Chapter 3

Linear Equations

3.1 Linear Equation in One Variable

A Linear Equation in one variable is an equation that can be written in the form

ax + b = 0

Where a = 0

A linear function can be written as f(x) = ax + b

Examples of Linear equation.2x –1 = 0, -5x = 10 + x, and 3x + 8 = 5

Properties of Equality

Addition Property of EqualityIf a, b, c are real numbers, thena = b is equivalent to a+ c = b + c.

Multiplication Property of EqualityIf a, b, c are real numbers with c = 0, thena = b is equivalent to ac = bc.

Example 8 Solving a Linear equation graphicallyusing graphing calculator

[ -6, 6, 1] by [-4, 4, 1]

Standard form of a line ( pg 158 - 159 )

An equation for a line is in standard form when itis written as ax + by = c, where a, b, c are constantsWith a, b and c are constants with a and b notboth 0

To find x-intercept of a line, let y = 0 in the equation and solve for x

To find y-intercept of a line, let x = 0 in the equation and solve for x

Example 115 ( Pg -163 )

[ 1984, 1991, 1] by [0, 350, 50] in 1987

In 1987 CD and LP record sales were both 107 million

3.2- Introduction to problem solving

Step1: Read the problems carefully to be sure that youunderstand it. ( You may need to read the problem more thanonce. ) Assign a variable to what you are being asked to find.If necessary, write other quantities in terms of this variable.

Step 2: Write an equation that relates the quantities described in the

problem. You may need to sketch a diagram, make a table, or refer to

known formulas.

Step 3: Solve the equation and determine the solution.

Step 4: Look back and check your answer. Does it seem reasonable? Did you find the required information ?

3.2 Solving a problemThe sum of three consecutive even integers is 96. Findthe three numbers. Solution Step 1: n is the smallest of the three integersn + 2 : next consecutive even integern + 4 : larges of the three even integers

Step 2: Write an equation that relates these unknown quantities. Asthe sum of these three even integers is 108The equation isn + (n + 2) + (n + 4) = 96

Step 3 : Solve the equation in Step 2n + (n + 2) + (n + 4) = 96( n + n +n) + (2 +4) = 963n + 6 = 963n + 6 = 963n = 90n = 30smallest of the three numbers is 30, so the three numbers are 30, 32, 34

Step 4: Check your answer. The sum of these three even integers is 30 + 32 + 34= 96The answer checks.

Mixing acid ( Ex 12 pg 172)

Step 1 x: liters of 60% sulphuric acid x + 2: Liters of 50% sulphuric acid

Step 2 Concentration Solution Amount Pure Acid

0.20 (20%) 2 0.20(2) 0.60 (60%) x 0.60x 0.50(50%) x + 2 0.50(x + 2)

Equation 0.20(2) + 0.60x = 0.50(x + 2) (pure acid in 20% sol.) (pure acid in 60% sol.) (pure acid in 50% sol.)

Step 3 Solve for x 0.20(2) + 0.60x = 0.50(x + 2) 2 (2) + 6x = 5(x + 2) Multiply by 10 4 + 6x = 5x + 10 (Distributive Property) Subtract 5x and 4 from each side x = 6Six liters of the 60% acid solution should be added to the 2 liters of 20% acid solution.

Step 4 If 6 liters of 60% acid solution are added to 2 liters of 20% solution, then therewill be 8 liters of acid solution containing 0.60(6) + 0.20(2) = 4 liters of pure acid.

Check The mixture represents a 4/8 = 0.50 or 50% mixture

20 %2 liters

60% x liters 50%

x + 2 liters+ =

Ex 50 (Pg 175) Anti freeze mixture

A radiator holds 4 gallons of fluidx represents the amount of antifreeze that is drained andreplacedThe remaining amount is 4 – x 20 % of + 70 % of solution of = 50% of solution 4 – x gallons x gallons 4 gallons

0.20(4-x) + 0.70x = 0.50(4)0.8 – 0.2x + 0.7x = 20.5x = 1.2x = 2.4The amount of antifreeze to be drained and replaces is 2.4 gallons

Geometric Formulas

Perimeter of triangle P = a+b+c unitArea of triangleA = ½ bh sq.unit

Area of Rectangle =LW sq.unitPerimeter P = 2(L + W) unit

Area of ParallelogramA = bh sq.unit

Area & Volume of cylinderA = 2 rh sq.unit

V = r2 h cu.unit

Area & Volume of a cubeA = 6a2 sq.unit

V = a3 cu.unit

h

hb

r

h

a a a

a

L

W

c

b

3.3 Linear Inequality in One Variable

A linear inequality in one variable is an inequalitythat can be written in the formax + b > 0, where a = 0. ( The symbol > may bereplaced with >, <, or < )

There are similarities among linear functions,equations, and inequalities. A linear function isgiven by f(x) = ax + b, a linear equation by ax + b= 0, and a linear inequality by ax + b > 0.

3.3 Properties of Inequalities

Let a, b, c be real numbers. a < b and a+c < b+c are equivalent

( The same number may be added to or subtracted from both sides of an inequality.)

If c > 0, then a < b and ac < bc are equivalent.(Both sides of an inequality may be multiplied or divided by the

same positive number)

If c< 0, then a < b and ac > bc are equivalent.Each side of an inequality may be multiplied or divided by the same

negative number provided the inequality symbol is reversed.

