Chapter 1-MATH 101

8/17/2019 Chapter 1-MATH 101

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Chapter 1

Linear Equations and Graphs

Section 1

Linear Equations and Inequalities

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Linear Equations, Standard Form

0=+bax

53

)3(23 −=+− x x

where a is not equal to zero. This is called the standard form

of the linear equation.

or e!a"#le$ the equation

is a linear equation %ecause it can %e con&erted to standard

for" %' clearin of fractions and si"#lif'in.

In eneral$ a first-degree, or linear, equation in one &aria%le

is an' equation that can %e written in the for"

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Equivalent Equations

Two equations are equivalent if one can %e transfor"ed

into the other %' #erfor"in a series of o#erations

which are one of two t'#es

1. The sa"e quantit' is added to or su%tracted

fro" each side of a i&en equation.

2. Each side of a i&en equation is "ulti#lied %'

or di&ided %' the sa"e nonzero quantit'.To solve a linear equation$ we #erfor" these o#erations

on the equation to o%tain si"#ler equi&alent for"s$ until

we o%tain an equation with an o%&ious solution.

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Example of Solving a

Linear Equation

Example: Sol&e 532

2=−

+ x x

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Example of Solving a

Linear Equation

2*

30+302+3

302)2(3

5+32

2+

==+=−+

=−+

⋅=

−+

x

x

x x

x x

x x

Example: Sol&e

Solution: Since the L,- of 2 and 3

is +$ we "ulti#l' %oth sides of the

equation %' + to clear of fractions.

,ancel the + with the 2 to o%tain a

factor of 3$ and cancel the + with

the 3 to o%tain a factor of 2.

-istri%ute the 3.

,o"%ine lie ter"s.

532

2=−

+ x x

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Solving a Formula for a

Particular Variale

Example: Sol&e M =Nt +Nr for N .

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Solving a Formula for a

Particular Variale

Example: Sol&e M=Nt+Nr for N .

( ) M N t r

M N

t r

= +

=+

actor out N

-i&ide %oth sides

%' (t + r )

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Linear !nequalities

If the equalit' s'"%ol / in a linear equation is re#laced %'

an inequalit' s'"%ol ($ $ $ or )$ the resultin e!#ression

is called a first-degree, or linear, inequality. or e!a"#le

is a linear inequalit'.

( )5 1 3 22

x x≤ − +

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Solving Linear !nequalities

4e can #erfor" the sa"e o#erations on inequalities that we

#erfor" on equations$ e!ce#t that the sense of the inequality

reverses if we multiply or divide both sides by a negative

number. or e!a"#le$ if we start with the true state"ent 2 6

and "ulti#l' %oth sides %' 3$ we o%tain

+ 27.

The sense of the inequalit' re"ains the sa"e.

If we "ulti#l' %oth sides %' 83 instead$ we "ust write

+ 27

to ha&e a true state"ent. The sense of the inequalit' re&erses.

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Example for Solving a

Linear !nequalit"

Sol&e the inequalit' 3( x 1) 5( x 9 2) 5

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Example for Solving a

Linear !nequalit"

Sol&e the inequalit' 3( x 1) 5( x 9 2) 5

Solution:

3( x 1) 5( x 9 2) 53 x 3 5 x 9 10 5 -istri%ute the 3 and the 5

3 x 3 5 x 9 5 ,o"%ine lie ter"s.

2 x : Su%tract 5! fro" %oth sides$ and add 3 to %oth sides

x 8* ;otice that the sense of the inequalit'

re&erses when we di&ide %oth sides %' 82.

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!nterval and !nequalit" #otation

Interval Inequality Interval Inequality

<a,b= a ≤ x ≤ b (> ,a= x ≤ a

<a,b) a ≤ x < b (> ,a) x < a

(a,b= a < x ≤ b <b,>) x ≥ b

(a,b) a < x < b (b,>) x > b

If a < b$ the double inequality a < x < b "eans that a < x and

x < b. That is$ x is %etween a and b.

Interval notation is also used to descri%e sets defined %' sinle

or dou%le inequalities$ as shown in the followin ta%le.

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and Line Graphs

(?) 4rite <5$ 2) as a dou%le inequalit' and ra#h .

(@) 4rite ! 2 in inter&al notation and ra#h.

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and Line Graphs

(?) 4rite <5$ 2) as a dou%le inequalit' and ra#h .

(@) 4rite ! 2 in inter&al notation and ra#h.

(?) <5$ 2) is equi&alent to 5 x 2

< ) x85 2

(@) x 2 is equi&alent to <2$ >)

< x82

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Procedure for Solving

$ord Prolems1. Aead the #ro%le" carefull' and introduce a &aria%le to

re#resent an unnown quantit' in the #ro%le".

2. Identif' other quantities in the #ro%le" (nown or

unnown) and e!#ress unnown quantities in ter"s of the&aria%le 'ou introduced in the first ste#.

3. 4rite a &er%al state"ent usin the conditions stated in the

#ro%le" and then write an equi&alent "athe"atical

state"ent (equation or inequalit'.)*. Sol&e the equation or inequalit' and answer the questions

#osed in the #ro%le".

5. ,hec the solutions in the oriinal #ro%le".

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Example% &rea'(Even )nal"sis

? recordin co"#an' #roduces co"#act dis (,-s). Bne8ti"e

fi!ed costs for a #articular ,- are C2*$000D this includes costs

such as recordin$ al%u" desin$ and #ro"otion. aria%le

costs a"ount to C+.20 #er ,- and include the "anufacturin$distri%ution$ and ro'alt' costs for each dis actuall'

"anufactured and sold to a retailer. The ,- is sold to retail

outlets at C:.70 each. Fow "an' ,-s "ust %e "anufactured

and sold for the co"#an' to %rea e&enG

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&rea'(Even )nal"sis

*continued+

Solution

Step 1. Let x / the nu"%er of ,-s "anufactured and sold.

Step . i!ed costs / C2*$000

aria%le costs / C+.20 x

C / cost of #roducin x ,-s

/ fi!ed costs 9 &aria%le costs/ C2*$000 9 C+.20 x

R / re&enue (return) on sales of x ,-s

/ C:.70 x

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&rea'(Even )nal"sis

*continued+

Step !. The co"#an' %reas e&en if R = C $ that is if

C:.70 x / C2*$000 9 C+.20 x

Step ". :.7 x / 2*$000 9 +.2 x Su%tract +.2 x fro" %oth sides

2.5 x / 2*$000 -i&ide %oth sides %' 2.5

x / 6$+00

The co"#an' "ust "ae and sell 6$+00 ,-s

to %rea e&en.

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&rea'(Even )nal"sis

*continued+Step #. ,hec

,osts / C2*$000 9 C+.2 H 6$+00 / C:3$520

Ae&enue / C:.7 H 6$+00 / C:3$520

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Chapter 1-MATH 101