chapter 0 algebra_l4 iry12(1)

7/26/2019 Chapter 0 Algebra_L4 IRY12(1)

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Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1

SOLVING EQUATIONS

SOLVING INEQUALITIES

TRANSPOSING FORMULA

Algebra




Part 1Solving Equations

• Solving Linear, Quadratic, Cubic Equations

• Rational Equations• Equations with Radicals



1.7 - 3


Check your understanding




Equations

An equation is a statement that twoexpressions are equal.

x + 2 =9 11 x = 5 x + 6 x x 2 – 2 x – 1 = 0

To solve an equation means to find all numbersthat make the equation a true statement. Thesenumbers are the solutions, or roots, of theequation. A number that is a solution of an

equation is said to satisfy the equation, and thesolutions of an equation make up its solution set. Equations with the same solution set areequivalent equations.



1.7 - 5


Addition and MultiplicationProperties of Equality

Let a, b, and c represent real numbers.

If a = b, then a + c = b + c .

That is, the same number may be

added to each side of an equationwithout changing the solution set.



1.7 - 6


Addition and Multiplication

Properties of Equality

Let a, b, and c represent real numbers.

If a = b and c ≠ 0, then ac = bc.

That is, each side of an equation may be

multiplied by the same nonzero numberwithout changing the solution set.


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Linear Equations

A linear equation is also called a first-

degree equation since the greatest degree

of the variable is 1.

3 2 0 x

2 5 x

312

4 x

1 8 x

0.5( 3) 2 6 x x

2 3 0.2 0 x x

Linear

equations

Nonlinear

equations



Solution

SOLVING A LINEAR EQUATION

Solve 3(2 4) 7 ( 5). x x 3(2 4) 7 ( 5) x x

6 12 7 5 x x

6 12 2 x x 6 12 2 x x x x 7 12 2 x

1212 2 27 1 x

7 14 x

7

7 7

14,

x 2 x

Distributive property

Combine like terms. Add x to each side.

Combine like terms.

Add 12 to each side.

Combine like terms.

Divide each side

by 7.

Be carefulwith signs.

Example 1



Exercise: Solve the equation.

1)

Answer: (1) x = 6



2 4 1 1 7

3 21

42 12

3

x x x


WITH FRACTIONS

Solve 2 4 1 1 7 .3 2 4 3

x x x

Solution Multiply by 12, the

LCD of the fractions.

4(2 4) 6 3 28 x x x


Distribute the 12 to all

terms within

parentheses.

Example 2

Multiply.

2 4 1 1 73 2 4 3

12 12 12 12 x x x




WITH FRACTIONS

Solve

Solution

14 16 3 28t t Combine like terms.

11 44 x Subtract 3 x ; subtract 16.

4 x Divide each side by 11.

Example 2

8 16 6 3 28 x x x Distributive property

2 4 1 1 7 .3 2 4 3

x x x



Quadratic Equations

• Solving Quadratic Equations by Factorization

• Solving Quadratic Equations Using the

Quadratic Formula• Solving Quadratic Equations by Completing

the Square



Quadratic Equations

Method When the Method is Beneficial

1. Factoring Use when the quadratic equationcan be easily factored.

2. Completing the square Rarely the best method, but

important for future topics.

3. Quadratic formula Use when factoring is not easy, or possible, with integer coefficients.

Methods for Solving Quadratic Equations



Example: Solving Quadratic Equations by

Factorization

2 3 18 0 x x

6 0 x

3 x

6 x 3 x

2 3 18 x x

6 x 0

3 0 x



Solving Quadratic Equations Using the

Quadratic Formula

Using the Quadratic Formula

To solve a quadratic equation in the form

ax2

+bx

+c = 0, where

a0, use the quadraticformula:

2 4

2

b b ac x

a



Example

• Solve using formula

2 4

2

b b ac x

a

22 7 3 0 x x

Solution The equation is in the form ax2 + bx + c = 0,

where a = 2, b = – 7, and c = 3.

27 7 2 34

22 x

7 49 24

4

x

7 254

x

7 5

4 x

7 5

4 x

7 5

4 x

3 x 1

2

x



Exercise

• Solve using formula

Answer:

2 2 11 0 x x

2 48

2 x

2 4

2

b b ac x

a



Solving Quadratic Equations by Completing

the Square

To solve a quadratic equation by completing the square:

1. Write the equation in the form x2 + bx = c.

2. Complete the square by adding (b/2)2 to both sides.

3. Write the completed square in factored form andsimplify the right side,

4. Use the square root principle to eliminate the

square.

