Linear equations inequalities and applications

1
MATHS PROJECT
Rachna Nathawat
Class IXth-c
Roll no. 29

2
Math 131 - Chapter 2
Linear Equations, Inequalities, and
Applications
• 2-1 Linear Equations in One Variable
• 2-2 Formulae
• 2-3 Applications
• 2-5 Linear Inequalities in One Variable

3
2-1 Linear Equations in One Variable
A linear equation in one variable can be
written in the form Ax + B = C, where A,B,
and C are real numbers with A  0.
• Example 1: x + 4 = -2
Note: x + 4 by itself is called an algebraic expression.
• Example 2: 2k + 5 = 10
Note: A linear equation is also called a first degree equation since the
highest power of x is 1.
• The following are called non-linear equations:
y = 3x2 +10 x  2

4
Decide whether a number is a solution of
a linear equation.
• Example 3: 8 is a solution of the equation x – 3 = 5
An equation is solved by finding its “solution set”.
The solution set of x – 3 = 5 is {8}.
Note: Equivalent equations are equations that have the same
solution set:
• Example 4: 5x +2 = 17, 5x =15 and x =3 are equivalent.
The solution set is {3}.

5
Addition Properties of Equality
• For all real numbers A,B, and C, the
equations A = B and A + C = B + C are
equivalent.
Note: The same number can be added to each side of an expression
without changing its solution set.
• Example 5: Solution of the equation x – 3 = 5
x – 3 +3 = 5 + 3
x = 8
The solution set of x – 3 = 5 is {8}.
• Example 6: 5x +2 = 17, 5x + 2 + (-2) =17 + (-2) and 5x =15
are equivalent.

6
Multiplication Properties of Equality
• For all real numbers A,B, and for
C  0, the equations A = B and AC = BC
are equivalent.
Note: Each side of an equation may be multiplied by the same non-zero
number without changing its solution set.
• Example 7: Solution of the equation
The solution set of 5x = 15 is {3}.
5x = 15
1 1
( ) 5 x 
( ) 15
5 5
x = 3

7
Example 8: Solve: -7 + 3x – 9x = 12x -5
-7 + 7 –6x = 12x –5 + 7
-6x –12x = 12x –12x +2
-18x = 2
1 1
 x  
( ) (-18) ( ) 2
18 18
1
x = -
9
1
-
9
The solution set is { }.

8
Solving a Linear Equation in One
Variable:
• Step 1: Clear fractions. (Eliminate any fractions by multiplying
each side by the least common denominator.)
• Step 2: Simplify each side separately. (Use the distributive
property to clear parenthesis and combine like terms.)
• Step 3: Isolate the variable terms on the left side of the equal
sign. (Use the addition property to get all the variable terms on
the left side and all the numbers on the right side.)
• Step 4: Isolate the variable. (Use the multiplication property to
get an equation where the coefficient of the variable is 1.)
• Step 5: Check. (Substitute the proposed solution into the original
equation and verify both sides of the equal sign are equivalent.)

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Example 8: Solve: 6 – (4 +x) = 8x – 2(3x + 5)
Step 1: Does not apply
Step 2: 6 – 4 –x = 8x – 6x –10
-x +2 = 2x – 10
Step 3: -x + (-2x) +2 +(-2) = 2x + (-2x) –10 + (-2)
-3x = -12
Step 4:
1 1
 x  
( ) (-3) ( ) (-12)
3 3
Step 5: 6 – (4 + 4) = 8(4) – 2(3(4) + 5)
6 – 8 = 32 - 34
-2 = -2
x = 4

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Example 9: Solve:
Step 1:
Step 2:
Step 3:
Step 4:
x x
 
 
x x
 
 
x x
 
 
x   x
 
Step 5:
The solution set is {-1}.
1 3 1
2 4 2
1 3 1
4( ) 4( )
2 4 2
1 3 1
4( ) 4( ) 4( )
2 4 2
2( 1) ( 3) 2
x x
2  2   3 
2
3 5 2
 
3  5  (  5)  2  ( 
5)
3 3
x
x
x
 
1 1
( ) 3 ( )( 3)
3 3
1
x
x
1 1 1 3 1 2 1 1 1
:
2 4 2 4 2 2 2
Check
 
 
   
