Transcript Topic 3 Applications Using Linear Expressionsx

Olympic College Topic 3 – Applications Using Linear Expressions
Topic 3 Applications Using Linear Expressions
Introduction:
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.
Page | 1
Olympic College Topic 3 – Applications Using Linear Expressions
1. Simple word problems
These 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:
Step 3:
Step 4:
x =
4 + 3x =
3x =
=
x =
missing number
40
36
Subtract 4 from Both Sides.
Divide Both Sides by 3.
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:
Step 3:
Step 4:
x =
x–5=
x =
missing number
20
Add 5 to Both Sides.
25
Step 5: The solution to the problem is that the missing number is 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:
Step 3:
Step 4:
n
3(n + 4)
3n + 12
3n
n
=
=
=
=
=
=
missing number
24
24
Subtract 12 from Both Sides.
12
Divide Both Sides by 3.
4
Step 5: The solution to the problem is that the missing number is 4.
Step 6: We check the solution by using n = 4 in 3(n + 4) = 3(4 + 4) = 3(8) = 24
Page | 2
Olympic College Topic 3 – Applications Using Linear Expressions
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:
Step 3:
Step 4:
x
3x – 10
2x – 10
3x
x
=
=
=
=
=
missing number
x–5
–5
Subtract x from Both Sides.
5
Add 10 to Both Sides
Divide Both Sides by 3
=
Step 5: The solution to the problem is that the missing number is .
Step 6: We check the solution by using x = in
3
3x – 10 =
x–5
– 10
=
( –5
=
–5
– 10
=
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
Page | 3
Olympic College Topic 3 – Applications Using Linear Expressions
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
= 2Width – 1 =
Length
2x – 1
Step 3: We now make an equation using x
Perimeter
=
2Length + 2Width
2(2x – 1) + 2x
=
=
70
70
Step 4: We now solve this equation to find the value of x.
2(2x – 1) + 2x
4x – 2 + 2x
4x + 2x – 2
6x – 2
6x
x
=
=
=
=
=
=
=
70
70
70
70
72
Rearrange like terms.
Add 2 to both sides.
Divide both sides by 6.
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 =
2
2(2x – 1) + 2x =
+ 2(12)
=
2
+ 24 =
=
70
70
70
70
Since both sides agree we have shown that this is the correct solution.
Page | 4
Olympic College Topic 3 – Applications Using Linear Expressions
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 + 2
Width of Rectangle = 3
Step 3: We make an equation using x
Perimeter of Square
4 Side
4x
4x
4x
=
=
=
=
=
Peimeter of rectangle
2Length + 2Width
2(x + 2) + 2(3)
2x + 2 + 6
2x + 8
Step 4: We now solve this equation to find the value of x.
4x =
2x =
=
2x + 8
8
x
4
=
Subtract 2x from both sides.
Divide Both Sides by 2.
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 = 4cm
Length of Rectangle = x + 2 = 4 + 2 = 6cm
Width of Rectangle = 3cm
Step 6: We check the solution by using x = 4
4x =
4(4) =
=
2x + 8
2(4) + 8
16
Since both sides agree we have shown that this is the correct solution.
Page | 5
Olympic College Topic 3 – Applications Using Linear Expressions
Example 3: The perimeter of this rectangle is 27 cm.
Find the value of x and the Length and Width
2x – 8
x+5
Solution:
Step 2: Using the information supplied in the problem to express the Length and Width in terms
of the variable x.
Length = x + 5 and Width = 2x – 8
Step 3: We make an equation using x
Perimeter
=
2Length + 2Width
2(x + 5) + 2(2x – 8)
=
=
27
27
Step 4: We now solve this equation to find the value of x.
2(x + 5) + 2(2x – 8)
2x + 10 + 4x – 16
2x + 4x + 10 – 16
6x – 6
6x
x
=
=
=
=
=
=
27
27
27
27
33
Collect Like Terms.
Add 6 to Both Sides.
Divide Both Sides by 6.
=
Step 5: The solution to the problem is the Width and Length of this rectangle.
Length = x + 5 =
ft and Width = 2x – 8 = 2(
Step 6: We check the solution by using x =
2(x + 5) + 2(2x – 8)
2( + 5) + 2(2( ) – 8)
= 11 – 8 = 3 ft
=
=
27
27
2( ) + 2(11 – 8) =
27
21 +2
= 27
= 27
Since both sides agree we have shown that this is the correct solution.
Page | 6
Olympic College Topic 3 – Applications Using Linear Expressions
Exercise 2A
Solve the following problems by first writing an equation and then solving it.
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
2x – 7
2x + 3
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
Page | 7
Olympic College Topic 3 – Applications Using Linear Expressions
3. Geometric word problems with Angles
These 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
Step 3: We now make an equation using x
Since we have a triangle Angle 1 + Angle 2 + Angle 3
=
2x + x – 30 + 3x =
180
180
Step 4: We now solve this equation to find the value of x.
