Do Now 1/25/12Do Now 1/25/12 Take out HW from last night.Take out HW from last night.

Mid-Term Review worksheet #1Mid-Term Review worksheet #1 Mid-Term Review worksheet #2Mid-Term Review worksheet #2

Copy HW in your planner.Copy HW in your planner. Mid-Term tomorrow! Mid-Term tomorrow! Go to bed by 8:00 Go to bed by 8:00 Eat breakfast in the morningEat breakfast in the morning

Algebra Mid-Term Review Worksheet #1Algebra Mid-Term Review Worksheet #1

1) -1621) -162 2) 3x2) 3x² - 5y + 1² - 5y + 1 3) 103) 10 4) x = b – (c/a)4) x = b – (c/a) 5) -25) -2 6) -16) -1 7) y = 8 7) y = 8 8) x = -68) x = -6 9) no solution9) no solution 10) no solution10) no solution 11) vertical11) vertical 12) y = -5x – 2212) y = -5x – 22 13) -15, 913) -15, 9 14) 2 < z < 514) 2 < z < 5 15) undefined / 015) undefined / 0 16) x < 1.616) x < 1.6 17) x > 9 or x < 917) x > 9 or x < 9

18) y 18) y ≤ 2 and y ≥ -3≤ 2 and y ≥ -3 19) x > -1 or x < -319) x > -1 or x < -3 20) -120) -1 21) whole, rational, integer, real 21) whole, rational, integer, real 22) 75, 80, 8522) 75, 80, 85 23) m = -5/2; y-int = 523) m = -5/2; y-int = 5 24) y – 2 = 7(x + 4)24) y – 2 = 7(x + 4) 25) t ≤ -225) t ≤ -2 26) empty set / no solution; x = 426) empty set / no solution; x = 4 27) 2x + y = 027) 2x + y = 0 28) identity/ all real numbers28) identity/ all real numbers 29) B29) B 30) 30)

ObjectiveObjective

SWBAT review Chapters 4-7 concepts for SWBAT review Chapters 4-7 concepts for Mid-TermMid-Term

Chapter 4Chapter 4Graphing Linear Graphing Linear

Equations and FunctionsEquations and Functions

Section 4.1 “Coordinate Plane”Section 4.1 “Coordinate Plane”

y-axisy-axis

x-axisx-axis

OriginOrigin (0,0)(0,0)

Quadrant IQuadrant I(+,+)(+,+)

Quadrant IIQuadrant II(-,+)(-,+)

Quadrant IIIQuadrant III(-,-)(-,-)

Quadrant IVQuadrant IV(+,-)(+,-)

Section 4.2 “Graph Linear Equations”Section 4.2 “Graph Linear Equations”

42 xy

xy 24

Solve the equation for y.

STEPSTEP 11

SOLUTIONSOLUTION

Graph the equation Graph the equation y + 2x = 4y + 2x = 4..

STEPSTEP 22

Make a table by choosing a few values for x and then finding values for y.

STEPSTEP 33

Plot the points. Notice the points appear on a line. Connect the points drawing a line through them.

xx -2-2 -1-1 00 11 22

yy 88 66 44 22 00

Section 4.3 “Graph Using Intercepts”Section 4.3 “Graph Using Intercepts”

Graph the equation Graph the equation 6x + 7y = 42 6x + 7y = 42..

6x + 7y = 42

x = x-intercept7

Find the x-intercept

6x + 7(0)=42 6(0) + 7y = 42

y = y-intercept6

6x + 7y = 42

Find the y-intercept

Plot points. The x-intercept is 7, so plot the point (7, 0). The y- intercept is 6, so plot the point (0, 6). Draw a line through the points.

(0, 6) and (5, –4)

m =y2 – y1

x2 – x1

Let (x1, y1) = (0, 6) and (x2, y2) = (5, – 4).

– 4 – 6 5 – 0

=

Write formula for slope.

Substitute.