5. Isolate the variable.



Example:

Solve by completing the square.

Solution We first write the equation in the form

x2 + bx = c.

2 10 16 0 x x

2 10 16 x x 2 10 25 16 25 x x

25 41 x

5 41 x

5 41 x

Add 16 to both sides to get the form

x 2 + bx = c .

Complete the square by adding 25 to

both sides.

Factor.

Use the square root principle.

Subtract 5 from both sides to isolate x .




Exercise:

Solve by completing the square2 12 9 13 x x

Answer: 6 2 10 x




Solving cubic equations.

Example:




Long division of polynomials.

Example:




Exercise: Solve the equations.

Answer: x = 1, -0.382, -2.618-




Rational Equations

A rational equation is an equation that has arational expression for one or more terms. To

solve a rational equation, multiply each side by

the least common denominator (LCD) of theterms of the equation and then solve the

resulting equation. Because a rational

expression is not defined when its denominator

is 0,

proposed solutions for which any

denominator equals 0 are excluded from the

solution set.




Example 3 SOLVING RATIONAL EQUATIONS THAT

LEAD TO LINEAR EQUATIONS

Solve each equation.

Solution

3 1 2

3 1

x x x

x

The least common denominator is 3( x – 1),

which is equal to 0 if x = 1. Therefore, 1

cannot possibly be a solution of thisequation.






Solution 3 1 23 1

x x x x

3( 1) 3( 1) 3( 13

)1 2

3 1

x x x x

x

x x

Multiply by the LCD, 3( x – 1),

where x ≠ 1.

( 1)(3 1) 3(2 ) 3 ( 1) x x x x x

2 23 4 1 6 3 3 x x x x x 1 10 3 x x

1 7 x

Divide out common

factors.

Multiply.

Subtract 3 x 2; combine like

terms.






Solution 1

7 x

Proposed solution

1The solution set is .

7

The proposed solution meets the requirement

that x ≠ 1 and does not cause any denominator

to equal 0. Substitute to check for correctalgebra.







Solution

22

2 2

x

x x

( 2) ( 2 2

) ( 2)22 2

x

x x x x x

Multiply by the LCD, x – 2, where x ≠ 2.

2 2( 2) x x Divide out common factors.






Solution 2 2 4 x x Distributive property

2 x Solve the linear equation.

2 x Proposed solution.

The proposed solution is 2. However, the variable is

restricted to real numbers except 2. If x = 2, then not only

does it cause a zero denominator, but multiplying by x – 2 in

the first step is multiplying both sides by 0, which is not

valid. Thus, the solution set is .0





LEAD TO QUADRATIC EQUATIONS


Solution

2

3 2 1 2

2 2

x

x x x x

( 2)

3 2 1 2

2

x

x x x x

Factor the last

denominator.

3 2 1 22 ( 2)

( 2) ( 2) ( 2) x

x x x x x x x x

x x

Multiply by x ( x – 2), x ≠ 0, 2.





LEAD TO QUADRATIC EQUATIONS

Solution (3 2) ( 2) 2 x x x

23 2 2 2 x x x Distributive property

Divide out common factors.

23 3 0 x x Standard form

3 ( 1) 0 x x Factor.

3 0 x

or 1 0 x Zero-factor property

0 x or 1 x Proposed solutions

Because of the restriction x ≠ 0, the only valid solution is – 1.

The solution set is { – 1}.

Set each

factor equal

to 0.





2) a) b)

Answer: (2) a) No solution, ø b) m = -4 (m = -1 cannot accept as

answer)



1.7 - 34Copyright © 2013, 2009, 2005 Pearson Education, Inc. 34

Solving an Equation Involving

RadicalsStep 1 Isolate the radical on one side of the equation.

Step 2 Raise each side of the equation to a power that

is the same as the index of the radical so that

the radical is eliminated.

If the equation st i l l co ntains a radical, repeat

Steps 1 and 2.

Step 3 Solve the resulting equation.

Step 4 Check each proposed solution in the original

equation.




