Download MAT0024-L2-Sections 9-4-5-6

Problem Solving
General Guidelines for Problem Solving
1. Understand the problem.
Read the problem carefully.
Identify the unknown and select a variable.
Construct a drawing if necessary.
2. Translate the information to an equation.
3. Solve the equation and check the solution.
4. Interpret the solution.
Problem Solving
Example 1:
Three times the difference of a number and 5 is the same as
twice the number decreased by 3. Find the number.
k is the number
The
Three difference
times of a number
and 5
3
 k  5
3  k  5  2k  3
is
twice
the
number

2k
decreased
by 3
3
Problem Solving
Example 1:
3  k  5  2k  3
3k 15  2k  3
3k 15 15  2k  3 15
3k  2k 12
3k  2k  2k  2k  12
k  12
Problem Solving
Example 1:
Check:
3 12  5  2 12  3
3  7   24  3
21  21
Problem Solving
Example 2:
The difference between two positive integers is 42. One
integer is three time as great as the other. Find the integers.
x = one integer
3x = the other integer
The difference
between two
positive
integers
is
42
 3x  x 

42
3x  x   42
Problem Solving
Example 2:
3x  x   42
Check:
2x  42
3  21  21  42
2 x 42

2
2
 63  21  42
x  21
3 x  3  21  63
42  42
Problem Solving
Example 3:
A 22-foot pipe is cut into two pieces. The shorter piece is 7
feet shorter than the longer piece. What is the length of the
longer piece?
Longer piece = m
Shorter piece = m – 7
Longer piece plus Shorter piece
is
22 feet


22
m
m  m  7  22
m7
Problem Solving
Example 3:
m  m  7  22
2m  7  22
2m  7  7  22  7
2m  29
2m 29

2
2
m  14.5 feet
Problem Solving
Example 3:
Check:
14.5 14.5  7  22
29  7  22
22  22
Problem Solving
Example 4:
A college graduating class is made up of 450 students. There
are 206 more females than males. How many males are in the
class?
Males = h
Females = h + 206
Males
h
plus

Females
h  206
h  h  206  450
is

450 students
450
Problem Solving
Example 4:
h  h  206  450
2h  206  450
2h  206  206  450  206
2h  244
2h 244

2
2
h  122 males
Problem Solving
Example 4:
Check:
122 122  206  450
244  206  450
450  450
Problem Solving
Example 5:
A triangle has three angles A, B, and C. Angle C is 18
degrees greater than angle B. Angle A is 4 times angle B.
What is the measure of each angle? Reminder – the sum of
the angles in a triangle is: mA  mB  mC  180
mB  B
mC  B  18
mA  4B
mA
plus
mB
plus
mC
is
180
4B

B

B  18

180
4B  B  B  18  180
Problem Solving
Example 5:
4B  B  B  18  180
6B 18  180
6B 18 18  180 18
6B  162
6 B 162

6
6
B  27
mB  27
Problem Solving
Example 5:
Other angles:
mA  4  27
mA  108
mC  27 18
mC  45
Check:
mA  mB  mC  180
108  27  45  180
180  180
Formulas and Problem Solving
General Guidelines for Solving for a Specific
Variable in a Formula
1. Eliminate fractions from the formula.
2. Remove parentheses from the formula using
the distributive property.
3. Simplify like terms.
4. Get all terms containing the specified variable on
one side of the equation.
5. Use the multiplicative inverse property to get
the specified variable’s coefficient to one.
6. Simplify the results if necessary.
Formulas and Problem Solving
Using the given values, solve for the variable in each
formula that was not assigned a value.
Example 1:
Check:
d  rt t  9, d  63
63  r  9
63  r  9
63  7  9
63 r  9

9
9
63  63
7r
r 7
Formulas and Problem Solving
Example 2: Volume of a Pyramid
1
V  Bh V  40, h  8
3
1

1

3  40   3  B  8 
40  B  8
3
3

120 B  8

120  B  8

8
8
15  B
B  15
Formulas and Problem Solving
Example 2: Volume of a Pyramid
1
V  Bh V  40, h  8
3
Check:
1
40   1 5  8
3
40  5  8
40  40
Formulas and Problem Solving
Example 3: Solve for the requested variable.
Area of a Triangle – solve for b
1
A   bh
2
2A  bh
2A bh

h
h
2A
b
h

1 
2  A   2  bh 
2 
Formulas and Problem Solving
Example 4: Solve for the requested variable.
Celsius to Fahrenheit – solve for C
9
F   C  32
5

