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Olympic College Topic 3 – Applications

Topic 3 ApplicationsIntroduction:

In this section we will look at how to take a “word problem” written in English and to translate it into a linear equation that will then be solved and so we will find a solution to a the problem.

There are many kinds of “word problems” but there is a common strategy that you can use that will lead you through the process from initial problem to its final solution.

Step 1: Read the problem carefully and figure out what it is asking you to find. Usually, but not always, you can find this information at the end of the problem.

Step 2: Assign a variable the quantity you are trying to find.We usually use the variable x, but sometimes it makes more sense to use a variable that has the same letter as the first letter of the quantity we are being asked to find. So if its time we might use e variable t, if it’s the weight we could use the variable w etc.

Step 3: Read the problem again then translate the information that was given from English into its equivalent mathematics.In particular write down an equation in the given variable that represents the information given in the problem.

Step 4: Solve the equation that you created in Step 3.

Step 5: Answer in English the original problem.In order to do this you need to look back at the problem and in particular find out what the variable that was used in your equation represents.

Step 6: Check your solution.Your solution should make sense in the context of the original problem as well As being a solution to the equation formed in step 3.

The following examples show the variety of word problems and the method we use to solve them.


1. Simple word problemsThese word problems are essentially just puzzles which contain a “find the missing number”.These problems will typically have a single statement that will turn into an equation.We then solve this equation we get the value of the “missing number”.

Example 1: When 4 is added to three times a number, the result is 40. Find the number.

Solution: Step 2: x = missing numberStep 3: 4 + 3x = 40Step 4: 3x = 36 Subtract 4 from Both Sides.

3x3 =

363 Divide Both Sides by 3.

x = 12

Step 5: The solution to the problem is that the missing number is 12.

Step 6: We check the solution by using x = 12 in 4 + 3x = 4 + 3(12) = 40

Example 2: Five is subtracted from a number and the result is 20. What is the value of this number?

Solution: Step 2: x = missing numberStep 3: x – 5 = 20 Add 5 to Both Sides.Step 4: x = 25


Step 6: We check the solution by using x = 25 in x – 5 = 25 – 5 = 20

Example 3: The sum of a number and 4 is multiplied by 3 and the answer is 24. Find the number.

Solution: Step 2: n = missing numberStep 3: 3(n + 4) = 24Step 4: 3n + 12 = 24 Subtract 12 from Both Sides.

3n = 123n3 =


n = 4


Step 6: We check the solution by using n = 4 in 3(n + 4) = 3(4 + 4) = 3(8) = 24


Example 4: If 10 is subtracted from three times a number, the result is 5 less than the number. Find the missing number

Solution: Step 2: x = missing numberStep 3: 3x – 10 = x – 5

2x – 10 = – 5 Subtract x from Both Sides.Step 4: 3x = 5 Add 10 to Both Sides

3x3 =

53 Divide Both Sides by 3

x =53

Step 5: The solution to the problem is that the missing number is 53 .

Step 6: We check the solution by using x = 53 in 3x – 10 = x – 5

3(53) – 10 = (

53¿– 5

153 – 10 =

53– 5

−153 =

−153

Exercise 1A Solve the following problems by first writing an equation and then solving it.

1. A number plus 12 is equal to 25. Find the number.

2. A number times 5 is equal to 100. Find the number.

3. When 4 is added to three times a number, the result is 40. Find the number.

4. When you subtract 8 from a number the answer is 60. Find the number.

5. Four is subtracted from a number and the result is 50. What is the value of this number?

6. The sum of a number and 12 is multiplied by 5 and the answer is 20. Find the number.

7. The sum of a number and 4 is doubled and the answer is 30. Find the number.

8. A number times 5 has 10 subtracted from it and is equal to 10.


What is the value of this number?

9. If 10 is subtracted from three times a number, the result is 5 less than the number. Find the missing number

2. Geometric word problems with Rectangles.These word problems contain rectangles or squares and they will typically ask you to find the Length and Width of a Rectangle or Square.

Example 1: The length of a rectangle is 1 foot less than twice its width. If its perimeter is 70 feet find its dimensions.