Example 6 SOLVING AN EQUATION CONTAINING

A RADICAL (SQUARE ROOT)

Solve

Solution

15 2 0. x x

15 2 x x Isolate the radical.

2

2 15 2 x x Square each side.

2 15 2 x x Solve the quadratic equation.

2 2 15 0 x x




Example 6 SOLVING AN EQUATION CONTAINING

A RADICAL (SQUARE ROOT)

Solve

Solution

15 2 0. x x

( 5)( 3) 0 x x Factor.

5 0 x or 3 0 x Zero-factor property

5 x or 3 x Proposed solutions

Only 3 is a solution, so the solution set is {3}.




Example 7 SOLVING AN EQUATION

CONTAINING TWO RADICALS

Solve

Solution When an equation contains tworadicals, begin by isolating one of the radicals

on one side of the equation.

2 3 1 1. x x

2 3 1 1 x x

2 3 1 1 x x Square each side.

Isolate 2 3. x

2 2

12 3 1 x x






Solve

Solution

2 3 1 1. x x

2 1)3 1 (2 1 x x x

Don’t forget this

term when squaring.

Be careful!

1 2 1 x x Isolate the

remaining

radical.

22

1 2 1 x x Square again.






Solution2 2 1 4( 1) x x x

2 2 1 4 4 x x x 2 2 3 0 x x Solve the quadratic equation.

( 3)( 1) 0 x x

3 0 x

1 0 x

or3 x 1 x or

Proposed solutions

Apply the exponents.


Factor

Zero-factor property

Check each proposed solution in the original equation. Both 3 and – 1 are

solutions of the original equation, so { – 1,3} is the solution set.




Exercises: Solve the equations.

3)

Answer: (3) x = 5




Part 2 Inequalities• Linear Inequalities

• Three-Part Inequalities• Quadratic Inequalities

• Rational Inequalities

• Absolute Value Equations and Inequalities




Properties of Inequality

Let a, b and c represent real numbers.

1. If a < b , then a + c < b + c .

2. If a < b and if c > 0, then ac < bc .

3. If a < b and if c < 0, then ac > bc .

Replacing < with >, ≤ , or ≥ results in similar properties.




Linear Inequality in OneVariable

A linear inequality in one variable is

an inequality that can be written in theform

where a and b are real numbers, with

a ≠ 0. (Any of the symbols ≥, <, and ≤may also be used.)

0,ax b




Example 8 SOLVING A LINEAR INEQUALITY

Solve

Solution

3 5 7. x

3 5 7 x

53 7 55 x Subtract 5.

3 12 x

3 12

3 3

x

Divide by – 3. Reverse

the direction of theinequality symbol

when multiplying or

dividing by a negative

number.

Don’t forget toreverse the

symbol here.

4 x

Combine like terms.




Type ofInterval Set IntervalNotation Graph

Open

interval

{ x x > a}

{ x a < x < b}

{ x x < b}

(a, )

(a, b)

( – , b)

Other

intervals

{ x x ≥ a}

{ x a < x ≤ b}

{ x a ≤ x < b}

{ x x≤ b}

[a, )

(a, b]

[a, b)

( – , b]

(

( (

(

a

a b

b

[a( ]a

b[ )a b

]b




Type of

Interval

Set Interval

Notation

Graph

Closed

interval { x a ≤ x ≤ b} [a, b]

Disjoint

interval { x x < a or x > b} ( – , a)

(b, )

All realnumbers { x x is a real number} ( – , )

[ ]

b

a b

a

( (




Example 9 SOLVING A LINEAR INEQUALITY

Solve 4 – 3 x ≤ 7 + 2 x. Give the solutionset in interval notation.

Solution 4 3 7 2 x x

4 44 3 7 2 x x Subtract 4.

3 3 2 x x

3 32 22 x x x x Subtract 2 x .

5 3 x

Combine like terms.

Combine like terms.




Example 9

Solve 4 – 3 x ≤ 7 + 2 x. Give the solutionset in interval notation.

Solution

Divide by –5. Reverse thedirection of the inequality

symbol. 55

53

x

x

3

5 x

3In interval notation the solution set is , .

5

03

5

[

SOLVING A LINEAR INEQUALITY

S f




Exercises: Solve the following

1)

SOLVING A THREE PART




Example 10 SOLVING A THREE-PART

INEQUALITY

Solution Solve –

2 < 5 + 3 x < 20.2 5 3 20 x

52 5 55 3 20 x Subtract 5 from

each part.

