BMayer@ChabotCollege.edu MTH55_Lec-19_sec_4-4_2Var_InEqualities.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical &

BMayer@ChabotCollege.edu • MTH55_Lec-19_sec_4-4_2Var_InEqualities.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Chabot Mathematics

§4.4 2-Var§4.4 2-VarInEqualitiesInEqualities

Review §Review §

Any QUESTIONS About• §4.3b → Absolute Value InEqualities

Any QUESTIONS About HomeWork• §4.3b → HW-13

4.3 MTH 55

Graphing InEqualitiesGraphing InEqualities

The graph of a linear equation is a straight line. The graph of a linear inequalityinequality is a half-planehalf-plane, with a boundaryboundary that is a straight line.

To find the equation of the boundary line, we simply replace the inequality sign with an equals sign.

Example Example Graph Graph yy ≥ ≥ xx

SOLUTION First graph the

boundary y = x. Since the inequality is greater than or equal to, the line is drawn solid and is part of the graph of the Solution

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• Note that in the graph each ordered pair on the half-plane above y = x contains ay-coordinate that is greater than thex-coordinate. It turns out that any point on the same side as (–2, 2) is also a solution. Thus, if one point in a half- plane is a solution, then all points in that half-plane are solutions.

x -5 -4 -3 -2 -1 1 2 3 4 5

• Finish drawing the solution set by shading the half-plane above y = x.

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For any point here, y > x.

For any point here, y = x.

• The complete solution set consists of the shaded half-plane as well as the boundary itself whichis drawnsolid

Example Example Graph Graph yy < 3 < 3 − 8− 8xx SOLUTION Since the inequality

sign is < , points on the line y = 3 – 8x do not represent solutions of the inequality, so the line is dashed.

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y = 3 – 8x

(3, 1)

Using (3, 1) as a test point, we see that it is NOT a solution:

Thus points in the other ½-plane are solns trueNOT 24313831

Graphing Linear InEqualitiesGraphing Linear InEqualities

1. Replace the inequality sign with an equals sign and graph this line as the boundary. If the inequality symbol is < or >, draw the line dashed. If the inequality symbol is ≥ or ≤, draw the line solid.

2. The graph of the inequality consists of a half-plane on one side of the line and, if the line is solid, the line is part of the Solution as well

Graphing Linear InEqualitiesGraphing Linear InEqualities

3. Shade Above or Below the Line• If the inequality is of the form y < mx + b

or y ≤ mx + b shade below the line.• If the inequality is of the form y > mx + b

or y ≥ mx + b shade above the line.

4. If y is not isolated, either solve for y and graph as in step-3 or simply graph the boundary and use a test point. If the test point is a solution, shade the half-plane containing the point. If it is not a solution, shade the other half-plane

Example Example Graph Graph

Draw Graph and test (3,3) = (xtest, ytest)

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y = (1/6)x – 1-4

Check Location of Test Value• ytest > (1/6)·xtest − 1 ¿?

• 3 > (1/6)(3) − 1 ¿?

• 3 > 2 − 1 Since 3 > 1 the pt

(3,3) IS a Soln, so shade on that side

Example Example Graph Graph xx ≥ −3≥ −3

Draw Graph

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Test (4,−2) & (1, 3)

(4,−2)

(1,3) Since both 4 & 1 are

greater than −3, thenpoints to the right ofthe line are solutions

Systems of Linear EquationsSystems of Linear Equations

To graph a system of equations, we graph the individual equations and then find the intersection of the individual graphs. We do the same thing for a system of inequalities, that is, we graph each inequality and find the intersection of the individual Half-Plane graphs.

Example Example xx + + yy > 3 > 3 && xx − − yy ≤ 3 ≤ 3

SOLUTION First graph

x + y > 3 in red.

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y > −x + 3

SOLUTION Next graph

x − y ≤ 3in blue

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y ≥ x − 3

SOLUTION Now find the

intersection of the regions

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The Solution is the OverLappingRegion• CLOSED dot

indicates that theIntersection is Part of the Soln

Solution set to the system

Example Example Graph Graph −1 <−1 < yy < 5 < 5

SOLUTION Break into Two

Inequalities andGraph• −1 < y

• y < 5

The Solution is the OverLappingRegion

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Solution set

Intersection of Two InequalitiesIntersection of Two Inequalities

Graph 3x + 4y ≥ 12 and y > 2 Graph Each InEquality Separately

Intersection of Two InequalitiesIntersection of Two Inequalities

Graph 3x+4y≥12 and y>2

Shade Region(s) common to BOTH

Union of Two InequalitiesUnion of Two Inequalities

Graph 3x + 4y ≥ 12 or y > 2 Again Graph Each InEquality Separately

Union of Two InequalitiesUnion of Two Inequalities

Graph 3x+4y≥12 or y>2

Shade Region(s) covered by EITHER soln

Graphing a System of InEqualsGraphing a System of InEquals

A system of inequalities may have a graph that consists of a polygon and its interior.