Simplify. 10

5= – = – 2

Find the slope of the line that passes through the points

Section 4.4 “Find Slope and Rate of Change”Section 4.4 “Find Slope and Rate of Change”

Section 4.5 “Section 4.5 “Graph Using Slope-Intercept FormGraph Using Slope-Intercept Form””

SLOPE-INTERCEPT FORM-SLOPE-INTERCEPT FORM-

a linear equation written in the forma linear equation written in the form

y = mx + by = mx + b

slope y-intercept

y-coordinate x-coordinate

Determine which of the lines are parallelDetermine which of the lines are parallel..

Find the slope Find the slope of each lineof each line.. Line Line aa:: mm == – – 11 – – 00

– – 11 – – 22

– – 33 – – ((––11 ))00 – – 55

==–– 11–– 33

1133==

Line Line b:b: m m == –– 22–– 55== 22

55==

Line Line c:c: m m == – – 55 – – ((––3)3) – – 2 – 2 – 44

–– 22–– 66==

11 33==

Line Line aa and line and line cc have the same slope, so they are parallelhave the same slope, so they are parallel..

Section 4.7 “Section 4.7 “Graph Linear FunctionsGraph Linear Functions””

Function Notation-Function Notation-a linear function written in the form a linear function written in the form y = mx y = mx + b+ b where y is written as a function where y is written as a function f.f.

f(x) = mx + bf(x) = mx + bslopeslope y-intercepty-intercept

x-coordinatex-coordinate

f(x) is another name for y.f(x) is another name for y.It means “the value of f at x.”It means “the value of f at x.”g(x) or h(x) can also be used to name functions g(x) or h(x) can also be used to name functions

This is read This is read as ‘f of x’as ‘f of x’

Graph a FunctionGraph a Function

32)( xxf

xx -2-2 -1-1 00 11 22

f(x)f(x) -7-7 -5-5 -3-3 -1-1 11

STEPSTEP 11

SOLUTION

Graph the Function Graph the Function f(x) = 2x – 3 f(x) = 2x – 3

STEPSTEP 22

Make a table by choosing a few values for x and then finding values for y.

STEPSTEP 33

Plot the points. Notice the points appear on a line. Connect the points drawing a line through them.

32)( xxf

The domain and range are not restricted therefore, you do not have to identify.

Compare graphs with the graph f(x) = x.Compare graphs with the graph f(x) = x.

Graph the function g(x) = x + 3, then compare it to Graph the function g(x) = x + 3, then compare it to the parent function f(x) = x. the parent function f(x) = x.

xx f(x)f(x)

-5-5 -2-2

-2-2 11

00 33

11 44

33 66

f(x) = xf(x) = xf(x) = x

xx f(x)f(x)

-5-5 -5-5

-2-2 -2-2

00 00

11 11

33 33

g(x) = x + 3g(x) = x + 3g(x) = x + 3

The graphs of g(x) and f(x) have the same slope of 1.

Chapter 5Chapter 5Writing Linear EquationsWriting Linear Equations

Section 5.1 & 5.2 Section 5.1 & 5.2 “Write Equations in Slope-Intercept Form”“Write Equations in Slope-Intercept Form”

SLOPE-INTERCEPT FORM-SLOPE-INTERCEPT FORM-

a linear equation written in the forma linear equation written in the form

y = mx + by = mx + b

slope y-intercept

y-coordinate x-coordinate

Section 5.3 Section 5.3 “Write Linear Equations in Point-Slope Form”“Write Linear Equations in Point-Slope Form”

POINT-SLOPE FORM-POINT-SLOPE FORM-

of a linear equation is written as:of a linear equation is written as:

slope

y-coordinatepoint 1

x-coordinatepoint 1

)( 11 xxmyy

y

x

),( 11 yx

),( yx

runrun

riserise

Section 5.4 Section 5.4 “Write Linear Equations in Standard Form”“Write Linear Equations in Standard Form”

(General Form) (General Form)

The The STANDARD FORMSTANDARD FORM of a of a linear equation is represented aslinear equation is represented as

AAx x ++ BBy y = C= Cwhere A, B, and C are real where A, B, and C are real

numbersnumbers

Section 5.5 “Write Equations of Parallel and Section 5.5 “Write Equations of Parallel and Perpendicular Lines”Perpendicular Lines”

PARALLEL LINESPARALLEL LINES If two nonvertical lines in the same plane have If two nonvertical lines in the same plane have

the the same slopesame slope, then they are , then they are parallelparallel..