7 3 15 x 7 1

3 3

3 5

3

x

Divide each

part by 3.

75

3 x

7The solution set is the interval ,5 .

3




Quadratic Inequalities

A quadratic inequality is an inequality

that can be written in the form2 0ax bx c

for real numbers a, b, and c , with a ≠ 0.(The symbol < can be replaced with >,≤, or ≥.)




Solving a Quadratic Inequality

Step 1 Solve the corresponding quadratic

equation.Step 2 Identify the intervals determined by

the solutions of the equation.

Step 3 Use a test value from each interval

to determine which intervals formthe solution set.

SOLVING A QUADRATIC




Example 11 SOLVING A QUADRATIC

INEQULITY

Solve

Solution

2

12 0. x x

Step 1 Find the values of x that satisfy x 2 – x – 12 = 0.

2 12 0 x x Corresponding quadratic

equation

( 3)( 4) 0 x x Factor.

3 0 x 4 0 x or Zero-factor property

3 x 4 x or Solve each equation.

SOLVING A QUADRATIC





INEQULITY

Step 2 The two numbers – 3 and 4 dividethe number line into three intervals. The

expression x 2 – x – 12 will take on a value

that is either less than zero or greater than zero on each of these intervals.

– 3 0 4

Interval A Interval B Interval C( –, – 3) ( – 3, 4) (4, )

-+ +

E l 11 SOLVING A QUADRATIC





INEQULITY

IntervalTest

Value

Is x 2 – x – 12 < 0

True or False?

A: ( – , – 3) – 4 ( – 4)2 – ( – 4) – 12 < 0 ?

8 < 0 False

B: ( – 3, 4) 0 02 – 0 – 12 < 0 ?

– 12 < 0 True

C: (4, ) 5 52 – 5 – 12 < 0 ?

8 < 0 False

Step 3 Choose a test value from each interval.

Since the values in Interval B make the inequality

true, the solution set is ( – 3, 4).






INEQUALITY

Solve

Solution

2

2 5 12 0. x x

Step 1 Find the values of x that satisfy

22 5 12 0 x x Corresponding

quadratic equation

(2 3)( 4) 0 x x Factor.

2 3 0 x or 4 0 x Zero-factor property

22 5 12 0. x x






INEQUALITY

SolveSolution

2

2 5 12 0. x x

Step 12 3 0 x or 4 0 x

32

x or 4 x

Step 2 The values form the intervals on the

number line.

– 40

3/2

Interval A Interval B Interval C( –, – 4) ( – 4, 3/2) (3/2, )

+ +-






INEQUALITY

Step3 Choose a test value from each interval.

IntervalTest

Value

Is 2 x 2 + 5 x – 12 ≥ 0True or False?

A: ( – , – 4) – 5 2( – 5)2 +5( – 5) – 12 ≥ 0 ?

13 ≥ 0 True

B: ( – 4, 3/2) 0 2(0)2 +5(0) – 12 ≥ 0 ?

– 12 ≥ 0 False

C: (3/2, ) 2 2(2)2

+ 5(2) – 12 ≥ 0 ?6 ≥ 0 True

The values in Intervals A and C make the inequality true, so

the solution set is the union of the intervals 3

, 4 , .2

E l 13 FINDING PROJECTILE HEIGHT




Example 13 FINDING PROJECTILE HEIGHT

If a projectile is launched from ground levelwith an initial velocity of 96 ft per sec, its

height s in feet t seconds after launching is

given by the following equation.216 96s t t

When will the projectile be greater than 80 ft

above ground level?






216 96 80t t Set s greater than

80.

Solution

216 96 80 0t t Subtract 80.

2 6 5 0t t Divide by – 16.

Reverse the

direction of the

inequalitysymbol.

Now solve thecorresponding equation.






2 6 5 0t t

Solution

( 1)( 5) 0t t Factor.

1t

Zero-factor property

or 5t Solve each equation.

or1 0t 5 0t

E ample 13 FINDING PROJECTILE HEIGHT





Solution

Interval C

10 5

Interval A Interval B

( – , 1) (1, 5) (5, )

Use the procedure of Examples 5 and 6 to

determine that values in Interval B, (1, 5) ,

satisfy the inequality. The projectile is

greater than 80 ft above ground level

between 1 and 5 sec after it is launched.




