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**Inequalities and Systems**

Regents Review #4 Inequalities and Systems

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Simple Inequalities Solve inequalities like you would solve an equation (use inverse operations and properties of equality to isolate the variable). When multiplying or dividing both sides of an inequality by a negative number, reverse (flip) the inequality sign. Graph the solution set on a number line.

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**-3x – 4 > 8 -3x > 12 -3 -3 x < - 4**

Simple Inequalities -3x – 4 > 8 -3x > x < - 4 3(3x – 1) + 3x > 4(2x + 1) 9x – 3 + 3x > 8x x – 3 > 8x + 4 4x > x > (1.75)

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**Simple Inequalities Words to Symbols Example At Least Minimum**

Cannot Exceed At Most Maximum Example In order to go to the movies, Connie and Stan decide to put all their money together. Connie has three times as much as Stan. Together, they have more than $17. What is the least amount of money each of them can have? Check $4.26 +$12.78 $17.04 Let x = Stan’s money Let 3x = Connie’s money x + 3x > 17 4x > 17 x > 4.25 Since Stan has to have more than $4.25, the least amount of money he can have is $4.26. Since Connie has three times (3 x 4.26) as much as Stan, she has $12.78.

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Simple Inequalities Erik and Julie, an engaged couple, are trying to decide which venue to use to hold their wedding reception. Venue A charges a $2500 site fee in addition to $45 per person. Venue B charges a $3200 site fee plus $40 per person. Using an inequality statement, determine the minimum number of people who must attend the wedding in order for venue B to be more cost effective than venue A. x: # of people Venue A: x Venue B: x # of people (X) Venue A (Y1 ) Venue B (Y2 ) 2500 3200 139 8755 8760 140 8800 141 8845 8840 Venue B < Venue A x < x 40x < x -5x < -700 x > 140 At least 141 people must attend the wedding for venue B to be more cost effective.

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**Compound Inequalities**

A compound inequality is a sentence with two inequality statements joined either by the word “OR” or by the word “AND” “AND” Graph the solutions that both inequalities have in common Solutions must make both inequalities true “OR” Graph the combination of both solution sets Solutions only need to make one inequality true

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**Compound Inequalities “AND”**

The temperature today will be 42⁰ plus or minus 5⁰. Write and graph a compound inequality to represent all the temperatures of the day. Solve -12 2x < -8 2x -12 and 2x < -8 x -6 and x < x < - 4 Let x = the temperatures for the day. 42 – 5 < x < 37⁰ < x < 47⁰

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**Compound Inequalities “OR”**

In order to participate in the big buddy/little buddy bowling league, you must be at least 18 years old or under 10 years of age. Write and graph a compound inequality to represent all the ages of people who participate in the program. Solve the inequality 2x + 5 < 11 or 3x > 15 2x < 6 or x > 5 x < 3 x < 3 or x > 5 Let x = ages of people in the program x < 10 or x > 18

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Linear Inequalities Graph Linear Inequalities in two variables the same way you graph Linear Equations in two variables but… Use a dashed line (----) if the signs are < or > Use a solid line ( ) if the signs are or Shade above the line if the signs are > or Shade below the line if the signs are < or SEE FLIP #8 ON HALGEBRA.ORG

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Linear Inequalities Graph -2y > 2x – 4 -2y > 2x – y < - x + 2 m = b = 2 (0,2) -2y > 2x - 4 Test point (0,0) -2y > 2x – 4 -2(0) > 2(0) – 4 0 > 0 – 4 0 > - 4 True

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Systems A "system" of equations is a collection of equations in with the same variables. When solving Linear Systems, there are three types of outcomes… Infinite Solutions y = 2x + 3 3y = 6x + 9 No Solution y = 2x + 5 y = 2x – 4 One Solution y = -2x + 4 y = 3x - 2

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**Systems There are two ways to solve a Linear System**

Graphically-graph both lines and determine the common solution (point of intersection) Algebraically -Substitution Method -Elimination Method