   

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Identification of
Conditional Equations,
Contradictions, and
Identities
• Some equations that
appear to be linear
have no solutions, while
others have an infinite
number of solutions.
Note: Identification
requires knowledge of
the information in the
table.
Type of
Equation
Number of
Solutions
Indication
when solving
Conditional One Final line is:
x = a number
Contradiction None;
solution set 0
or null
Final line is
false, such as
0=1
Identity Infinite;
solution set
is all real
numbers
Final line is
true, such as
0=0

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Example 9: Solve each equation, Decide whether it is conditional,
identity, or contradiction.
A) 5x – 9 = 4(x-3)
5x – 9 = 4x –12
x = -3  conditional
The solution set is {-3}
B) 5x –15 = 5(x-3)
5x – 15 = 5x –15
0 = 0  identity
The solution set is {all real numbers}
C) 5x –15 = 5(x-4)
5x – 15 = 5x –20
0 = -5  contradiction
The solution set is {} or 0 (null set)

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2-2 Formulae
A “mathematical model” is an equation or inequality that
describes a real situation. Models for many applied problems
already exist and are called “formulas”.
A “formula” is a mathematical equation in which variables
are used to describe a relationship.
Common formulae that will be used are:
d = rt distance = (rate)(time)
I = prt Interest = (principal)(rate)(time)
P = 2L + 2 W Perimeter = 2(Length) + 2(Width)
amount
percent percent
 100%   100%

base
a
b
Additional formulae are included on the inside covers of your text.

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2-2 Formulae
Solving for a specified variable:
Step 1: Get all terms containing the specified variable on
the left side of the equal sign and all other terms on the
right side.
Step 2: If necessary, use the distributive property to
combine the terms with the specified variable.
Step 3: Isolate the variable. (Use the multiplication (division)
property to get an equation where the coefficient of the variable
is 1.)
• Example 1: Solve m = 2k + 3b for k
2k = m –3b
k = (m – 3b)/2

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2-2 Formulae
• Example 2: Solve
• Example 3: Solve
1
( 3) for
2
y  x  x
1 3
( )
2 2
y x
 
1 3
( x )
 y

2 2
1 3
x y
2( ( )) 2( )
 
2 2
2 3
x y
 
A  2HW  2LW  2LH for W
2 HW 2 LW A 2
LH
W (2 H 2 L ) A 2
LH
  
  
A 
LH
( 2 )
(2 2 )
W
H L



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2-2 Formulae
• Example 4: Given a distance of 500 miles and a rate (speed)
of 25 mph, find the time for the trip.
d rt solve for t
rt d
d
d miles
t   
hours
r miles
• Example 5: A 20 oz mixture of gasoline and oil contains 1
oz of oil. What percent of the mixture is oil?
t
r



500
20
25
hour
100%
amount
percent
base
 
1
100% .05 100% 5%
20
   

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2-3 Applications of Linear Equations
Translating words to mathematical
expressions.
Note: There are usually key words or phrases that translate into
mathematical expressions.
Translating from Words to Mathematical Expressions
x and y represent numbers
Verbal Expression Mathematical Expression
Additions
The sum of a number and 7 x + 7
6 more than a number x + 6
3 plus a number 3 + x
24 added to a number x + 24
A number increased by 5 x + 5
The sum of two numbers x + y

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expressions.
x and y represent the numbers
Subtractions
2 less than a number x - 2
3 minus a number 3 - x
A number decreased by 5 x - 5
A number subtracted from 10 10 - x
The difference between two numbers x - y

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expressions.
x and y represent the numbers
Multiplication
2 times a number 2  x or 2x
A number multiplied by 6 x  6 or 6x
2
of a number
3
2 2

x or x
3 3
Twice a number 2  x or 2x
The product of two numbers x  y or xy

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Translating words in to equations.
Note: The symbol for equality, = , is often indicated by the word “is”.
Translating from Words to Mathematical Equations
Verbal Sentence Mathematical Equation
Twice a number, decreased by 3, is 42 2 x - 3 = 42
The product of a number and 12,
decreased by 7, is 105 12x - 7 = 105
The quotient of a number and the number
plus 4 is 28
The quotient of a number and 4, plus
the number, is 10
28
4
x
x


x
x
10
4

21
Distinguishing between expressions and
equations.
Note: An expression is a phrase. An equation includes the = symbol
and translates the sentence.
• Example 1: Decide whether each is an expression or an equation.
A) 5x – 3(x+2) = 7 Equation
B) 5x – 3(x+2) Expression

