T2 Equations and Inequalities v1

7/29/2019 T2 Equations and Inequalities v1

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Topic 2

Equations and

Inequalities

Sections 2.12.8


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2.1 Linear Equations

2.2 Applications and Modeling with

Linear Equations

2.3 Quadratic Equations

Equations and Inequalities2


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2.4 Applications and Modeling withQuadratic Equations

2.5 Other Types of Equations andApplications

2.6 Inequalities

Equations and Inequalities2


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Linear Equations2.1 Basic Terminology of Equations Solving Linear Equations Identities, Conditional Equations, and Contradictions Solving

for a Specified Variable (Literal Equations)


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An equation is a statement that two expressionsare equal.

x + 2 = 9, 11x = 5x + 6x, x2 2x 1 = 0

To solve an equation means to find all numbersthat make the equation a true statement thesenumbers are called solutions or roots.

A linear equation is also called a first degreeequation since the greatest degree of thevariable is one.


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An equation satisfied by every number thatis a meaningful replacement for thevariable is called an identity for example,

3(x+1) = 3x + 3. An equation that is satisfied by some

numbers but not others, such as 2x = 4, iscalled a conditional equation.

An equation that has no solution, such asx = x + 1, is called a contradiction.



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2-7

Solve .

2.1Example 1 Solving a Linear Equation

Solution set: {6}

Distributive property

Combine terms.

Add 4 to both sides.Add 12xto both sides.Combine terms.Divide both sides by 4.


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2-8

Solve .

2.1Example 2 Clearing Fractions Before Solving a LinearEquation

Solution set: {10}

Multiply by 10, the LCD of

all the fractions.


Combine terms.

Add4sand6 to bothsides. Combine terms.

Divide both sides by6.


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Decide whether the equation is an identity, a

conditional equation, or a contradiction. Give thesolution set.

2.1Example 3(a) Identifying Types of Equations

This is a conditionalequation.Solution set: {11}

Add4xand 9 to bothsides. Combine terms.

Divide both sides by 2.


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2.1Example 3(b) Identifying Types of Equations(page 86)

This is a contradiction.Solution set:


Subtract 14xfrom bothsides.


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2.1Example 3(c) Identifying Types of Equations(page 86)

This is an identity.Solution set: {all real numbers}


Combine terms.

Add xand3 to both

sides.


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Solve for the specified variable.

2.1Example 4 Solving for a Specified Variable (page 87)


(a) d = rt, fort

(b) , fork

Divide both sides by r.

Factor out k.

Divide both sides by


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2.1Example 4 Solving for a Specified Variable (cont.)


(c) , fory


Subtract 8yand 8 from both

sides.Divide both sides by 3.


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2.2Solving Applied Problems Geometry Problems Motion

Problems Mixture Problems Modeling with Linear Equations

Applications and Modeling with LinearEquations


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2.2Example 3 Solving a Mixture Problem (page 93)

How many liters of a 25% anti-freeze solutionshould be added to 5 L of a 10% solution to obtain a15% solution?

Let x= the amount of 25% solution

The number of gallons of pure antifreeze in the25% solution plus the number of gallons of pure

antifreeze in the 10% solution must equal thenumber of gallons of pure antifreeze in the 15%solution.


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2.2Example 3 Solving a Mixture Problem (cont.)

Create a table to show the relationships in the problem.

Write an equation:


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2.2Example 3 Solving a Mixture Problem (cont.)

2.5 liters of the 25% solution should be added.

Distributive property.

Subtract .15xand .5.

Divide by .1.


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2.2Example 4 Solving an Investment Problem (page 94)

Last year, Owen earned a total of $1456 in interestfrom two investments. He invested a total of$28,000, part at 4.8% and the rest at 5.5%. Howmuch did he invest at each rate?

Let x= amount invested at 4.8%.Then 28,000 x= amount invested at 5.5%.


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2.2Example 4 Solving an Investment Problem (cont.)

Create a table to show the relationships in the problem.

The amount of interest from the 4.8% account

plus the amount of interest from the 5.5%account must equal the total amount ofinterest.


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2.2Example 4 Solving an Investment Problem (cont.)

Owen invested $12,000 at 4.8% and$28,000 $12,000 = $16,000 at 5.5%.


Combine terms.

Subtract 1540.

Divide by.007.


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Quadratic Equations12.3Solving a Quadratic Equation Completing the Square

The Quadratic Formula Solving for a Specified Variable

The Discriminant


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2.1 Quadratic Equations

An equation that can be written in the form

ax2 + bx + c = 0

Where a, b, c are real numbers with a 0, is aquadratic equation.

A quadratic equation is a second-degreeequation that is, an equation with squaredvariable term and no terms of greater degree.


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2.1 Quadratic Formula

The solutions of the quadratic equation ax2 + bx +c = 0, where a 0, are


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2.4Example 1 Using the Zero-Factor Property

Solve .

Factor.

Set each factorequal to 0 and thensolve forx.

or

or

or


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2.4Example 2 Using the Square Root Property

Solve each quadratic equation.