If two nonvertical lines in the same plane are If two nonvertical lines in the same plane are parallelparallel, then they have the , then they have the same slope.same slope.

Write an equation of the line that passes through Write an equation of the line that passes through (–2,11)(–2,11) and is parallel to the lineand is parallel to the line yy = -= -xx + 5 + 5..

STEP 1Identify the slope. The graph of the given equation has a slope of -1. So, the parallel line through (– 2, 11) has a slope of -1.

STEP 2Find the y-intercept. Use the slope and the given point.

yy = = mxmx + + bb

1111 = -1(–2) + = -1(–2) + bb9 = 9 = bb

Write slope-intercept formWrite slope-intercept form..

SubstituteSubstitute -1-1 forfor mm,, - -22 for for xx,, andand 11 11 forfor yy..Solve forSolve for bb..

STEP 3

Write an equation. Use y = mx + b.

y = -x + 9 SubstituteSubstitute -1-1 forfor mm andand 99 forfor bb..

Section 5.5Section 5.5 “Write Equations of Parallel and “Write Equations of Parallel and

Perpendicular Lines”Perpendicular Lines”

PERPENDICULAR LINESPERPENDICULAR LINES If two nonvertical lines in the same plane have If two nonvertical lines in the same plane have

slopes that are slopes that are negative reciprocalsnegative reciprocals, then the lines , then the lines are are perpendicularperpendicular..

If two nonvertical lines in the same plane are If two nonvertical lines in the same plane are perpendicularperpendicular, then their slopes are , then their slopes are negative negative reciprocalsreciprocals

½ and -2 are negative reciprocals.½ and -2 are negative reciprocals.

3 and -1/3 are negative reciprocals.3 and -1/3 are negative reciprocals.

Modeling DataModeling DataWhen data show a positive or negative When data show a positive or negative

correlation, you can model the trend in the correlation, you can model the trend in the data using a data using a LINE OF FITLINE OF FIT

Using a Line of Fit to Model DataUsing a Line of Fit to Model Data

1)1) ..

2)2) ..

3)3) ..

4)4) ..

Decide whether the data can be a modeled by a line. (Does Decide whether the data can be a modeled by a line. (Does it have positive or negative correlation?)it have positive or negative correlation?)

Make a scatter plot of the data.Make a scatter plot of the data.

Draw a line that appears to fit the data closely.Draw a line that appears to fit the data closely.

Write an equation using two points on the line. Write an equation using two points on the line. (The points do not have to be actual data pairs, but (The points do not have to be actual data pairs, but they do have to be on the line.)they do have to be on the line.)

Draw a line of fit for the scatter plot. Write an equation that Draw a line of fit for the scatter plot. Write an equation that models the number of ounces of water left in the water models the number of ounces of water left in the water

bottle as a function of minutes on the treadmill.bottle as a function of minutes on the treadmill.Minutes on Minutes on

treadmilltreadmill00 55 1010 1515 2020 2525 3030 3535

Ounces of Ounces of water in water in

water bottlewater bottle

1212 1212 1010 99 77 77 55 33

44

88

00 55 1010 1515 2020 2525 3030

y-axisy-axis

x-axisx-axis3535

Write an equation using two Write an equation using two points on the line. points on the line. 1212

Ounces of water in water bottle

Minutes on the treadmill

Use the points (5,12) and (30,4). Use the points (5,12) and (30,4).