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Solving Applied Problems.
Note: The following six steps are useful in developing a
technique for solving applied problems.
• Step 1: Read the problem carefully until you understand what
is given and what is to be found.
• Step 2: Assign a variable to represent the unknown value, using
diagrams or tables as needed.
• Step 3: Write an equation using the variable expressions.
• Step 4: Solve the equation. (Isolate the variable on the left side of
the equals sign)
• Step 5: State the answer with appropriate units. (Does it seem
reasonable?)
• Step 6: Check the answer.

23
At the end of the 1999 baseball season, Sammy Sosa and Mark
McGwire had a lifetime total of 858 homeruns. McGwire had 186
more than Sosa. How many runs did each player have?
• Step 1: Total runs = 858 Sosa has 186 less than McGwire
• Step 2: Let x = Sosa’s runs, then x + 186 = McGwire’s runs
• Step 3: x + (x + 186) = 858
• Step 4: 2x + 186 = 858
2x = 672
x = 336 x + 186 = 522
• Step 5: Sosa had 336 homeruns. McGwire had 522 homeruns.
• Step 6: 336 + 522 = 858
858 = 858
Checked:

24
A woman has $34,000 to invest. She invests some at 17% and the
balance at 20%. Her total annual interest income is $6425. Find the
amount invested at each rate.
• Step 1: Total investment = $34,000 Total interest earned is $6245
• Step 2: Let x = amount at 17%, then $34,000 - x = amount at 20%
• Step 3: .17x + .20(34,000 - x) = 6245
• Step 4: .17x + 6800 - .20x = 6245
-.03x = -555
x = 18,500 34,000 – 18,500 = 15,500
• Step 5: $18,500 invested at 17% $15,500 invested at 20%
• Step 6: .17(18,500) + .20(15,5000 = 6245
3145 + 3100 = 6245
6245 = 6245
Checked:

25
How many pounds of candy worth $8 per lb should be mixed with 100
lb of candy worth $4 per lb to get a mixture that can be sold for $7 per
lb?
• Step 1: Total mixture should sell for $7/lb. Given 100 lbs at $4/lb and
an unknown amount at $8/lb.
• Step 2: Let x = amount at $8 then x + 100 should sell at $7/lb
• Step 3: 8x + 4(100) = 7(x + 100)
• Step 4: 8x + 400 = 7x + 700
x = 300
• Step 5: 300 lb at $8/lb
• Step 6: 8(300) + 4(100) = 7(300 + 100)
2400 + 400 = 7(400)
2800 = 2800
Checked:

26
How much water must be added to 20 L of 50% antifreeze solution to
reduce it to a 40% antifreeze solution?
• Step 1: Total mixture should be 40% antifreeze. Given 20 L at 50%
produces 10 L of water and 10 L of antifreeze. Add an unknown
amount of water.
• Step 2: Let x = amount of water added, then x + 10 will be the water in
the final solution and the amount of antifreeze is 10 L.
• Step 3:
• Step 4:
amount
percent percent
 100%   100%

base
a
b
10 L of antifreeze
 100% 
40
10 L of antifreeze + (x+10) of water
1000  40(10  ( 
10))
1000  400  40 
400
40 
200
5
x
x
x
x

10 L of antifreeze
• Step 5: 5 L of water
• Step 6: Check:
100% 40%
10 L of antifreeze + (5+10) of water
10
100% 40%
25
40% 40%
 
 


27
2-5 Linear Inequalities in One Variable
Interval Notation is used to write solution sets of
inequalities.
Note: A parenthesis is used to indicate an endpoint in not included. A
square bracket indicates the endpoint is included.
Interval Notation
Type of Interval Set Interval Notation Graph
{x a  x } (a, )
{x  a  x  b} (a, b)
{x  x  b} (-, b)
{x  x is a real number} (-, )
Open Interval
(
a
)
b
(
a
)
b