Solution:

Step 1: We are being asked to find the Length and Width of the rectangle. (It’s Dimensions)

Step 2: Using the information supplied in the problem to express the Length and Width in terms of the variable x.

Width = x and the Length = 2Width – 1 = 2x – 1

Step 3: We now make an equation using x

Perimeter = 2Length + 2Width = 702(2x – 1) + 2x = 70

Step 4: We now solve this equation to find the value of x.

2(2x – 1) + 2x = 704x – 2 + 2x = 704x + 2x – 2 = 70 Rearrange like terms.

6x – 2 = 706x = 72 Add 2 to both sides.

6 x6 =

726 Divide both sides by 6.

x = 12

Step 5: The solution to the problem is the Width and Length of this rectangle.

Width = x = 12 ft and that the Length = 2x – 1 = 2(12) – 1 = 23 ft

Step 6: We check the solution by using x = 12 2(2x – 1) + 2x = 702(2 (12 )−1) + 2(12) = 70

2(24−1) + 24 = 7070 = 70


Since both sides agree we have shown that this is the correct solution.

Example 2: The perimeter of a square and the perimeter of a rectangle are equal. The length of the rectangle is 2 cm longer than the side of the square and the width of the rectangle is 3 cm. Find the dimensions of the Square and rectangle.

Solution:

Step 2: Using the information supplied in the problem we express the Side of the Square and Length of the Rectangle in terms of the variable x

Side of Square = x Length of rectangle = Side of square + 2 = x + 2Width of Rectangle = 3

Step 3: We make an equation using x

Perimeter of Square = Peimeter of rectangle4 Side = 2Length + 2Width

4x = 2(x + 2) + 2(3)4x = 2x + 2 + 64x = 2x + 8


4x = 2x + 82x = 8 Subtract 2x from both sides.2x2 =


x = 4

Step 5: The solution to the problem is the length of the Side of the Square and the Length and Width of the rectangle.

Side of Square = x = 4cmLength of Rectangle = x + 2 = 4 + 2 = 6cmWidth of Rectangle = 3cm

Step 6: We check the solution by using x = 4 4x = 2x + 84(4) = 2(4) + 816 = 16


x + 5

2x – 8


Example 3: The perimeter of this rectangle is 27 cm. Find the value of x and the Length and Width

Solution:

Step 2: Using the information supplied in the problem to express the Length and Width in termsof the variable x.

Length = x + 5 and Width = 2x – 8

Step 3: We make an equation using x

Perimeter = 2Length + 2Width = 272(x + 5) + 2(2x – 8) = 27


2(x + 5) + 2(2x – 8) = 272x + 10 + 4x – 16 = 272x + 4x + 10 – 16 = 27 Collect Like Terms.

6x – 6 = 276x = 33 Add 6 to Both Sides.6 x6 =


x =112

Step 5: The solution to the problem is the Width and Length of this rectangle.

Length = x + 5 = 112

+5=212 ft and Width = 2x – 8 = 2(

112

¿−8 = 11 – 8 = 3 ft

Step 6: We check the solution by using x = 112 2(x + 5) + 2(2x – 8) = 27

2(112 + 5) + 2(2(

112 ) – 8) = 27

2x + 3

2x – 7


2(212 ) + 2(11 – 8) = 27

21 +2(3) = 2727 = 27



1. The Perimeter of a Square is 100 ft, What is the size of one side of the square.

2. The length of a rectangle is twice the width. The perimeter is 42 cm. Find the length and width of this rectangle.

3. The length of a rectangle is 8 cm longer than the width. The perimeter is 36 cm. Find the length and the width of this rectangle.

4. The length of a rectangle is 2 cm shorter than the width. The perimeter is 26 cm. Find the length and the width of this rectangle.

5. The length of a rectangular map is 15 inches and the perimeter is 50 inches. Find the width.

6. The perimeter of a square and the perimeter of a rectangle are equal. The length of the rectangle is 5 cm less than the side of the square and the width of the rectangle is 15 cm. Find the dimensions of the Square and rectangle.