12 – 412 – 4 = _ = _8_8_ 5 – 30 -255 – 30 -25

Find the y-intercept. Use (5,12).Find the y-intercept. Use (5,12).

y = mx + by = mx + b

12 =12 = 8 ( 8 (5) + b b = 5) + b b = 6868 -25 5-25 5

y = y = 8 8 x + x + 6868 -25 5-25 5

Chapter 6Chapter 6Solving and Graphing Solving and Graphing

Linear InequalitiesLinear Inequalities

Writing Equations with InequalitiesWriting Equations with Inequalities

SymbolSymbol MeaningMeaning Key phrasesKey phrases

== Is equal toIs equal to The same asThe same as

<< Is less thanIs less than Fewer thanFewer than

≤≤ Is less than or equal Is less than or equal to to

At most, no more At most, no more thanthan

>> Is great thanIs great than More thanMore than

≥≥ Is greater than or Is greater than or equal toequal to

At least, no less At least, no less thanthan

On a number line, the On a number line, the GRAPH OF AN GRAPH OF AN INEQUALITYINEQUALITY in one variable is the set of in one variable is the set of points that represent all solutions of the points that represent all solutions of the inequality.inequality.

Graph x < 3Graph x < 3

00 11-1-1 22-2-2 33

Graph x ≥ -1Graph x ≥ -1

00 11-1-1 22-2-2 33

““Less than” and “greaterLess than” and “greaterthan” are representedthan” are representedwith an open circle.with an open circle.

““Less than or equal to” Less than or equal to” and “greater than or equaland “greater than or equalto” are represented with to” are represented with closed circle.closed circle.

Multiplying and/or dividing each Multiplying and/or dividing each side of an inequality by a NEGATIVE side of an inequality by a NEGATIVE number only produces an equivalent number only produces an equivalent inequality inequality IF the inequality sign is IF the inequality sign is REVERSED!!REVERSED!!

Solve Inequalities When Multiplying and Dividing Solve Inequalities When Multiplying and Dividing by a NEGATIVE”by a NEGATIVE”

1414xx + 5 < 7(2+ 5 < 7(2x x –– 3)3) Write original inequality.Write original inequality.

1414xx + 5 < 14+ 5 < 14x x –– 2121 Distributive propertyDistributive property

5 < –5 < – 2121 Subtract Subtract 1414xx from each side.from each side.

There are no solutions because There are no solutions because 55 << –– 2121 is false.is false.

Solve: Solve: 1414xx + 5 < 7(2+ 5 < 7(2x x –– 3)3)

**HINT****HINT**If an inequality is equivalent to an inequality that is false, If an inequality is equivalent to an inequality that is false, such as 5 < -21, then the solution of the inequality has such as 5 < -21, then the solution of the inequality has NO SOLUTION. NO SOLUTION.

1212x x –– 1 > 6(21 > 6(2x x – 1)– 1) Write original inequality.Write original inequality.

Distributive propertyDistributive property

Subtract Subtract 1212xx from each side.from each side.

1212x x –– 1 > 121 > 12x x – 6– 6

–– 1 > –1 > – 66

All real numbers are solutions because All real numbers are solutions because –– 11 > > –– 66 is true. is true.

1212x x –– 1 > 6(21 > 6(2x x – 1)– 1)

**HINT****HINT**If an inequality is equivalent to an inequality that is true, If an inequality is equivalent to an inequality that is true, such as -1 > -6, then the solutions of the inequality aresuch as -1 > -6, then the solutions of the inequality areALL REAL NUMBERS . ALL REAL NUMBERS .

SolveSolve

Solve a Compound Inequality with ANDSolve a Compound Inequality with AND

––7 < 7 < x x – 5 < -1– 5 < -1.. Graph your solutionGraph your solution..

Separate the compound inequality into two Separate the compound inequality into two inequalities. Then solve each inequality separately.inequalities. Then solve each inequality separately.

––7 < 7 < x x – 5– 5 Write two inequalities.Write two inequalities.

––7 7 + 5+ 5 < < x x –5 –5 + 5+ 5 Add Add 55 to each side. to each side.

––2 < 2 < xx Simplify.Simplify.

The compound inequality can be written as The compound inequality can be written as – 2 < – 2 < x x < 4.< 4.

andand x x – 5 < -1– 5 < -1

x x –– 5 5 + 5+ 5 < -1 < -1 + 5+ 5

x x < 4< 4

andand

andand

– – 3 – 2 – 1 0 1 2 3 4 5 3 – 2 – 1 0 1 2 3 4 5 Graph:Graph:

Write original Write original inequality.inequality.

