# Ch 3 Sec 3: Slide #1 Columbus State Community College Chapter 3 Section 3 Solving Application Problems with One Unknown Quantity.

## Presentation on theme: "Ch 3 Sec 3: Slide #1 Columbus State Community College Chapter 3 Section 3 Solving Application Problems with One Unknown Quantity."— Presentation transcript:

Ch 3 Sec 3: Slide #1 Columbus State Community College Chapter 3 Section 3 Solving Application Problems with One Unknown Quantity

Ch 3 Sec 3: Slide #2 Solving Application Problems with One Unknown Quantity 1.Translate word phrases into algebraic expressions. 2.Translate sentences into equations. 3.Solve application problems with one unknown quantity.

Ch 3 Sec 3: Slide #3 Translating Word Phrases into Algebraic Expressions Write each phrase as an algebraic expression. Use x as the variable. EXAMPLE 1 Translating Word Phrases WordsAlgebraic Expression a) A number plus 7 x + 7 or 7 + x b) The sum of 3 and a number 3 + x or x + 3 c) 6 more than a number x + 6 or 6 + x d) – 15 added to a number – 15 + x or x + – 15 e) A number increased by 2 x + 2 or 2 + x Two correct ways to write each addition expression.

Ch 3 Sec 3: Slide #4 Translating Word Phrases into Algebraic Expressions Write each phrase as an algebraic expression. Use x as the variable. EXAMPLE 1 Translating Word Phrases WordsAlgebraic Expression f) 8 less than a number x – 8 g) A number subtracted from 1 1 – x h) 6 subtracted from a number x – 6 i) A number decreased by 4 x – 4 j) 9 minus a number 9 – x Only one correct way to write each subtraction expression.

Ch 3 Sec 3: Slide #5 The Order of Terms Recall that addition can be done in any order, so x + 4 gives the same result as 4 + x. This is not true in subtraction, so be careful. 3 – x does not give the same result as x – 3. CAUTION

Ch 3 Sec 3: Slide #6 Translating Word Phrases into Algebraic Expressions Write each phrase as an algebraic expression. Use x as the variable. EXAMPLE 2 Translating Word Phrases WordsAlgebraic Expression a) 8 times a number 8x8x b) The product of 32 and a number 32x c) Double a number (meaning 2 times)2x2x d) The quotient of – 7 and a number e) A number divided by 4 f) 14 subtracted from 3 times a number –7–7 x x 4 3x – 14

Ch 3 Sec 3: Slide #7 Translating Word Phrases into Algebraic Expressions Write each phrase as an algebraic expression. Use x as the variable. EXAMPLE 2 Translating Word Phrases WordsAlgebraic Expression g) The result is =

Ch 3 Sec 3: Slide #8 Translating a Sentence into an Equation If 4 times a number is added to 15, the result is 11. Find the number. EXAMPLE 3 Translating a Sentence into an Equation Let x represent the unknown number. 4x4x 4 times a number + added to 15 = is 11 4x + 15 = 11 – 15 4x + 15 = – 4 4 4 x + 15 = – 1

Ch 3 Sec 3: Slide #9 Translating a Sentence into an Equation If 4 times a number is added to 15, the result is 11. Find the number. EXAMPLE 3 Translating Sentence into an Equation 4 – 1 If 4 times a number is added to 15, the result is 11. +15=11 Check Go back to the words of the original problem. Does 4 – 1 + 15 really equal 11? Yes. 4 – 1 + 15 = – 4 + 15 = 11. So – 1 is the correct solution because it works when you put it back into the original problem.

Ch 3 Sec 3: Slide #10 Solving an Application Problem Step 1 Read the problem once to see what it is about. Read it carefully a second time. As you read, make a sketch or write word phrases that identify the known and the unknown parts of the problem. Step 2 (a) If there is one unknown quantity, assign a variable to represent it. Write down what your variable represents. Step 2 (b) If there is more than one unknown quantity, assign a variable to represent the thing you know the least about. Then write variable expression(s), using the same variable, to show the relationship of the other unknown quantities to the first one. continued… Solving an Application Problem

Ch 3 Sec 3: Slide #11 Step 3 Write an equation, using your sketch or word phrases as a guide. Step 4 Solve the equation. Step 5 State the answer to the question in the problem and label your answer. Step 6 Check whether your answer fits all the facts given in the original statement of the problem. If it does, you are done. If it doesnt, start again at Step 1. Solving an Application Problem (continued) Solving an Application Problem

Ch 3 Sec 3: Slide #12 Solving an Application Problem: One Unknown Quantity EXAMPLE 4 Application Problems: One Unknown Quantity Mike gained 8 pounds over the winter. He went on a diet and lost 12 pounds. Then he regained 15 pounds and weighed 192 pounds. How much did he weigh originally? Step 1 Read the problem once. It is about Mikes weight. Read it a second time and write word phrases. Step 2(a) There is only one unknown quantity, so assign a variable to represent it. Let w represent Mikes original weight. Step 3 Write an equation, using the phrases you wrote as a guide. Start Weight w Lost 12 – 12 Gained 15 + 15 Ending Weight = 192 Gained 8 + 8

Ch 3 Sec 3: Slide #13 Step 4 Solve the equation. w + 8 – 12 + 15 = 192 w + 11 = 192 – 11 w = 181 Solving an Application Problem: One Unknown Quantity EXAMPLE 4 Application Problems: One Unknown Quantity Mike gained 8 pounds over the winter. He went on a diet and lost 12 pounds. Then he regained 15 pounds and weighed 192 pounds. How much did he weigh originally?