7. The perimeter of a square and the perimeter of a rectangle are equal. The length of the rectangle is 7 cm longer than the side of the square and the width of the rectangle is 2 cm. Find the dimensions of the Square and rectangle.

8. The perimeter of a square and the perimeter of a rectangle are equal. The length of the rectangle is 10 in longer than the side of the square and the width of the rectangle is 6 in. Find the dimensions of the Square and rectangle.

9. The length of a rectangle is 3 feet less than twice its width. If its perimeter is 90 feet, find its dimensions.

10. The perimeter of this rectangle is 32 cm. Find the value of x and the Length and Width


11. The length of a rectangle is 5 centimeters less than twice its width. The perimeter of the rectangle is 26 cm. What are the dimensions of the rectangle?

A. Length = 5 cm; Width = 5 cm B. Length = 6 cm; Width = 7 cm

C. Length = 7 cm; Width = 6 cm D. Length = 4 cm; Width = 9 cm

3. Geometric word problems with AnglesThese word problems contain triangles or angle pictures and will typically ask you to find the Length of a side or the size of an angle.

Example 1: The three angles of a triangle are 2x, x – 30 and 4x degrees. Find the value of x.

Solution:

Step 1: We are being asked to find the sizes of the 3 angles of the triangle

Step 2: Using the information supplied in the problem to express the three angles in terms of the variable x.

Angle 1 = 2x Angle 2 = x – 30 Angle 3 = 3x


Since we have a triangle Angle 1 + Angle 2 + Angle 3 = 1802x + x – 30 + 3x = 180


2x + x – 30 + 3x = 1802x + x + 3x – 30 = 180 Rearrange like terms. 6x – 30 = 180

6x = 210 Add 30 to both sides.6 x6 =


x = 350

Step 5: The solution to the problem is the size of the three angles.

Angle 1 = 2x = 2(35) = 700 Angle 2 = x – 30 = 35 – 30 = 50

Angle 3 = 3x = 3(35) = 1050

x + 2

2x – 7

10 – x


Step 6: We check the solution by using x = 35 2x + x – 30 + 3x = 1802(35) + 35 – 30 + 3(35) = 180

70 + 35 – 30 + 105 = 180180 = 180


Example 2: The perimeter of this triangle is 15 ft.

What are the lengths of its three sides?

Solution:

Step 1: We are being asked to find the sizes of the 3 sides of the triangle.

Step 2: Using the information supplied in the problem to express the three sides in terms of the variable x.

Side 1 = 2x – 7 Side 2 = x + 2 Side 3 = 10 – x


Perimeter = 152x – 7 + x + 2 + 10 – x = 15


2x – 7 + x + 2 + 10 – x = 152x + x –x – 7 + 2 + 10 = 15 Rearrange like terms.

2x + 5 = 152x = 10 Subtract 15 from both sides.

2x2 =


x = 5

Step 5: The solution to the problem is the size of the three angles.

Side 1 = 2x – 7 = 2(5) – 7 = 3 ft Side 2 = x + 2 = 5 + 2 = 7 ftSide 3 = 10 – x = 10 – 5 = 5 ft

2x – 103x + 5


Step 6: We check the solution by using x = 5 2x – 7 + x + 2 + 10 – x = 152(5) – 7 + 5 + 2 + 10 – 5 = 1510 – 7 + 5 + 2 + 10 – 5 = 15

15 = 15


Example 3: What are the sizes of the two Supplementary angles.

Solution:

Step 1: We are being asked to find the sizes of the 2 missing angles.

Step 2: Using the information supplied in the problem to express the three sides in terms of the variable x.

Angle 1 = 3x + 5 Angle 2 = 2x – 10

Step 3: We now make an equation using x.Since we have two supplementary angles they will add up to 1800

Angle 1 + Angle 2 = 1803x + 5 + 2x – 10 = 180


3x + 5 + 2x – 10 = 1803x + 2x + 5 – 10 = 180 Rearrange like terms.

5x – 5 = 1805x = 185 Add 5 to 15 both sides.5x5 =


x = 370

Step 5: The solution to the problem is the size of the two angles.