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inequalities.
Interval Notation
{x a  x} [a, )
{x  a  x  b} (a, b]
{x  a  x  b} [a, b)
{x  x  b} (-, b]
Half-open
Interval
[
a
]
b
]
b
(
a
)
b
[
a

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inequalities.
Interval Notation
Closed Interval {x a  x  b} [a, b]
]
b
[
a

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A linear inequality in on variable can be
written in the form Ax + B = C, where A, B,
and C are real numbers with A  0.
Note: The next examples include definitions and rules for < , >,
, and  .
• Examples of linear inequalities:
x + 5 < 2 x – 3 > 5 2k +4  10

31
Solving linear inequalities using the Addition Property:
For all real numbers A, B, and C, the inequalities A < B and
A + C < B + Ca are equivalent.
Note: As with equations, the addition property can be used to
add negative values or to subtract the same number from each
side .
• Example 1: Solve k – 5 > 1
k – 5 +5 > 1 + 5
k > 6
Solution set: (6,)
• Example 2: Solve 5x + 3  4x – 1 and graph the solution set.
5x – 4x  -1 –3
x  -4
Solution set: [-4,)
[
-4

32
Solving linear inequalities using the Multiplication Property:
For all real numbers A, B, and C, with C  0 ,
1) if C > 0, then the inequalities A < B and AC < BC are equivalent.
2) if C < 0, then the inequalities A < B and AC > BC are equivalent.
Note: Multiplying or Dividing by a negative number requires the
inequality sign be reversed.
• Example 1: Solve -2x < 10
x > -5
Solution set: (-5,)
• Example 2: Solve 2x < -10
x < -5
Solution set: (-,-5)
Note: The first example requires a symbol change because both sides
are multiplied by (-1/2). The second example does not because both
sides are multiplied by (1/2)

33
Solving linear inequalities using the Multiplication
Property:
• Example 3: Solve –9m < -81 and graph the solution set
m > 9
Solution set: (9, ) (
9

34
Solving a Linear Inequality:
Step 1: Simplify each side separately. If necessary, use the distributive
property to clear parenthesis and combine like terms.
Step 2: Use the addition property to get all terms containing the
specified variable on the left side of the inequality sign and all other
terms on the right side.
Step 3: Isolate the variable. (Use the multiplication (division) property
to get an inequality where the coefficient of the variable is 1.)
• Example 4: Solve 6(x-1) + 3x  -x –3(x + 2) and graph the solution set
Step 1: 6x-6 + 3x  -x –3x – 6
9x - 6  –4x – 6
Step 2: 13x  0
Step 3: x  0
Solution set: [0, )
[
0

35
Solving a Linear Inequality:
• Example 5: Solve and graph the
solution set
Step 1:
Step 2:
Step 3:
1 3
( 3) 2 ( 8)
4 4
Solution set: [ , )
m   m
m m
m m
[
-13/2
1 3 3
2 6
4 4 4
1 11 3 24
4 4 4 4
1 3 24 11
m m
4 4 4 4
2 13
4 4
13
2
m
m
   
  
  
 
 
13
2


36
Solving applied problems using linear inequalities:
• Example 6: Solve 5 < 3x – 4 < 9 and graph the solution set
5 + 4 < 3x < 9 + 4
9 < 3x < 13
3 < x < (13/3)
Solution set: ( 3, ) 13
2 ) (
3 13
2

37
Note: Expressions for inequalities sometimes appear as indicated in the table:
Word Expression Interpretation
• Example 7: Teresa has been saving dimes and nickels. She has three
times as many nickels as dimes and she has at least 48 coins. What is
the smallest number of nickels she might have?
Let: x equal the number of dimes
Then: 3x is the number of nickels
and: 3x + x  48
4x  48
x  12  36 nickels
a is at least b
a is no less than b
a is at most b
a is no more than b
a  b
a  b
a  b
a  b

38
• Example 8: Michael has scores of 92, 90, and 84 on his first
three tests. What score must he get on the 4th test in order
to maintain an average of at least 90?
Let: x equal the 4th score
92 90 84
: 90
4
266
90
4
266 360
360 266
94
x
Then
x
x
x
x
  



 
 


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