Title: Math 131 - Chapter 2 Linear Equations, Inequalities, and Applications
1Math 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
22-1 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
-
32-1 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.
42-1 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.
52-1 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 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.
62-1 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 .
72-1 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.)
82-1 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.
92-1 Linear Equations in One Variable
- Example 9 Solve
- Step 1
-
-
- Step 2
-
- Step 3
-
- Step 4
-
-
- Step 5
- The solution set is -1.
102-1 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. -
112-1 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)
-
122-2 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. -
132-2 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
-
142-2 Formulae
- Example 2 Solve
- Example 3 Solve
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152-2 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? -
162-3 Applications of Linear Equations
- Translating words to mathematical expressions.
- Note There are usually key words or phrases
that translate into mathematical expressions. -
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172-3 Applications of Linear Equations
- Translating words to mathematical expressions.
- Note There are usually key words or phrases
that translate into mathematical expressions. -
-
182-3 Applications of Linear Equations
- Translating words to mathematical expressions.
- Note There are usually key words or phrases
that translate into mathematical expressions. -
-
192-3 Applications of Linear Equations
- Translating words in to equations.
- Note The symbol for equality, , is often
indicated by the word is. -
-
202-3 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(x2) 7 Equation
- B) 5x 3(x2) Expression
-
212-3 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.
222-3 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 Sosas runs, then x 186
McGwires 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
232-3 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 6245
- Checked
242-3 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 2800
- Checked
252-3 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
- Checked
262-5 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.
272-5 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.
282-5 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.
292-5 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 lt , gt, ?, and ? . - Examples of linear inequalities
- x 5 lt 2 x 3 gt 5 2k 4 ? 10
302-5 Linear Inequalities in One Variable
- Solving linear inequalities using the Addition
Property - For all real numbers A, B, and C, the
inequalities A lt B and A C lt 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 gt 1
- k 5 5 gt 1 5
- k gt 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,?)
312-5 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 gt 0, then the inequalities A lt B and AC
lt BC are equivalent. - 2) if C lt 0, then the inequalities A lt B and AC
gt BC are equivalent. - Note Multiplying or Dividing by a negative
number requires the inequality sign be reversed. - Example 1 Solve -2x lt 10
- x gt -5
- Solution set (-5,?)
- Example 2 Solve 2x lt -10
- x lt -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)
322-5 Linear Inequalities in One Variable
- Solving linear inequalities using the
Multiplication Property - Example 3 Solve 9m lt -81 and graph the solution
set - m gt 9
- Solution set (9, ?)
332-5 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, ?)
342-5 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 , ?)
352-5 Linear Inequalities in One Variable
- Solving applied problems using linear
inequalities - Example 6 Solve 5 lt 3x 4 lt 9 and graph the
solution set - 5 4 lt 3x lt 9 4
- 9 lt 3x lt 13
- 3 lt x lt (13/3)
- Solution set ( 3, )
362-5 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
372-5 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
-