Angle 1 = 3x + 5 = 3(37) + 5 = 111 + 5 = 1160

Angle 2 = 2x – 10 = 2(37) – 10 = 74 – 10 = 640

x + 20

2x – 10 3x – 10


Step 6: We check the solution by using x = 37 3x + 5 + 2x – 10 = 1803(37) + 5 + 2(37) – 10 = 180

111 + 5 + 74 – 10 = 180180 = 180



1. The three angles of a triangle are 2x, x – 10 and 40 degrees. Find the size of the missing angles?

2. The three angles of a triangle are x + 20, 4x – 10 and 5x + 10 degrees. Find the size of the missing angles?

3. The three angles of a triangle are 3x + 20, 2x + 10 and 50 degrees. Find the size of the missing angles?

4. The three angles of a triangle are 2x , 3x – 10 and 50 – x degrees. Find the size of the missing angles?

5. A triangle has three equal angles. The total of the angles in the triangle is 180 degrees. What is the size of each angle?

6. What are the sizes of the three angles in 7. The perimeter of this triangle is 16 ft. this triangle? What are the lengths of its three sides?

8. For each of the diagrams below, write down an equation for x and solve it.

2x + 2

2x – 4

10 – 2x


9. The angles on a straight line add up to 180o. Write down and solve the equation for each diagram shown below.

4. Complex Word problems.This category of word problems are very varied and do not easily fall into any specific form.The solution of these problems involve the following the main solution strategy of all word problems.

Example 1: On an algebra test, the highest grade was 42 points higher than the lowest grade. The sum of the two grades was 138. Find the lowest grade.

Solution:

Step 1: We are being asked to find the highest and lowest grade.

Step 2: Using the information supplied in the problem to express the highest and lowest grade in terms of the variable x.

Lowest grade = x

Highest Grade = lowest grade + 42 = x + 42


Since we have sum of two grades = 138x + x + 42 = 138

2x + 42 = 138



2x + 42 = 1382x = 96 Subtract 42 to both sides.

2x2 =


x = 48

Step 5: The solution to the problem is the Lowest and Highest grades.

Lowest grade = x = 48 Highest Grade = x + 42 = 48 + 42 = 90

Step 6: We check the solution by using x = 48 2x + 42 = 1382(48) + 42 = 138

96 + 42 = 138138 = 138


Example 2: Donald has a board that is 50 inches long. He wishes to cut it into two pieces so that one piece will be 12 inches longer than the other. How long should the shorter piece be?

Solution:

Step 1: We are being asked to find the lengths of the two pieces.

Step 2: Using the information supplied in the problem to express the shortest and longest piece of wood in terms of the variable x.

Shortest piece = x

Longest piece = Shortest piece + 12 = x + 12


Since we have sum of two pieces = 50x + x + 12 = 50

2x + 12 = 50


2x + 12 = 50

2x = 38 Subtract 12 to both sides.


2x2 =


x = 19

Step 5: The solution to the problem is the Lowest and Highest grades.

Shortest piece = x = 19 inches

Longest piece = x + 12 = 19 + 12 = 31 inches


38 + 12 = 5050 = 50


Example 3: Two consecutive numbers are two numbers which are one after the other. So 5 and 6 or 11 and 12 or 27 and 28 or 33 and 34

are all examples of consecutive numbers.Find two consecutive numbers that add up to 55.

Solution:

Step 1: We are being asked to find the value of two consecutive numbers.

Step 2: Using the information supplied in the problem to express the smallest and larger consecutive numbers in terms of the variable x.

Smallest number = x

Largest number = x + 1


Since we have sum of the numbers = 55x + x + 1 = 55

2x + 1 = 55


2x + 1 = 55



2x2 =


x = 27

Step 5: The solution to the problem is the smallest and largest consecutive numbers.

Smallest number = x = 27

Largest number = x + 1 = 27 + 1 = 28


54 + 1 = 5555 = 55


Example 4: The sum of three consecutive odd integers is 33. What are the three odd integers?

Solution:

Step 1: We are being asked to find the value of three consecutive odd numbers.

Step 2: Using the information supplied in the problem to express the three consecutive numbers in terms of the variable x.