Generalized square root property

(a)

(b)


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2.4Example 5 Using the Quadratic Formula (Real Solutions)

Solve .

a= 1,b= 6,c=3

Write the equation in standard form.

Quadratic formula

2 4 Example 5 Using the Quadratic Formula (Real Solutions)


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2.4Example 5 Using the Quadratic Formula (Real Solutions)(cont.)

Solution set:

2 4 E l 6 U i th Q d ti F l


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2.4Example 6 Using the Quadratic Formula(Nonreal Complex Solutions)

Solve .

a= 4,b=3,c= 5

Write the equation in standard form.

Quadratic formula

2 4 Example 8(a) Solving for a Quadratic Variable in a


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2.4Example 8(a)Solving for a Quadratic Variable in aFormula

Solve forr. Use when taking squareroots.

Goal: Isolate r.

Multiply by 3.

Divide by h.

Square root property

Rationalize thedenominator.

Simplify.

2 4 Example 8(b) Solving for a Quadratic Variable in a


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2.4Example 8(b)Solving for a Quadratic Variable in aFormula

Solve fory. Use whentaking square roots.

Write in standard form.

Use the quadraticformula with a= 2m,b=n, c=3p.

Simplify.


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2.4Example 9(a) Using the Discriminant

Determine the number of distinct solutions, and tellwhether they are rational, irrational, ornonrealcomplexnumbers.

a= 4, b=12, c= 9

There is one distinct rational solution.


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2.4Example 9(b) Using the Discriminant


a= 3, b= 1, c= 5

There are two distinct nonreal complex solutions.



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2.4Example 9(c) Using the Discriminant


a= 2, b=6, c=7

There are two distinct irrational solutions.



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Applications and Modeling withQuadratic Equations2.4Sets of Numbers and the Number Line Exponents Order of

Operations Properties of Real Numbers Order on the Number

Line Absolute Value

2 5 Example 4(a) Analyzing Sport Utility Vehicle (SUV) Sales


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2.5Example 4(a)Analyzing Sport Utility Vehicle (SUV) Sales

Based on figures from 19902001, the equation

models sales of SUVs from 1990 to 2001,where S

represents sales in millions, and x= 0 represents1990, x= 1 represents 1991, etc.

Use the model to determine sales in 2000 and 2001.

Compare the results to the actual figures of 3.6million and 3.7 million.

2 5 Example 4(a) Analyzing Sport Utility Vehicle (SUV) Sales


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2.5Example 4(a)Analyzing Sport Utility Vehicle (SUV) Sales(cont.)

For 2000, x= 10.

For 2001, x= 11.

million

million

For 2000, the prediction is equal to the actual figureof 3.6 million.

For 2001, the prediction is greater than the actualfigure of 3.7 million.

2 5 Example 4(b) Analyzing Sport Utility Vehicle


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2.5Example 4(b)Analyzing Sport Utility Vehicle(SUV) Sales

According to the model, in what year did sales reach3 million? (Round down to the nearest year.)

Let S= 3, then solve forx.

Quadratic formula

2 5 Example 4(b) Analyzing Sport Utility Vehicle


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2.5Example 4(b)Analyzing Sport Utility Vehicle(SUV) Sales (cont.)

Reject the negative solution, and round 8.5 up to 9.

The year 1999 corresponds to x= 9.

SUV sales reached 3 million in 1999.


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Other Types of Equations andApplications2.5Rational Equations Work Rate Problems Equations with

Radicals Equations Quadratic in Form

2 6 Example 1(a) Solving Rational Equations that Lead to


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2.6Example 1(a) Solving Rational Equations that Lead toLinear Equations

Solve .

The least common denominator is 2(x+ 1), which equals0 when x= 1. Therefore, 1 cannot be a solution of theequation.



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2.6Example 1(a) Solving Rational Equations that Lead toLinear Equations(cont.)

Multiply by the LCD,2(x+ 1).

Simplify.

Multiply.

Combine terms.

The restriction, x1, does not affect this result.



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2.6Example 1(a) Solving Rational Equations that Lead toLinear Equations(cont.)

Now check.

Solution set:

2 6 Example 1(b) Solving Rational Equations that Lead to


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2.6Example 1(b) Solving Rational Equations that Lead toLinear Equations

Solve .

The least common denominator is x 5, whichequals 0 when x= 5. Therefore, 5 cannot be asolution of the equation.

2.6 Example 1(b) Solving Rational Equations that Lead to


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2.6Example 1(b) Solving Rational Equations that Lead toLinear Equations (cont.)

The only possible solution is 5. However, thevariable is restricted to real numbers except 5.

Multiply by the LCD,x 5.

Simplify.

Solution set:

2.6 Example 2(a) Solving Rational Equations that Lead to


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2.6Example 2(a) Solving Rational Equations that Lead toQuadratic Equations

Solve .

The least common denominator is ,which equals 0 when x= 0 orx= 3. Therefore, 0 and 3cannot be solutions of the equation.



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2.6Example 2(a) Solving Rational Equations that Lead toQuadratic Equations (cont.)

Multiply by theLCD, x(x 3).