To construct the PolyGon we find the CoOrdinates for the corners, or vertices (singular vertex), of such a graph

Example Example Graph of System Graph of System

Graph System2,

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(3, 5)

(3, –3)

(–1, 1 )

Draw Graph• 3 Lines

Intersecting at 3 locations

Graph System2,

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The Solution isthe EnclosedRegion; a PolyGon• A TriAngle in this case

– Check that, say, (2, 2) works in all threeof the InEqualities

Example Example Find Vertices Find Vertices

Graph the following system of inequalities and find the coordinates of any vertices formed:

Graph the related equations using solid lines.

Shade the region common to all three solution sets.

To find the vertices, we solve three systems of 2-equations.

The system of equations from inequalities (1) and (2)• y + 2 = 0 & −x + y = 2

Solving find Vertex pt (−4, −2) The system of equations from

inequalities (1) and (3):• y + 2 = 0 & x + y = 0

The Vertex for The system of equations from inequalities (1) & (3): (2, −2)

The system of equations from inequalities (2) and (3):• −x + y = 2 & x + y = 0

The Peak Vertex Point is (−1, 1)

(−4,−2)

(2,−2)

(−1,−1)

Graph the following system. Find the coordinates of any vertices formed.

Graph by Lines The CoOrd of the

vertices are: (0, 3), (0, 4), (3, 4) and (3, 1)

Types of Eqns & InEqualsTypes of Eqns & InEquals

Type Example Solution

Linear Equations in one variable

2x – 8 = 3(x + 5) A number in One Variable

Linear InEqualities in one variable

–3x + 5 > 2 A set of numbers;

an interval

Linear Equations in two variable

2x + y = 7 A set of ordered

pairs; a line

Linear InEqualities in two variable

x + y ? 4 A set of ordered

pairs; a half-Plane

System of Equations in two variables

x + y = 3 5x - y = -27

An ordered pair or a (possibly empty)

set of ordered pairs

System of two variables

6x – 2y ? 12 y – 3 ? 0 x + 7 ? 0

A set of ordered inequalities in

pairs; a region of a plane

≤≤≥

Example Example PopCorn Revenue PopCorn Revenue A popcorn stand in an amusement park sells

two sizes of popcorn. The large size sells for $4.00 and the smaller for $3.00 The park management feels that the stand needs to have a total revenue from popcorn sales of at least $400 each day to be profitablea) Write an inequality that describes the amount of

revenue the stand must make to be profitable.b) Graph the inequality.c) Find two combinations of large and small

popcorns that must be sold to be profitable

Example Example PopCorn Revenue PopCorn Revenue

Translate by Tabulation

Category Price Number Sold Revenue

Large 4.00 x 4x

Small 3.00 y 3y

a) The total revenue would be found by the expression 4x + 3y. If that total revenue must be at least $400, then we can write the following inequality:

4x + 3y ≥ 400

b) Graph 4x + 3y ≥ 400

c) We assume that fractions of a particular size are not sold, so we will only consider whole number combinations.

• One combination is 100 large and 0 small popcorns which is exactly $400.

• A second combination is 130 large and 40 small, which gives a total revenue of $640.

WhiteBoard WorkWhiteBoard Work

Problems From §4.4 Exercise Set• 46 (ppt), 62

PopCornBag & BucketSizes

P4.4-46 Graph SystemP4.4-46 Graph System

Graph2x + y ≤ 6

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M55_§JBerland_Graphs_0806.xls

Test (0,0)• 2(0)+0 ≤ 6?

• 0 ≤ 6 Shade

BELOWLine

Graphx + y ≥ 2

Test (0,0)• 0+0 ≥ 2?

• 0 ≥ 2 Shade

ABOVELine

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Graph1 ≤ x ≤ 2

Test (0,0)• 1 ≤ 0 ≤ 2

Test (1.5,0)• 1 ≤ 1.5 ≤ 2

ShadeBETWEENLines

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Graphy ≤ 3

Test (0,0)• 0 ≤ 3

ShadeBELOWLine

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Now CheckFor OverLapRegion

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Found One;a five sidedPolyGon

Thus Solution

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All Done for TodayAll Done for Today

HealthyHeart

WorkOut

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Chabot Mathematics

AppendiAppendixx

srsrsr 22

Graph Graph yy = | = |xx||

Make T-tablex y = |x |

-6 6-5 5-4 4-3 3-2 2-1 10 01 12 23 34 45 56 6

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

file =XY_Plot_0211.xls

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BMayer@ChabotCollege.edu MTH55_Lec-19_sec_4-4_2Var_InEqualities.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical &

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