First odd number = x

Second odd number = x + 2

Third odd number = x + 4


Since we have sum of the three odd numbers = 33x + x + 2 + x + 4 = 33x + x + x + 2 + 4 = 33

3x + 6 = 33


3x + 6 = 33



3x3 =


x = 9

Step 5: The solution to the problem is the smallest and largest consecutive numbers.

First odd number = x = 9Second odd number = x + 2 = 9 + 2 = 11Third odd number = x + 4 = 9 + 4 = 13

Step 6: We check the solution by using x = 9 3x + 6 = 333(9) + 6 = 3327 + 6 = 3333 = 33


Example 5: The Smiths invested $2000, part at 5% simple interest and part at 4% simple interest. If they received a total annual interest of $85, how much did they invest at each rate?

Solution:

Step 1: We are being asked to find the amount invested at 5% and at 4%.

Step 2: Using the information supplied in the problem to express the amount of money invested at 5% and at 4% in terms of the variable x.

Amount of money invested at 5% = x

Amount of money invested at 4% = 2100 – x


Since we have total interest = 850.05x + 0.04(2000 – x) = 85


0.05x + 0.04(2000 – x) = 850.05x + 80 – 0.04x = 85 Use the distributive law.0.05x – 0.04x + 80 = 85 Collect like terms.

0.01x + 80 = 85 Subtract 80 to both sides.0.01x = 5


0.01 x0.01 =

50.01 Divide both sides by 0.01.

x = 500

Step 5: The solution to the problem is the amount of money invested at 5% and at 4%

Amount of money invested at 5% = x = $500

Amount of money invested at 4% = 2000 – x = 2000 – 500 = $1500

Step 6: We check the solution by using x = 500

0.05x + 0.04(2000 – x) = 850.05(500) + 0.04(2000 – 500) = 85

25 + 0.04(1500) = 8525 + 60 = 85

85 = 85


Example 5: Glasgow is 500 miles from London. Donald left Glasgow, traveling toward London at an average speed of 50 miles per hour. Mary also left London but two hours later and traveling toward Glasgow at an average speed of 30 miles per hour. How many hours does it take until they meet?

Solution:

Step 1: We are being asked to find the time until they meet

Step 2: Using the information supplied in the problem to express the time in terms of x.

Time taken by Donald = xDistance travelled by Donald = 50xDistance travelled by Mary = 30(x – 2)


Total Distance = Distance travelled by Donald + Distance travelled by Mary 500 = 50x + 30(x – 2)500 = 50x + 30x – 60500 = 80x – 60


80x – 60 = 50080x = 560 Add 60 to both sides.


80 x80 =

56080 Divide both sides by 80

x = 7

Step 5: The solution to the problem is the time until they meet

Time taken by Donald = x = 7 hours

Time taken by Mary = x – 2 = 7 – 2 = 5 hours

Step 6: We check the solution by using x = 7

80x – 60 = 50080(7) – 60 = 500560 – 60 = 500

500 = 500



1. On an algebra test, the highest grade was 30 points higher than the lowest grade. The sum of the two grades was 150. Find the lowest grade.

2. Dylan cuts 25cm from a length of wood. If this leaves him with 25cm, how long was the piece of wood?

3. Brendan cuts 25cm from a length of wood. If this leaves him with 35cm, how long was the piece of wood?

4. Jose has a board that is 44 inches long. He wishes to cut it into two pieces so that one piece will be 6 inches longer than the other. How long should the shorter piece be?

5. Two consecutive numbers are two numbers which are one after the other. So 5 and 6 or 11 and 12 or 27 and 28 or 33 and 34 are all examples of consecutive numbers.Find two consecutive numbers that add up to 25.

6. Find two consecutive numbers that add up to 99.

7. The sum of three consecutive even numbers is 18. Find the three consecutive integers?

8. The sum of three consecutive odd numbers is 21. Find the three consecutive integers?


9. The sum of three consecutive integers is 24. Find the three consecutive integers?

10. The sum of three even integers is 42. What are the three even integers?

11. The sum of three consecutive odd numbers is 213. Find the three consecutive integers?

12. There are 376 stones in three piles. The second pile has 24 more stones than the first . The third pile has twice as many stones as the second. How many stones in each pile ?