Distributiveproperty

Standard form

Factor.

The restrictions x 0 and x 3 do not affect the result.



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2.6Example 2(a) Solving Rational Equations that Lead toQuadratic Equations (cont.)

Now check.

Solution set: {2}

2.6 Example 2(b) Solving Rational Equations that Lead to


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2.6Example 2(b) Solving Rational Equations that Lead toQuadratic Equations (page 134)

Solve .

The least common denominator is ,which equals 0 when x= 5. Therefore, 5 cannot besolutions of the equation.

2.6Example 2(b) Solving Rational Equations that Lead to


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p ( ) g qQuadratic Equations (cont.)

The possible solutions are 5. However, the variableis restricted to real numbers except 5.

Solution set:


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Inequalities2.6 Linear Inequalities Three-Part Inequalities Rational Inequalities

2 6 Inequalities


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2.6 Inequalities

An inequalities says that one expression is greaterthan (>), greater than or equal to (>), less than(


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Subtract 7.

Divide by2. Reverse thedirection of the inequality

symbol when multiplying ordividing by a negative number.

Solution set: {x|x> 6}

2.7Example 1 Solving a Linear Inequality (page 146)

2 7 Example 2 Solving a Linear Inequality (page 147)


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2.7Example 2 Solving a Linear Inequality (page 147)

Subtract 8.

Add 4x.

Divide by 6.

Solution set:Write the solution setin interval notationand graph it.

2 7 Example 4 Solving the Break Even Point (page 148)


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2.7Example 4 Solving the Break-Even Point (page 148)

If the revenue and cost of a certain product are given

by R= 45xand C= 30x+ 5250, where xis the numberof units produced and sold, at what production leveldoes Rat least equal C?

Set R Cand solve forx.

Subtract 30x.

Divide by 15.

The break-even point is at x= 350.

This product will at least break even only if the numberof units produced and sold is in the interval .

2 7 Example 5 Solving a Quadratic Inequality (page 149)


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2.7Example 5 Solving a Quadratic Inequality (page 149)

Solve .

Step 1: Find the values ofxthat satisfy .

oror

Step 2: The two numbers divide a number line into threeregions.

Use closed dotssince theinequality symbolincludes equality.

2 7 Example 5 Solving a Quadratic Inequality (cont )


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2.7Example 5 Solving a Quadratic Inequality (cont.)

Step 3: Choose a test value to see if it satisfies the inequality.

The values in interval B make the inequality true.

Solution set: [3, 5]

2 7 Example 6 Solving a Quadratic Inequality


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2.7Example 6 Solving a Quadratic Inequality

Solve .

or


Use open dotssince theinequality symboldoes not include

equality.

Step 1: Find the values ofxthat satisfy .

or

2 7 Example 6 Solving a Quadratic Inequality (cont )


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2.7Example 6 Solving a Quadratic Inequality (cont.)


The values in intervals A and C make the inequality true.

Solution set:

2.7Example 7 Solving a Problem Involving the Height of a


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p g g gProjectile

If an object is launched from ground level with aninitial velocity of 144 ft per sec, its height in feet tseconds after launching is sfeet, where

When will the object be greater than 128 ft aboveground level?

Set sgreater than 128.

Subtract 128.

Divide by16.

2.7Example 7 Solving a Problem Involving the Height of a


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g g gProjectile (cont.)

Step 1: Solve the corresponding equation.

Factor.

Zero-factor propertyor


Use open dots

since theinequality symboldoes not includeequality.

2.7Example 7 Solving a Problem Involving the Height of aP j til


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Projectile (cont.)


The values in interval B make the inequality true, so thesolution set is (1, 8).

The object will be greater than 128 ft above ground levelbetween 1 and 8 seconds after it is launched.

2 7 Example 8 Solving a Rational Inequality


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2.7Example 8 Solving a Rational Inequality

Solve .

Step 1: Subtract 4.

x 3 is the commondenominator.

Write as a singlefraction.

Combine terms in thenumerator.

2 7 Example 8 Solving a Rational Inequality (cont )


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2.7Example 8 Solving a Rational Inequality (cont.)

or

Step 2: The quotient changes sign only where x-values make

the numerator or denominator 0.

or

The values and 3 divide the number line into three regions.Use an open circle on 3 because it makes the denominatorequal 0.



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2.7Example 8 Solving a Rational Inequality(cont.)


The values in interval B make the inequality true.

Solution set:

2 7 Example 9 Solving a Rational Inequality (page 153)


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2.7Example 9 Solving a Rational Inequality (page 153)

Solve .

Step 1: Subtract 4.

2x 3 is the commondenominator.

Write as a singlefraction.

Combine terms in thenumerator.



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2.7Example 9 Solving a Rational Inequality (cont.)

or

Step 2: The quotient changes sign only where x-values make

the numerator or denominator 0.

or

The values and divide the number line into three regions.Use open circles because equality is not included.



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2.7Example 9 Solving a Rational Inequality(cont.)


The values in intervals A and C make the inequality true.

Solution set:

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T2 Equations and Inequalities v1