Solving a Rational Inequality

Step 1 Rewrite the inequality, if necessary, so that

0 is on one side and there is a single fraction on the

other side.

Step 2 Determine the values that will cause either

the numerator or the denominator of the rational

expression to equal 0. These values determine the

intervals of the number line to consider.




Solving a Rational Inequality

Step 3 Use a test value from each interval to

determine which intervals form the solution set.

A value causing the denominator to equal zero will

never be included in the solution set. If the

inequality is strict, any value causing the numerator

to equal zero will be excluded. If the inequality isnonstrict, any such value will be included.




Caution Solving a rational inequality such as5

14 x

by multiplying each side by x + 4 to obtain5 ≥ x + 4 requires considering two cases, since

the sign of x + 4 depends on the value of x . If

x + 4 is negative, then the inequality symbol

must be reversed. The procedure used in thenext two examples eliminates the need for

considering separate cases.

Example 14 SOLVING A RATIONAL




Example 14

Solve

Solution

51.4 x

Step 1

5

1 04 x Subtract 1 so that 0 ison one side.

450

4 4 x x

x

Use x + 4 as the

common denominator.

( )5 40

4

x

x

Write as a single fraction.

Note the

careful use

of

parentheses.

SOLVING A RATIONAL

INEQUALITY





Example 14

Solution

Step 1 14

0 x

x

Combine terms in thenumerator, being careful

with signs.

Step 2 The quotient possibly changes sign only where x -

values make the numerator or denominator 0. This occurs at1 0 x or 4 0 x

1 x or 4 x

SOLVING A RATIONAL

INEQUALITY

– 40

1

1- x

X + 4

+- -

+- ++ -+





Example 14

The values in the interval ( –4, 1) satisfy the

original inequality. The value 1 makes the

nonstrict inequality true, so it must be included

in the solution set. Since –4 makes thedenominator 0, it must be excluded. The

solution set is ( –4, 1].

SOLVING A RATIONAL

INEQUALITY




Caution Be careful w ith the

endpo ints of the intervals when so lv ing

rat ional inequali t ies .






INEQUALITY

Solve

Solution

2 15.3 4

x

x

2 15 0

3 4

x

x

Subtract 5.

2 1 5(3 4)0

3 4 3 4

x x

x x

Common denominator

is 3 x + 4.

2 1 5(3 4)0

3 4

x x

x

Write as a single

fraction.





Example 15

Solve

Solution

2 15.3 4

x

x

013 21 x

Set the numerator and denominator equal to

0 and solve the resulting equations to findthe values of x where sign changes may

occur.

or 3 4 0 x 21

13 x or

4

3 x

SOLVING A RATIONAL

INEQUALITY





Example 15

Solution

-13x - 21

3x + 4

21

13

4

3

SOLVING A RATIONAL

INEQUALITY

Because of the < symbol, neither endpoint satisfies the inequality, so the

solution set is 21 4, , .

13 3

+-

-

--

+

-+

-

13 213

04 x

x






2)

Answer: (2) Solution set:





Example 16

Solve

Solution

SOLVING A RATIONAL

INEQUALITY





Example 16

Solution: Find all the possible values of x.

SOLVING A RATIONAL

INEQUALITY

x + 3x - 4

1 - x

-

--

-

+

++

-

-

- ++

+

+

+ -

Answer:






3)

Answer: (3) Solution set:




Hint You can wri te your answer in

inequal ity notat ion or in terval notat ion .

Example:





Lina Kristin

Who is correct? Explain your reasoning.


To solve inequations (Rational Functions)




To solve inequations (Rational Functions)

5

05

9

05

)5(4

015

4

15

4

CORRECT

x

x

x

x x

x

x

x

x

?????

54

15

4

WRONG

x x

x

x

Absolute Value




Absolute Value

For each equation or inequality in Cases 1-3 in the table,

assume that k > 0.

In Cases 2 and 3, the strict inequality may be replaced by its

nonstrict form. Additionally, if an absolute value equation takesthe form │a │= │b │, then a and b must be equal in value or

opposite in value.

Thus, the equivalent form of │a │= │b │ is a = b or a = –b.