Examples of linear inequalities are

2x + 1< 0, 1-x > 6, and 5x + 1 < 3 – 2xA solution to an inequality is a value of the variable thatmakes the statement true. The set of all solutions is calledthe solution set. Two inequalities are equivalent if theyhave the same solution set.

Inequalities frequently have infinitely many solutions. Forexample, the solution set to the inequality x- 5> 0 includesall real numbers greater than 5, which can be written as x > 5. Using set builder notation, we can write the solutionset as { x x > 5 }.

MeaningThis expression is read as “ the set of all real numbers x suchthat x is greater than 5. “

Ex -4 Graphical Solutions (Pg 182)

350

300 250 200 150 100 50

y1

y2 y2

y1 x = 2

0 1 2 3 4

Time ( hours )

Distances of two cars

Distance(miles)

y1 < y2 when x< 2 car 1 is closer to Chicago than car 2

y1 is below the graph of y2

y1 > y2 when x > 2 Car 1 is farther from Chicago than Car 2

Y1 above the graph of y2

When x = 2 , y1 = y2, ie car 1 and car 2 both are 150 miles

From Chicago

…ContinuedEx 5 (Pg 183) Solving an inequality graphically

Solve 5 – 3x < x – 3y1 = 5 – 3x and y2 = x – 3 Intersect at the point (2, -1)

-4 -3 -2 -1 0 1 2 3 4

X = 2

y1

y2

Thus 5 – 3x < x – 3 is satisfied when x > 2.The solution set is {x / x> 2 }

y1 = y2 when x = 2

y1 < y2 when x > 2 , y1 is below the graph of y2

Combining the above resulty1 < y2 when x > 2

-1-2-3

(2, -1)

54321

Using Graphing Calculator

[ -5, 5, 1] by [-5, 5, 1]

Hit window Hit GraphHit YEnter inequality Enter

Y1 = 5 – 3xY2 = x - 3

Ex 92 ( Pg 178) Sales of CD and LP recordsUsing Graphing Calculator

Hit Y , enter equations Enter table set Enter window Hit Table

Hit graph

1987 or after CD sales were greater than or equal to LP records

3.4 Compound inequalities

A compound inequality consists of two inequalitiesjoined by the words and or

In compound inequality contains word and, a solution must satisfyboth two inequalities.

2x > - 5 and 2x < 3 2(1) > -5 and 2(1) < 3 1 is a solution

True True

In compound inequality contains word or , a solution must satisfy atleast one of the two inequalities.

5 + 2 > 3 or 5 – 1 < -5 5 is a solution x + 2 > 3 or x – 1 < -5

True False

Example 1

Example 2

Symbolic Solutions and Number Lines

x < 6

x> - 4

x < 6 and x > - 4

-8 -6 -4 -2 0 2 4 6 8

-8 -6 -4 -2 0 2 4 6 8

-8 -6 -4 -2 0 2 4 6 8

- 4 < x < 6

(

]

](

Three-part inequality

Sometimes compound inequality containing theword and can be combined into a three part inequality.

For example, rather than writing

x > 4 and x < 10

We could write the three-part inequality4 < x < 10

-1 0 1 2 3 4 5 6 7 8 9 10

4 < x < 10

( ]Numberline

Compound Inequality

Example Solve x + 2 < -1 or x + 2 > 1 x < -3 or x > -1 ( subtract 2 )The solution set for the compound inequality results from taking theunion of the first two number lines. We can write the solution, usingSet builder notation, as { x x < - 3} U { x x > - 1 } or {x x < - 3 or x > - 1}

x < - 3 -

x > - 1 - x < - 3 or x > -1 -

- 4 -3 -2 -1 0 1 2 3 4

- 4 -3 -2 -1 0 1 2 3 4

- 4 -3 -2 -1 0 1 2 3 4

(

)

) (

Interval Notation

Inequality Interval Notation Number line Graph

- 1 < x < 3 ( - 1, 3)

- 3 < x < 2 ( - 3, 2]

- 2 < x < 2 [ - 2, 2 ]

x < - 1 or x > 2 ( - , - 1) U (2, ) -

x > - 1 ( - 1, )

x < 2 ( - , 2 ] -

-4 -3 -2 -1 0 1 2 3 4

-4 -3 -2 -1 0 1 2 3 4

-4 -3 -2 -1 0 1 2 3 4

-4 -3 -2 -1 0 1 2 3 4

-4 -3 -2 -1 0 1 2 3 4

-4 -3 -2 -1 0 1 2 3 4

]

][

()

(

]

( )

(

+

+

Example Solve graphically and numerically. Write your answer in interval notation

x + 1 < -1 or x + 1 > 1

Y1 = -1, Y2 = x + 1, Y3 = 1

x + 1 < - 1 or x + 1 > 1Solution in interval notation is ( - , - 2) U (0, )

Enter window [ -5, 5, 1] by [ -5, 5, 1]Enter y1, y2, y3

-1

+1

-

- 2

0Hit Graph

Solving a compound inequality numerically and graphically Using Technology Tution at private colleges and universities from 1980 to 1997 Can be modelled by f(x) = 575(x – 1980) + 3600 Estimate when average tution was between $8200 and $10,500.

[ 1980, 1997, 1] by [3000, 12000, 3000 ]

Example 6 (Page 195)

Hit Y and enter equation Hit Window and enter Hit 2nd and Table

Hit 2nd and calc and go to Intersect and enter 4 times to get intersection

School Enrollment

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