Ch 3 Sec 3: Slide #14 Step 5 State the answer to the question, How much did he weigh originally? Mike originally weighed 181 pounds. Solving an Application Problem: One Unknown Quantity EXAMPLE 4 Application Problems: One Unknown Quantity Mike gained 8 pounds over the winter. He went on a diet and lost 12 pounds. Then he regained 15 pounds and weighed 192 pounds. How much did he weigh originally?

Ch 3 Sec 3: Slide #15 Step 6 Check the solution by going back to the original problem and inserting the solution. Because 181 pounds works when you put it back into the original problem, you know it is the correct solution. Originally weighed 181 pounds. Gained 8 lbs, so 181 + 8 = 189 pounds Lost 12 lbs, so 189 – 12 = 177 pounds Gained 15 lbs, so 177 + 15 = 192 pounds Solving an Application Problem: One Unknown Quantity EXAMPLE 4 Application Problems: One Unknown Quantity Mike gained 8 pounds over the winter. He went on a diet and lost 12 pounds. Then he regained 15 pounds and weighed 192 pounds. How much did he weigh originally?

Ch 3 Sec 3: Slide #16 Joe purchased 6 boxes of calendars. He gave his secretary 12 calendars, his business partner 5 calendars, and his accountant 19 calendars. If Joe has 12 calendars remaining, how many were in each box he purchased? Step 1 Read the problem once. It is about 6 boxes of calendars. Unknown: the number of calendars in each box purchased. Known: 6 boxes; gave away 12, 5, and 19 calendars. Step 2(a) There is only one unknown quantity. Assign a variable, b, to represent the number of calendars in each box. Solving an Application Problem: One Unknown Quantity EXAMPLE 5 Application Problems: One Unknown Quantity

Ch 3 Sec 3: Slide #17 Step 3 Write an equation. Joe purchased 6 boxes of calendars. He gave his secretary 12 calendars, his business partner 5 calendars, and his accountant 19 calendars. If Joe has 12 calendars remaining, how many were in each box he purchased? Number of boxes 6 Number in each box b Gave away – 12 Amount remaining = 12 Gave away – 5 Solving an Application Problem: One Unknown Quantity EXAMPLE 5 Application Problems: One Unknown Quantity Gave away – 19

Ch 3 Sec 3: Slide #18 Step 4 Solve the equation. 6b – 12 – 5 – 19 = 12 6b – 36 = 12 + 36 6b = 48 6 6 b = 8 Joe purchased 6 boxes of calendars. He gave his secretary 12 calendars, his business partner 5 calendars, and his accountant 19 calendars. If Joe has 12 calendars remaining, how many were in each box he purchased? Solving an Application Problem: One Unknown Quantity EXAMPLE 5 Application Problems: One Unknown Quantity

Ch 3 Sec 3: Slide #19 Step 5 State the answer. There were 8 calendars in each box. Step 6 Check the solution using the original problem. 6 boxes each containing 8 calendars, so 6 8 = 48 calendars. Gave away 12, 5, and 19 calendars, so 48 – 12 – 5 – 19 = 12 calendars. 8 calendars in each box is the correct solution because it works. Joe purchased 6 boxes of calendars. He gave his secretary 12 calendars, his business partner 5 calendars, and his accountant 19 calendars. If Joe has 12 calendars remaining, how many were in each box he purchased? Solving an Application Problem: One Unknown Quantity EXAMPLE 5 Application Problems: One Unknown Quantity

Ch 3 Sec 3: Slide #20 Ty won 4 less than three times as many golf matches as Mike. If Ty won 11 matches, how many matches did Mike win? Step 1 Read the problem. It is about the number of golf matches won by two people. Step 2(a) Assign a variable, m, to represent the number of matches that Mike won. Step 3 Write an equation. The number of matches Ty won 11 4 less than three times the number of matches Mike won 3m – 4 is = Solving More Complex Application Problems: One Unknown Quantity EXAMPLE 6 Complex Application Problems: One Unknown

Ch 3 Sec 3: Slide #21 11 = 3m – 4 + 4 15 = 3m 3 3 5 = m Solving More Complex Application Problems: One Unknown Quantity EXAMPLE 6 Complex Application Problems: One Unknown Ty won 4 less than three times as many golf matches as Mike. If Ty won 11 matches, how many matches did Mike win? Step 4 Solve the equation

Ch 3 Sec 3: Slide #22 Step 6 Check the solution using the original problem. Three times the number of Mikess wins 3 5 = 15 Less 4 15 – 4 = 11 Ty won 11 matches. The correct solution is: Mike won 5 golf matches. Solving More Complex Application Problems: One Unknown Quantity EXAMPLE 6 Complex Application Problems: One Unknown Ty won 4 less than three times as many golf matches as Mike. If Ty won 11 matches, how many matches did Mike win? Step 5 State the answer. Mike won 5 golf matches.

Ch 3 Sec 3: Slide #23 Solving Application Problems with One Unknown Quantity Chapter 3 Section 3 – Completed Written by John T. Wallace

Similar presentations