Solve Solve 22x x + 3 < 9 + 3 < 9 oror 33x x – 6 > 12.– 6 > 12. Graph your solution. Graph your solution.

Solve the two inequalities separately.Solve the two inequalities separately.

22x x + 3 < 9+ 3 < 9 oror 33x x – 6 > 12– 6 > 12

22x x ++ 3 3 – 3– 3 < 9 < 9 – 3– 3 oror 33x x –– 6 6 + 6+ 6 > 12 > 12 + 6+ 6 Addition or Subtraction Addition or Subtraction property of inequalityproperty of inequality

22x x < 6< 6 oror 33x x > 18> 18 Simplify.Simplify.

Solve a Compound Inequality with ORSolve a Compound Inequality with OR

x x < 3< 3 oror x x > 6> 6 Simplify.Simplify.

The solutions are all real numbers less than The solutions are all real numbers less than 33 or or greatergreater than than 66..

Solving an Solving an Absolute Value EquationAbsolute Value Equation

The equation The equation |ax + b| = c|ax + b| = c where where c ≥ 0c ≥ 0,, is equivalent to the statement:is equivalent to the statement:

ax + b = cax + b = c or or ax + b = -cax + b = -c

Solve 4|t + 9| - 5 = 19Solve 4|t + 9| - 5 = 19

Solve an Absolute Value EquationSolve an Absolute Value Equation

FirstFirst,, rewrite the equation in the form rewrite the equation in the form ax ax + + bb = = cc..

Next, solve the absolute value equation.Next, solve the absolute value equation.

44 t + t + 9 – 5 = 19 9 – 5 = 19

4 4 tt + 9 = 24 + 9 = 24

t +t + 9 = 6 9 = 6

Write original equation.Write original equation.

Add Add 55 to each side. to each side.

Divide each side byDivide each side by 4.4.

tt + 9 = 6 + 9 = 6

tt + 9 = 6 + 9 = 6 oror t t + 9 = –6 + 9 = –6

tt = –3 = –3 or or t t = –15 = –15

Write absolute value equation.Write absolute value equation.

Rewrite as two equations.Rewrite as two equations.

Addition & subtraction to each Addition & subtraction to each sideside

Solving an Solving an Absolute Value InequalityAbsolute Value Inequality

The inequality The inequality |ax + b| > c|ax + b| > c where c > 0, will where c > 0, willresult in a compound OR inequality.result in a compound OR inequality. ax + b > cax + b > c or or ax + b < -cax + b < -c

The inequality The inequality |ax + b| < c|ax + b| < c where c > 0, will where c > 0, willresult in a compound AND inequality.result in a compound AND inequality.

-c < -c < ax + b < cax + b < c

Solve |x - 5| > 7. Graph your solutionSolve |x - 5| > 7. Graph your solution

Solve an Absolute Value InequalitySolve an Absolute Value Inequality

Write original inequality.Write original inequality.

Rewrite as compoundRewrite as compound inequality.inequality.

Add 5 to each side.Add 5 to each side.

x > 12 ORx > 12 OR x < -2 x < -2

|x - 5|x - 5| > 7| > 7

+5 +5x - 5x - 5 > 7 > 7 x - 5x - 5 < -7 < -7OR

Simplify.Simplify.

Graph of x > 12 or x < -2Graph of x > 12 or x < -2

66 8844 1010-2-2 121200 22

Solve |-4x - 5| +3 < 9. Graph your solutionSolve |-4x - 5| +3 < 9. Graph your solution

Solve an Absolute Value InequalitySolve an Absolute Value Inequality

Write original inequality.Write original inequality.

Rewrite as compoundRewrite as compound inequality.inequality.