13. The Smiths invested $2100, part at 9% simple interest and part at 6% simple interest. If they received a total annual interest of $168, how much did they invest at each rate?

14. A woman invested $200, part at 2% simple interest and part at 10% simple interest. If they received a total annual interest of $12. How much did they invest at each rate?

15. The Robertsons invested $5000, part at 5% simple interest and part at 15% simple interest.If they received a total annual interest of $350. How much did they invest at each rate?

16. A rocket is launched at noon and travels at 12,000 miles per hour. One hour later, a second rocket is launched in the same direction, traveling at 14,400 miles per hour. How long after the first rocket is launched will the rockets be the same distance from Earth

17. Salt Lake City is 500 miles from Denver. Dan left Denver, traveling toward Salt Lake City at an average speed of 60 miles per hour. One hour later Sally also left Salt Lake City, traveling toward Denver at an average speed of 50 miles per hour. How long after Dan left Denver will they meet?

18. Last week, Pat earned $60 less than Tony. If their combined earnings for the week were $500, how much did each earn?

19. Karin’s mom runs a dairy farm. Last year Betty the cow gave 375 gallons less than twice the amount from Bessie the cow. Together, Betty and Bessie produced 1464 gallons of milk. How many gallons did each cow give?

20. In a given amount of time, Jamie drove twice as far as Rhonda. Altogether they drove 90 miles. Find the number of miles driven by each.

21. Last week, Pat earned $60 less than Tony. If their combined earnings for the week were $500, how much did each earn?


22. A Geology book costs $14 more than a Math book. If the two books together cost $96, how much does each book cost?

23. Jill and Ben have the same number of pens. Jill has 3 full boxes and 2 loose pens. Ben has 2 full boxes and 14 loose pens. How many pens are in each box ?

24. Jack, Jo and Jim are sailors. They were shipwrecked on an island with a monkey and a crate of 191 bananas. Jack ate 5 more bananas than Jim. Jo ate 3 more bananas than Jim. Form an equation in x and work out how many bananas Jim ate.

25. John is 5 years older than Mike, the sum of their ages is 29. Find the ages of John and Mike.

26. At a restaurant, a pizza costs twice as much as a sub sandwich. One group bought 5 sub sandwiches and 2 pizzas and their bill was $135. Find the cost of a sub sandwich and the cost of a pizza.

Solutions:

Exercise 1A 1. 13 2. 20 3. 12 4. 68 5. 54 6. – 8 7. 11 8. 0 9. 52

Exercise 2A 1. Side = 50 ft2. Length = 14 cm Width = 7 cm3. Length = 13 cm Width = 5 cm

4. Length = 112 cm Width =

152 cm

5. Width = 10 cm6. Side = 10 cm L = 5cm W = 15 cm7. Side = 9 cm L = 16 cm W = 2 cm8. Side = 16 in L = 26 in W = 6 in9. Length = 29 ft Width = 16 ft10. Length = 13 cm Width = 3 cm11. C

Exercise 3A 1. 2200 ,1000 ,400

2. 360 ,540 ,900

3. 800 ,500 ,500

4. 700 ,950 ,150

5. 600 ,600 ,600

6. 800 ,500 ,500

7. 500 ,800 ,500


8. 10ft,4 ft, 2ft9. (a) 800 ,150 ,850

9.(b) 320 ,220 ,740,520

10. x = 700 x = 880 x = 1490 x = 980

Exercise 4A 1. 60 and 90 2. 50 cm3. 60 cm 4. 19 in5. 12 and 13 6. 49 and 507. 4,6 and 8 8. 5,7 and 99. 7,8 and 9 10. 12,14 and 1611. 69,71 and 73 12. 76,100 and 20013. $1400 and $700 14. $100 and $10015. $4000 and $1000 16. 6 hours17. 5 hours 18. $280 and $22019. 851 and 613 gallons 20. 30 and 60 miles21. $220 and $280 22. $41 and $5623. Box = 12 pens 24. Jim = 61 bananas25. 12 and 17 years 26. $15 and $30

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