9
F  32   C  32  32
5
9 
 5  F  32   5   C 
5 
5  F  32  9C

5  F  32  9C 
9
9
9
F  32   C
5
5  F  32 
C
9
or
5
 F  32   C
9
Formulas and Problem Solving
Example 4: Solve for the requested variable.
Celsius to Fahrenheit – solve for C –
Alternate Solution
9
F   C  32
5
9
 F  32   C  32  32
5
5
59 
9
 F  32     C 
F  32   C 
9
95 
5
5
 F  32   C
9
Formulas and Problem Solving
Formulas describe a known relationship among variables.
Most formulas are given as equations, so the guidelines
for problem solving are relatively the same.
Guidelines for Using Formulas in Problem Solving
1. Understand the problem.
Read the problem carefully.
Identify the known, unknown and the
variable(s).
Construct a drawing if necessary.
2. Translate the information to a known formula.
3. Solve the equation and check the solution.
4. Interpret the solution.
Formulas and Problem Solving
Example 1:
A pizza shop offers a 2-foot diameter round pizza and a 1.8foot square pizza for the same price. Which one is the better
deal?
Round Pizza
Square Pizza
Area  Sq.  s
s  1.8
Area   r
  3.14
r  1
d  2
2
Area   3.14 1
Area  3.14 ft
2
2
2
Area  Sq.  1.8 
2
Area  Sq.  3.24 ft
2
Formulas and Problem Solving
Example 1:
A certain species of fish requires 1.6 cubic feet of
water per fish. What is the maximum number of fish
that could be put into a tank that is 3 feet long by 2.4
feet wide by 2 feet deep?
Cubic feet is a unit of volume.
Volume for Fish
Volume of Tank
Number of times Required
volume per equals length*width*height
fish (f)
fish
f
*
1.6 CF /fish
1.6 f  l  w  h
=
3*2.4*2
 1.6 f  3  2.4  2
Example 1:
Formulas and Problem Solving
1.6 f  l  w  h
1.6 f  3  2.4  2
1.6 f  7.2  2
1.6 f  14.4
1.6 f 14.4

1.6
1.6
f  9 fish
Linear Inequalities and Problem Solving
Properties of Inequality
Addition Property of Inequality
If a, b, and c are real numbers, then
a  b and a  c  b  c or
a  b and a  c  b  c
(The property is also true for subtraction.)
Linear Inequalities and Problem Solving
Properties of Inequality
Multiplication Property of Inequality
1. If a, b, and c are real numbers and c is positive, then
a  b and ac  bc
are equivalent inequalities.
2. If a, b, and c are real numbers and c is negative, then
a  b and ac  bc
are equivalent inequalities.
Linear Inequalities and Problem Solving
Graphing an Inequality
y4
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
x  2
-6
-5
-4
-3

-2
-1
0
1
2
3
4
5
6
5  x  1
-6
-5
-4
-3
-2
-1
0

1
2
3
4
5
6
Linear Inequalities and Problem Solving
Guidelines for Solving a Linear Inequality
1. Eliminate fractions from the formula.
2. Remove parentheses from the formula using
the distributive property.
3. Simplify like terms.
4. Get all terms containing the specified variable on
one side of the equation using the addition property
of inequality.
5. Use the multiplication property of inequality to
get the specified variable’s coefficient to one.
Reverse the inequality sign when multiplying or
dividing by a negative value.
6. Simplify the results if necessary.
Linear Inequalities and Problem Solving
Solve each inequality and graph the solution.
Example 1:

x  7  12
x  7  7  12  7
x5
-6
-5
-4
-3
-2
-1
0
1
2
3
4

5
6
Linear Inequalities and Problem Solving
Solve each inequality and graph the solution.
Example 2:
8x  7  10x  4  8x  7  7  10x  4  7
8x  10x  3  8x 10x  10x 10x  3
2 x 3

2 x  3 
2 2
3
x

x  1.5
2

-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Linear Inequalities and Problem Solving
Solve each inequality and graph the solution.
18  2x  3x  24
Example 3:
18 18  2x  3x  24 18
2x  3x  6
2x  3x  3x  3x  6
x  6

-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Linear Inequalities and Problem Solving
Solve each inequality and graph the solution.
8
1
Example 4:
 x  2    x  3
21
7
 8

1

21  x  2    21  x  3 
 21

7

8  x  2  3 1 x  3  8x  16  3  x  3
8x 16  3x  9
8x 16 16  3x  9 16

8x  3x  7
Linear Inequalities and Problem Solving
8x  3x  7
Example 4:
8x  3x  3x  3x  7  5x  7
5 x 7


5
5
7
x
5
x  1.4
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6