Add 5 to both sides.Add 5 to both sides.

x > -2.75 ANDx > -2.75 AND x < 0.25 x < 0.25

|-4x - 5|-4x - 5| +3 < 9| +3 < 9

+5 +5-4x – 5 <-4x – 5 < 6 6 -4x - 5-4x - 5 > -6 > -6AND

Simplify.Simplify.

Graph of -2.75 < x < 0.25Graph of -2.75 < x < 0.25

11 2200 33-3-3 -2-2

Subtract 3 from each side.Subtract 3 from each side.-3 -3|-4x - 5|-4x - 5| < 6| < 6

Divide by -4 to each side.Divide by -4 to each side.-4x <-4x < 11 11 -4x -4x > -1> -1AND

-1-1

Graphing Linear Inequalities Graphing Linear Inequalities

Graphing Graphing Boundary LinesBoundary Lines::

Use a dashed line for < or >.Use a dashed line for < or >.

Use a solid line for Use a solid line for ≤ or ≥.≤ or ≥.

Graph an InequalityGraph an Inequality

Graph the equation

STEPSTEP 11

Graph the inequality Graph the inequality y > 4x - 3 y > 4x - 3..

STEPSTEP 22

Test (0,0) in the original inequality.

STEPSTEP 33

Shade the half-plane that contains the point (0,0), because (0,0) is a solution to the inequality.

34 xy 34 xy

3)0(40

True

Chapter 7Chapter 7Systems of Equations and Systems of Equations and

InequalitiesInequalities

Equation 1 Equation 1

x + 2y = 11x + 2y = 11 Equation 2Equation 2

y = 3x + 2 y = 3x + 2

““Solve Linear Systems by Substituting”Solve Linear Systems by Substituting”

x + 2y = 11x + 2y = 11 x + 2x + 2(3x + 2)(3x + 2) = 11 = 11 Substitute Substitute

x + x + 6x + 46x + 4 = 11 = 11 7x 7x + 4+ 4 = 11 = 11

x = 1x = 1

Equation 1 Equation 1 y = 3x + 2 y = 3x + 2 Substitute value forSubstitute value forx into the original x into the original equation equation y = 3(1) + 2y = 3(1) + 2

y = 5y = 5

The solution is the point (1,5). The solution is the point (1,5). Substitute (1,5) into both Substitute (1,5) into both

equations to check.equations to check.

(1) + 2(5) = 11(1) + 2(5) = 1111 = 1111 = 11

(5) = 3(1) + 2(5) = 3(1) + 25 = 5 5 = 5

Section 7.3 “Solve Linear Systems Section 7.3 “Solve Linear Systems by Adding or Subtracting”by Adding or Subtracting”

ELIMINATION-ELIMINATION-adding or subtracting equations to obtain a adding or subtracting equations to obtain a

new equation in one variable. new equation in one variable.

Solving Linear Systems Using Elimination

(1) Add or Subtract the equations to eliminate one variable.

(2) Solve the resulting equation for the other variable.

(3) Substitute in either original equation to find the value of the eliminated variable.

Equation 1 Equation 1

-3x + 2y = -9-3x + 2y = -9 Equation 2Equation 2

4x + 5y = 35 4x + 5y = 35

““Solve Linear Systems by Elimination Solve Linear Systems by Elimination Multiplying First!!”Multiplying First!!”

Equation 1 Equation 1 4x + 5y = 354x + 5y = 35 Substitute value forSubstitute value forx into either of the x into either of the original equations original equations 4(5) + 5y = 354(5) + 5y = 35

20 + 5y = 3520 + 5y = 35

The solution is the point (5,3). The solution is the point (5,3). Substitute (5,3) into both Substitute (5,3) into both

equations to check.equations to check.

4(5) + 5(3) = 354(5) + 5(3) = 3535 = 3535 = 35

-3(5) + 2(3) = -9-3(5) + 2(3) = -9-9 = -9-9 = -9

Multiply Multiply

FirstFirst

++ 23x = 11523x = 115

x = 5x = 5

y = 3y = 3

EliminatedEliminated

x (2)x (2)

15x - 10y = 4515x - 10y = 45 8x + 10y = 70 8x + 10y = 70

x (-5)x (-5)

Section 7.5 “Solve Special Types of Linear Section 7.5 “Solve Special Types of Linear Systems”Systems”

LINEAR SYSTEM-LINEAR SYSTEM-consists of two or more linear equations in the same consists of two or more linear equations in the same

variables.variables.