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**Systems Solution (2,7) y = 4x – 1 m =**

Solve the system y = 4x – 1 and 3x + 2y = 20 graphically Solution (2,7) y = 4x – m = b = -1 (0,-1) 3x + 2y = 20 2y = -3x + 20 y = m = - b = 10 (0,10) 3x + 2y = 20 Y = 4x – 1 Check (2, 7) y = 4x – x + 2y = 20 7= 4(2) – (2) + 2(7) = 20 7 = 8 – = 20 7 = = 20

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Systems Andy’s cab Service charges a $6 fee plus $0.50 per mile. His twin brother Randy starts a rival business where he charges $0.80 per mile, but does not charge a fee. Write a cost equation for each cab service in terms of the number of miles. Graph both cost equations. c) For what trip distances should a customer use Andy’s Cab Service? For what trip distances should a customer use Randy’s Cab Service? x = the number of miles C = the cost Andy’s C(x) = 0.5x + 6 Randy’s C(x) = 0.8x If the trip is less than 20 miles, use Randy’s cab service. If the trip is more than 20 miles, use Andy’s cab service. If the trip is exactly 20 miles, both cabs cost the same amount. [Check algebraically 0.5x + 6 = 0.8x]

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**Systems Solving Linear Systems Algebraically (Substitution)**

x + y = 7 3x = 17 + y Finding y 3x = 17 + y 3(7 – y) = 17 + y 21 – 3y = 17 + y -4y = -4 y = 1 Finding x x + y = 7 x + 1 = 7 x = 6 x = 7 – y Check x + y = 7 6 + 1 = 7 7 = 7 3x = 17 + y 3(6) = 18 = 18 Solution (6,1)

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**Systems Solving Linear Systems Algebraically (Elimination) +**

5x – 2y = 10 2x + y = 31 5x – 2y = 10 2[2x + y = 31] 5x – 2y = 10 4x + 2y = 62 Check 5x – 2y = 10 5(8) – 2(15) = 10 40 – 30 = 10 10 = 10 4x + 2y = 62 4(8) + 2(15) = 62 = 62 62 = 62 + Finding y 2x + y = 31 2(8) + y = 31 16 + y = 31 y = 15 9x + 0y = 72 9x = 72 x = 8 Solution (8, 15)

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**Using Systems to Solve Word Problems**

A discount movie theater charges $5 for an adult ticket and $2 for a child’s ticket. One Saturday, the theater sold 785 tickets for $ How many children’s tickets were sold? Finding x x + y = 785 x = 785 x = 570 Let x = the number of adult tickets Let y = the number of children tickets 5x + 2y = 3280 x + y = 785 5x + 2y = 3280 -5[x + y = 785] 5x + 2y = 3280 -5x – 5y = -3925 + 570 adult tickets 215 children tickets 0x – 3y = -645 -3y = -645 y = 215

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**Solving Linear-Quadratic Systems Graphically**

Two Solutions One Solution No Solution

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**Solving Linear-Quadratic Systems Graphically**

y = x2 – 4x – 2 y = x – 2 y = x2 – 4x – 2 x = y = x – 2 m = b = (0,-2) x y -1 3 -2 1 -5 2 -6 4 5 y = x2 – 4x – 2 y = x – 2 Solutions (0,-2) and (5,3)

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**Solving Systems of Linear Inequalities**

Solve the system and state one solution. Justify your choice. y < 3x y < -2x + 3 y < 3x m = 3/1 b = 0 (0,0) y < -2x m = -2/1 b = 3 (0,3) y x + 3 y < 3x Graph each inequality Label each inequality Label the solution region with S 4) Check with calculator or algebraically S A solution to the system is (1, -3). Justification: y < 3x y < -2x + 3 -3 < 3(1) < -2(1) + 3 -3 < < True < 1 True

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**Now it’s your turn to review on your own**

Now it’s your turn to review on your own! Using the information presented today and the study guide posted on halgebra.org, complete the practice problem set. Regents Review #5 Friday, May 30th BE THERE!

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