2x + x – 30 + 3x
2x + x + 3x – 30
6x – 30
6x
x
=
=
=
=
=
=
180
180
180
210
Rearrange like terms.
Add 30 to both sides.
Divide both sides by 7.
350
Step 5: The solution to the problem is the size of the three angles.
Angle 1 =
Angle 2 =
Angle 3 =
2x
x – 30
3x
=
=
=
2(35) =
35 – 30 =
3(35) =
Step 6: We check the solution by using x =
700
50
1050
2x + x – 30 + 3x =
2(35) + 35 – 30 + 3(35) =
70 + 35 – 30 + 105 =
=
180
180
180
180
Since both sides agree we have shown that this is the correct solution.
Page | 8
Olympic College Topic 3 – Applications Using Linear Expressions
Example 2: The perimeter of this triangle is 15 ft.
x+2
What are the lengths of its three sides?
Solution:
10 – x
2x – 7
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 3 = 10 – x
Side 2 = x + 2
Step 3: We now make an equation using x
Perimeter
2x – 7 + x + 2 + 10 – x
=
=
15
15
Step 4: We now solve this equation to find the value of x.
2x – 7 + x + 2 + 10 – x
2x + x –x – 7 + 2 + 10
2x + 5
2x
x
=
=
=
=
=
=
15
15
15
10
Rearrange like terms.
Subtract 15 from both sides.
Divide both sides by 2.
5
Step 5: The solution to the problem is the size of the three angles.
Side 1
Side 2
Side 3
=
=
=
2x – 7
x+2
10 – x
=
=
=
2(5) – 7 =
+2
=
10 –
=
Step 6: We check the solution by using x =
3 ft
7 ft
5 ft
2x – 7 + x + 2 + 10 – x
=
2(5) – 7 + 5 + 2 + 10 – 5 =
10 – 7 + 5 + 2 + 10 – 5
=
=
15
15
15
15
Since both sides agree we have shown that this is the correct solution.
Page | 9
Olympic College Topic 3 – Applications Using Linear Expressions
Example 3: What are the sizes of the two Supplementary angles.
3x + 5
2x – 10
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 = 180
3x + 5 + 2x – 10 = 180
Step 4: We now solve this equation to find the value of x.
3x + 5 + 2x – 10
3x + 2x + 5 – 10
5x – 5
5x
x
=
=
=
=
=
=
180
180
180
185
Rearrange like terms.
Add 5 to 15 both sides.
Divide both sides by 5.
370
Step 5: The solution to the problem is the size of the two angles.
Angle 1 =
Angle 2 =
3x + 5 =
2x – 10 =
3(37) + 5
2(37) – 10
Step 6: We check the solution by using x =
=
=
111 + 5 =
74 – 10 =
1160
640
3x + 5 + 2x – 10 =
3(37) + 5 + 2(37) – 10
=
111 + 5 + 74 – 10
=
=
180
180
180
180
Since both sides agree we have shown that this is the correct solution.
Page | 10
Olympic College Topic 3 – Applications Using Linear Expressions
Exercise 3A
Solve the following problems by first writing an equation and then solving it.
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
this triangle?
7.
The perimeter of this triangle is 16 ft.
What are the lengths of its three sides?
x + 20
10 – 2x
2x + 2
2x – 10
3x – 10
2x – 4
8. For each of the diagrams below, write down an equation for x and solve it.
9. The angles on a straight line add up to 180o. Write down and solve the equation for each
diagram shown below.
Page | 11
Olympic College Topic 3 – Applications Using Linear Expressions
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
Step 3: We now make an equation using x
Since we have sum of two grades =
x + x + 42 =
2x + 42 =
138
138
138
Step 4: We now solve this equation to find the value of x.
2x + 42 =
2x
=
=
x =
138
96
Subtract 42 to both sides.
Divide both sides by 2.
48
Step 5: The solution to the problem is the Lowest and Highest grades.
Lowest grade
Highest Grade
=
=
x =
x + 42
48
= 48 + 42
Step 6: We check the solution by using x =
=
90
2x + 42 =
2(48) + 42 =
96 + 42 =
=
138
138
138
138
Since both sides agree we have shown that this is the correct solution.
Page | 12
Olympic College Topic 3 – Applications Using Linear Expressions
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
Step 3: We now make an equation using x
Since we have sum of two pieces =
x + x + 12 =
2x + 12 =
50
50
50
Step 4: We now solve this equation to find the value of x.
2x + 12 =
2x
=
50
38
Subtract 12 to both sides.
=
x
=
Divide both sides by 2.
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
Step 6: We check the solution by using x =
=
31 inches
2x + 12 =
2(19) + 12 =
38 + 12 =
=
50
50
50
50
Since both sides agree we have shown that this is the correct solution.