Types of solutions:Types of solutions:

(1) (1) a a single pointsingle point of intersection of intersection – intersecting lines – intersecting lines

(2) (2) no solutionno solution – parallel lines – parallel lines

(3) (3) infinitely many solutionsinfinitely many solutions – when two equations – when two equations represent the same linerepresent the same line

Equation 1 Equation 1

3x + 2y = 2 3x + 2y = 2 Equation 2Equation 2

3x + 2y = 10 3x + 2y = 10

““Solve Linear Systems with No Solution”Solve Linear Systems with No Solution”

_

0 = 80 = 8

No SolutionNo Solution

EliminatedEliminated EliminatedEliminated

-3x + (-2y) = -2 -3x + (-2y) = -2 +This is a false statement,therefore the system has nosolution.

By looking at the graph, the lines are PARALLEL and therefore willnever intersect.

“ “InconsistentInconsistentSystem”System”

Equation 1 Equation 1

Equation 2Equation 2

x – 2y = -4 x – 2y = -4

““Solve Linear Systems with Infinitely Many Solutions”Solve Linear Systems with Infinitely Many Solutions”

-4 = -4-4 = -4

Infinitely Many SolutionsInfinitely Many Solutions

y = ½x + 2 y = ½x + 2

This is a true statement,therefore the system has infinitely many solutions.

By looking at the By looking at the graph, the lines are the graph, the lines are the SAME and therefore SAME and therefore intersect at every point, intersect at every point, INFINITELY!INFINITELY!

“ “ConsistentConsistentDependentDependent

System”System”

Equation 1 Equation 1 x – 2y = -4x – 2y = -4

Use ‘Substitution’ Use ‘Substitution’ because we know because we know what y is equals.what y is equals.

x – x – 4 = -4x – x – 4 = -4 x – x – 22(½x + 2)(½x + 2) = -4 = -4

Section 7.6 “Solve Systems of Linear Section 7.6 “Solve Systems of Linear Inequalities”Inequalities”

SYSTEM OF INEQUALITIES-SYSTEM OF INEQUALITIES-consists of two or more linear inequalities in consists of two or more linear inequalities in

the same variables.the same variables.

Inequality 1 Inequality 1

2x + y < 82x + y < 8 Inequality 2Inequality 2

x – y > 7 x – y > 7

A solution to a system of inequalities is an A solution to a system of inequalities is an ordered pair (a point) that is a solution to ordered pair (a point) that is a solution to both linear inequalities. both linear inequalities.

y < x – 4 y < x – 4 Inequality 1

y y ≥ -x + 3≥ -x + 3Inequality 2

Graph a System of InequalitiesGraph a System of Inequalities

00 < < 55 – 4 – 4 ??

00 < 1 < 1 00 ≥≥ -2 -2 00 ≥≥ -5-5 + 3 + 3 ??

Graph both inequalities in the same coordinate plane. The graph of the system is the intersection of the two half-planes, which is shown as the darker shade of blue.

(5,0)(5,0)

y y ≥≥ -1 -1 Inequality Inequality 1:1:

x > -2x > -2Inequality Inequality 2:2:

Graph a System of THREE InequalitiesGraph a System of THREE Inequalities

y y ≥≥ -1 -1 ??

00 ≥≥ -1 -1 00 > -2 > -2 xx > > -2-2 ??

Graph all three inequalities in the same coordinate plane. The graph of the system is the triangular region, which is shown as the darker shade of blue.x + 2y x + 2y ≤ 4≤ 4Inequality 3:Inequality 3:

xx + 2y + 2y ≤≤ 4 4 ??

00 + 0 + 0 ≤≤ 4 4 CheckCheck(0,0)(0,0)

Clock PartnersClock Partners

With your 3:00 partner,

complete Mid-Term

Review worksheet

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