Absolute Value Inequality




q y

Example 16 SOLVING ABSOLUTE VALUE




Example 16EQUATIONS

Solve each equation.(a) 5 3 12 x

Solution

5 3 12 x

5 3 12 x or 5 3 12 x Case 1

3 7 x or 3 17 x Subtract 5.

7

3 x or

17

3 x Divide by

– 3.The solution set is 7 17

, .3 3





Example 17EQUATIONS


(b) 4 3 6 x x

Solution

4 3 6 x x

or 4 3 ( 6) x x

3 9 x or 4 3 6 x x

3 x or 5 3 x

35

x

4 3 6 x x

3The solution set is ,3 .

5





a p e 8INEQUALITIES

Solution

(a)

Solve each inequality.

2 1 7 x

This inequality fits Case 2.

2 1 7 x 7 2 1 7 x Case 2

8 2 6 x Subtract 1 from each part.

4 3 x Divide each part by 2.

The final inequality gives the solution set ( – 4, 3).





pINEQUALITIES

Solution This inequality fits Case 3.

(b)Solve each inequality.2 1 7 x

2 1 7 x

2 1 7 x Case 3

Subtract 1 from

each side.

Divide each part by 2.

or 2 1 7 x

2 8 x

or 2 6 x 4 x or 3 x

The solution set is ( , 4) (3, ).

Example 19 SOLVING AN ABSOLUTE VALUE

INEQUALITY




pINEQUALITY

SolveSolution

2 7 1 4. x

2 7 1 4 x

2 7 5 x

Add 1 to each side.2 7 5 x or 2 7 5 x Case 3

7 7 x or 7 3 x

1 x or3

7 x

Subtract 2.

Divide by – 7. Reverse

the direction of each

inequality.

3

The solution set is , 1, .7

Example 20 SOLVING SPECIAL CASES




p

Solve each equation or inequality.

Solution Since the absolute value of anumber is always nonnegative, the inequality

is always true. The solution set includes allreal numbers, written ( –∞,∞).

2 5 4 x (a)

Solution There is no number whoseabsolute value is less than – 3 (or less thanany negative number). The solution set is .

(b) 4 7 3 x

0

Example 20 SOLVING SPECIAL CASES




p

Solve each equation or inequality.

Solution The absolute value of a number

will be 0 only if that number is 0. Therefore,

5 15 0 x (c)

5 15 0 x is equivalent to 5 15 0 x

which has solution set { – 3}. Check by

substituting into the original equation.





4)

Answer:(4) Solution set:




Part 3 Transposing formula• A formula may have several variables,

represented by symbols and we can solve onevariable if we know the values of the other

variables.

•Rearranging a formula to change the subject is

called transposition.

Example 21 Solve the value of unknown




p

The volume of a circular cylinder is given

by the formula V = r 2h. Given that r =

2.5 cm and h =15.75 cm, find V correct

to 2 decimal places (Note that, V is thesubject of the formula)

Ans: 309.25cm3

Example 22 Transposing formula




Solve for the specified variable. Use when

taking square roots if needed.

Solution

(a)2

, for4

d A d

2

4 A

d Goal: Isolate d ,

the specified

variable.

2

4 A d Multiply each side by 4.

24 Ad

Divide each side by .





Solution Square root property

4d A

4 A

d

Multiply by .

4d

A

2

d A

Multiply numerators.

Multiply denominators.

Simplify the radical.





The formula gives the time

of swing of a simple pendulum of

length L. Transpose the formula to

make L the subject.

g Lt 2





divided by 2 at bothsides

square both sides

multiply by g at both

sides

t g L 2

2

t

g

L

2

2

t

g

L

2

2

t g L

__

2 2 2

__2

g x x g

Solution

What if we transpose the formula to

k th bj t?




t g

L

)2(

22

2 t L

L g

2

2

g

make g the subject?

Square both sides22

Multiply both sides by g

Exercises:




Exercises:

Rearrange each of the following formulae tomake the quantity shown the subject.

Answers:




Answers:

REVIEW EXERCISE



1) Solve the rational inequality .01010

x x

Answer:

1) the solution is the set ( –, –10] (10, ).

2) the solution is the set (-4 -11/3] [0 )

2) Solve the rational inequality . p

p

p3

4

chapter 0 algebra_l4 iry12(1)

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