information is given? What are you asked to find? Step 2:Assign a variableto represent the unknown value. Use a sketch, diagram, or table, as needed. If necessary, express any other unknown values in terms of the variable. Step 3:Write an equationusing the variable expression(s). Step 4:Solvethe equation. Step 5:State the answer.Label it appropriately. Does it seem reasonable? Step 6:Check the answer in the words of the original problem. Slide 2.4-4
Objective 2 Solve problems involving unknown numbers. Slide 2.4-5
Learn the six steps for solving applied problems. (cont’d) 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 equals or same with an = sign. Other forms of the verb “to be,” such as is, are, was, andwere, also translate this way. The = sign leads to an equation to be solved. Slide 2.4-6
Finding the Value of an Unknown Number CLASSROOM EXAMPLE 1 If 5 is added to the product of 9 and a number, the result is 19 less than the number. Find the number. Equation: Let x =the number. Solve: The number is −3. When solving an equation, use solution set notation to write the answer. When solving an application, state the answer in a sentence. Slide 2.4-7
Objective 3 Solve problems involving sums of quantities. Slide 2.4-8
Solve problems involving sums of quantities. A common type of problem in elementary algebra involves finding two quantities when the sum of the quantities is known. PROBLEM-SOLVING HINT To solve problems involving sums of quantities, choose a variable to represent one of the unknowns. Then represent the other quantity in terms of the same variable, using information from the problem. Slide 2.4-9
Finding the Numbers of Olympic Medals CLASSROOM EXAMPLE 2 In the 2006 Winter Olympics in Torino, Italy, Canada won 5 more medals than Norway. The two countries won a total of 43 medals. How many medals did each country win? (Source: U.S. Olympic Committee.) Solution: Let x = the number of medals Norway won. Let x + 5 = the number of medals Canada won. Norway won 19 medals and Canada won 24 medals. Slide 2.4-10
The nature of the applied problem restricts the set of possible solutions. For example, an answer such as −33 medals or 25 medals should be recognized as inappropriate. Solve problems involving sums of quantities. (cont’d) The problem in Example 2 could also be solved by letting x represent the number of medals Canada won. Then x− 5 would represent the number of medals Norway won. The equation would be The solution to this equation is 24, which is the number of medals Canada won. The number of Norwegian medals would be 24 − 5 = 19. The answers are the same, whichever approach is used, even though the equation and its solution are different. Slide 2.4-11
Solution: Let x = the number of muffins. Then x = the number of croissants. Finding the Number of Orders for Tea CLASSROOM EXAMPLE 3 On that same day, the owner of Terry’s Coffeehouse found that the number of orders for croissants was the number of muffins. If the total number for the two breakfast rolls was 56, how many orders were placed for croissants? 8 croissants were ordered. Slide 2.4-12
Solve problems involving sums of quantities. (cont’d) PROBLEM SOLVING HINT In Example 3, it was easier to let the variable represent the quantity that was not specified. This required extra work in Step 5 to find the number of orders for croissants. In some cases, this approach is easier than letting the variable represent the quantity that we are asked to find. Slide 2.4-13
Analyzing a Gasoline-Oil Mixture CLASSROOM EXAMPLE 4 At a meeting of the local computer user group, each member brought two nonmembers. If a total of 27 people attended, how many were members and how many were nonmembers? Solution: Let x = number of members. Then 2x = number of nonmembers. There were 9 members and 18 nonmembers at the meeting. Slide 2.4-14
Solve problems involving sums of quantities. (cont’d) PROBLEM SOLVING HINT Sometimes it is necessary to find three unknown quantities. When the three unknowns are compared in pairs, let the variable represent the unknown found in both pairs. Slide 2.4-15
Dividing a Pipe into Pieces CLASSROOM EXAMPLE 5 A piece of pipe is 50 in. long. It is cut into three pieces. The longest piece is 10 in. more than the middle-sized piece, and the shortest piece measures 5 in. less than the middle-sized piece. Find the lengths of the three pieces. Solution: Let x = the length of the middle-sized piece, then x +10 = the longest piece, and x − 5 = the shortest piece. The shortest piece is 10 in., the middle-size piece is 15 in., and the longest is 25 in. Slide 2.4-16
Objective 4 Solve problems involving consecutive integers. Slide 2.4-17
Solve problems involving consecutive integers. Two integers that differ by 1 are called consecutive integers. For example, 3 and 4, 6 and 7, and −2 and −1 are pairs of consecutive integers. In general, if x represents an integer, x + 1 represents the next greater consecutive integer. Consecutive even integers, such as 8 and 10, differ by 2. Consecutive odd integers, such as 9 and 11, also differ by 2. In general if xrepresents an even integer, x + 2 represents the greater consecutive integer. The same holds true for odd integers; that is if xis an odd integer, x + 2 is the greater odd integer. Slide 2.4-18
PROBLEM SOLVING HINT If x = the lesser integer, then, for any two consecutive integers, use two consecutive even integers, use two consecutive odd integers, use Solve problems involving consecutive integers. (cont’d) Slide 2.4-19
It is a good idea to use parentheses around x + 1, (even though they are not necessary here). Finding Consecutive Integers CLASSROOM EXAMPLE 6 Solution: Let x = the lesser page number. Then x + 1= the greater page number. Two pages that face each other have 569 as the sum of their page numbers. What are the page numbers? The lesser page number is 284, and the greater page number is 285. Slide 2.4-20
Finding Consecutive Odd Integers CLASSROOM EXAMPLE 7 Find two consecutive even integers such that six times the lesser added to the greater gives a sum of 86. Solution: Let x = the lesser integer. Then x + 2 = the greater integer. The lesser integer is 12 and the greater integer is 14. Slide 2.4-21
Objective 5 Solve problems involving supplementary and complementary angles. Slide 2.4-22
An angle can be measured by a unit called the degree (°), which is of a complete rotation. Two angles whose sum is 90° are said to be complementary, or complements of each other. An angle that measures 90° is a right angle. Two angles who sum is 180° are said to be supplementary,or supplements of each other. One angle supplements the other to form a straight angle of 180°. Solve problems involving supplementary and complementary angles. Slide 2.4-23
Solve problems involving supplementary and complementary angles. (cont’d) PROBLEM-SOLVING HINT If x represents the degree measure of an angle, then 90 − xrepresents the degree measure of its complement, 180 − xrepresents the degree measure of is supplement. Slide 2.4-24
Finding the Measure of an Angle CLASSROOM EXAMPLE 8 Find the measure of an angle whose complement is eight times its measure. Solution: Let x = the degree measure of the angle. Then 90 −x = the degree measure of its complement. The measure of the angle is 10°. Slide 2.4-25
Finding the Measure of an Angle CLASSROOM EXAMPLE 9 Find the measure of an angle such that the sum of the measures of its complement and its supplement is 174°. Solution: Let x = the degree measure of the angle. Then 90 −x = the degree measure of its complement, and 180 −x = the degree measure of its supplement. The measure of the angle is 48°. Slide 2.4-26