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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 2x
4
2-1 Linear Equations in One Variable
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
2-1 Linear Equations in One Variable
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.
The solution set is {3}.
6
2-1 Linear Equations in One Variable
Multiplication Properties of Equality
• For all real numbers A,B, and for
C 0, the equations A = B and AC = BCare 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 ( ) 155 5
x = 3
x
7
2-1 Linear Equations in One Variable
Example 8: Solve: -7 + 3x – 9x = 12x -5
-7 + 7 –6x = 12x –5 + 7
-6x –12x = 12x –12x +2
-18x = 2
The solution set is { }.
1 1( ) (-18) ( ) 2
18 18
1x = -
9
x
1-
9
8
2-1 Linear Equations in One Variable
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.)
9
2-1 Linear Equations in One Variable
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:
Step 5: 6 – (4 + 4) = 8(4) – 2(3(4) + 5)
6 – 8 = 32 - 34
-2 = -2
The solution set is {4}.
1 1( ) (-3) ( ) (-12)
3 3
x = 4
x
10
2-1 Linear Equations in One Variable
Example 9: Solve:
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
The solution set is {-1}.
1 3 1
2 4 2
1 3 1 4( ) 4( )
2 4 2
1 3 14( ) 4( ) 4( )
2 4 2
2( 1) ( 3) 2
2 2 3 2
3 5 2
3 5 ( 5) 2 ( 5)
3 3
x x
x x
x x
x x
x x
x
x
x
1 1 ( ) 3 ( )( 3)
3 3
1
1 1 1 3 1 2 1 1 1:
2 4 2 4 2 2 2
x
x
Check
11
2-1 Linear Equations in One Variable
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
12
2-1 Linear Equations in One Variable
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)
13
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)
Additional formulae are included on the inside covers of your text.
amount100% 100%
base
apercent percent
b
14
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
15
2-2 Formulae
• Example 2: Solve
• Example 3: Solve
1( 3) for
2y x x
1 3( )
2 2
1 3( )
2 2
1 32( ( )) 2( )
2 2
2 3
y x
x y
x y
x y
2 2 2 for A HW LW LH W
2 2 2
(2 2 ) 2
( 2 )
(2 2 )
HW LW A LH
W H L A LH
A LHW
H L
16
2-2 Formulae
• Example 4: Given a distance of 500 miles and a rate (speed) of 25 mph, find the time for the trip.
• Example 5: A 20 oz mixture of gasoline and oil contains 1 oz of oil. What percent of the mixture is oil?
solve for d rt t
rt d
dt
r
500 20
25
d milest hours
milesr
hour
100%amount
percentbase
1
100% .05 100% 5%20
17
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
18
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 the numbers
Verbal Expression Mathematical Expression
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
19
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 the numbers
Verbal Expression Mathematical Expression
Multiplication
2 times a number 2 x or 2x
A number multiplied by 6 x 6 or 6x
Twice a number 2 x or 2x
The product of two numbers x y or xy
2 of a number
3
2 2 x or x
3 3
20
2-3 Applications of Linear Equations
Translating words in to equations.Note: The symbol for equality, = , is often indicated by the word “is”.
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
Translating from Words to Mathematical Equations
284
x
x
104
xx
21
2-3 Applications of Linear Equations
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
22
2-3 Applications of Linear Equations
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
2-3 Applications of Linear Equations
Solving Applied Problems.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 = 858Checked:
24
2-3 Applications of Linear Equations
Solving Applied Problems.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 = 6245Checked:
25
2-3 Applications of Linear Equations
Solving Applied Problems.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 = 2800Checked:
26
2-3 Applications of Linear Equations
Solving Applied Problems.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:
• Step 5: 5 L of water
• Step 6: Check:
amount100% 100%
base
apercent percent
b
10 L of antifreeze100% 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 antifreeze100% 40%
10 L of antifreeze + (5+10) of water
10100% 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.
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} (-, )
Interval Notation
Open Interval
(
a
)
b
(
a
)
b
)
b
28
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.
Type of Interval Set Interval Notation Graph
{x a x} [a, )
{x a x b} (a, b]
{x a x b} [a, b)
{x x b} (-, b]
Interval Notation
Half-open
Interval
[
a
]
b
]
b
(
a
)
b
[
a
29
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.
Type of Interval Set Interval Notation Graph
Closed Interval {x a x b} [a, b]
Interval Notation
]
b
[
a
30
2-5 Linear Inequalities in One Variable
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
2-5 Linear Inequalities in One Variable
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
2-5 Linear Inequalities in One Variable
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
2-5 Linear Inequalities in One Variable
Solving linear inequalities using the Multiplication Property:
• Example 3: Solve –9m < -81 and graph the solution set
m > 9
Solution set: (9, ) (
9
34
2-5 Linear Inequalities in One Variable
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
2-5 Linear Inequalities in One Variable
Solving a Linear Inequality:
• Example 5: Solve and graph the solution set
Step 1:
Step 2:
Step 3:
Solution set: [ , )[
-13/2
1 3( 3) 2 ( 8)
4 4m m
1 3 32 6
4 4 4
1 11 3 24
4 4 4 4
1 3 24 11
4 4 4 4
2 13
4 4
13
2
m m
m m
m m
m
m
13
2
36
2-5 Linear Inequalities in One Variable
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
2-5 Linear Inequalities in One Variable
Solving applied problems using linear inequalities:
Note: Expressions for inequalities sometimes appear as indicated in the table:
• 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
Word Expression Interpretation
a is at least b
a is no less than b
a is at most b
a is no more than b
a ba b
a ba b
38
2-5 Linear Inequalities in One Variable
Solving applied problems using linear inequalities:
• 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
26690
4
266 360
360 266
94
xThen
x
x
x
x