Page | 13
Olympic College Topic 3 – Applications Using Linear Expressions
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
Step 3: We now make an equation using x
Since we have sum of the numbers =
x+x+1
=
2x + 1 =
55
55
55
Step 4: We now solve this equation to find the value of x.
2x + 1
=
55
2x
=
54
=
x
=
Subtract 1 to both sides.
Divide both sides by 2.
27
Step 5: The solution to the problem is the smallest and largest consecutive numbers.
Smallest number = x
Largest number =
=
27
x + 1 = 27 + 1 = 28
Step 6: We check the solution by using x =
2x + 1
2(27) + 1
54 + 1
=
=
=
=
55
55
55
55
Since both sides agree we have shown that this is the correct solution.
Page | 14
Olympic College Topic 3 – Applications Using Linear Expressions
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
Step 3: We now make an equation using x
Since we have sum of the three odd numbers
x+x+2+x+4
x+x+x+2+4
3x + 6
=
=
=
=
33
33
33
33
Step 4: We now solve this equation to find the value of x.
3x + 6
=
33
3x
=
27
=
x
=
Subtract 6 to both sides.
Divide both sides by 3.
9
Step 5: The solution to the problem is the smallest and largest consecutive numbers.
First odd number = x = 9
Second odd number = x + 2 = 9 + 2 = 11
Third odd number = x + 4 = 9 + 4 = 13
Step 6: We check the solution by using x =
3x + 6 =
3(9) + 6 =
27 + 6 =
=
33
33
33
33
Since both sides agree we have shown that this is the correct solution.
Page | 15
Olympic College Topic 3 – Applications Using Linear Expressions
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
Step 3: We now make an equation using x
Since we have total interest
0.05x + 0.04(2000 – x)
=
=
85
85
Step 4: We now solve this equation to find the value of x.
0.05x + 0.04(2000 – x)
0.05x + 80 – 0.04x
0.05x – 0.04x + 80
0.01x + 80
0.01x
x
=
=
=
=
=
=
=
85
85
85
85
5
Use the distributive law.
Collect like terms.
Subtract 80 to both sides.
Divide both sides by 0.01.
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 =
0.05x + 0.04(2000 – x)
0.05(500) + 0.04(2000 – 500)
25 + 0.04(1500)
25 + 60
85
=
=
=
=
=
85
85
85
85
85
Since both sides agree we have shown that this is the correct solution.
Page | 16
Olympic College Topic 3 – Applications Using Linear Expressions
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 = x
Distance travelled by Donald = 50x
Distance travelled by Mary = 30(x – 2)
Step 3: We now make an equation using x
Total Distance
500
500
500
=
=
=
=
Distance travelled by Donald + Distance travelled by Mary
50x + 30(x – 2)
50x + 30x – 60
80x – 60
Step 4: We now solve this equation to find the value of x.
80x – 60
80x
x
=
=
=
=
500
560
Add 60 to both sides.
Divide both sides by 80
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 =
80x – 60
80(7) – 60
560 – 60
500
=
=
=
=
500
500
500
500
Since both sides agree we have shown that this is the correct solution.
Page | 17
Olympic College Topic 3 – Applications Using Linear Expressions
Exercise 4A
Solve the following problems by first writing an equation and then solving it.
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?
Page | 18
Olympic College Topic 3 – Applications Using Linear Expressions
15. The Robertson’s 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.
Page | 19
Olympic College Topic 3 – Applications Using Linear Expressions
Solutions:
Exercise 1A
1. 13
2. 20
3. 12
4. 68
5. 54
Exercise 2A
1. Side = 50 ft
2. Length = 14 cm Width = 7 cm
3. Length = 13 cm Width = 5 cm
4. Length = cm Width =
cm
5. Width = 10 cm
6. Side = 10 cm L = 5cm W = 15 cm
7. Side = 9 cm L = 16 cm W = 2 cm
8. Side = 16 in L = 26 in W = 6 in
9. Length = 29 ft Width = 16 ft
10. Length = 13 cm Width = 3 cm
11. 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, 2ft
9. (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
3. 60 cm
5. 12 and 13
7. 4,6 and 8
9. 7,8 and 9
11. 69,71 and 73
13. $1400 and $700
15. $4000 and $1000
17. 5 hours
19. 851 and 613 gallons
21. $220 and $280
23. Box = 12 pens
25. 12 and 17 years
6. – 8
7. 11
8. 0
9.
2. 50 cm
4. 19 in
6. 49 and 50
8. 5,7 and 9
10. 12,14 and 16
12. 76,100 and 200
14. $100 and $100
16. 6 hours
18. $280 and $220
20. 30 and 60 miles
22. $41 and $56
24. Jim = 61 bananas
26. $15 and $30
Page | 20