Solving Inequalities with variables on both sides of the Sign

Slides:



Advertisements
Similar presentations
Equations with Variables on Both Sides
Advertisements

Variables on Both Sides of the Equation
Do Now 10/30/09 Copy HW in your planner. Copy HW in your planner. –Text p. 134, #12-28 evens & #34 & #36 Did you turn in POTW#8?
Solving Equations with variables on both sides of the Equals
Solving Equations with the Variable on Both Sides Objectives: to solve equations with the variable on both sides.
3-5 Solving Equations with the variable on each side Objective: Students will solve equations with the variable on each side and equations with grouping.
Solving Equations with the Variable on Both Sides
Solve an equation with variables on both sides
Vocabulary Chapter 6.
Solving Linear Inequalities Chapter 1.6 Part 3. Properties of Inequality.
Solving Equations with variables on both sides of the Equals Chapter 3.5.
Solving Multi- Step Equations. And we don’t know “Y” either!!
Chapter 2 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 2-1 Solving Linear.
Solving Linear Inequalities Chapter 1.6 Part 3. Properties of Inequality 2.
Solving Open Sentences Involving Absolute Value
Practice 2.2 Solving Two Step Equations.
2.1 Solving One Step Equations. Addition Property of Equality For every real number a, b, and c, if a = b, then a + c = b + c. Example 8 = For every.
Lesson 1-8 Solving Addition and Subtraction Equations.
Warm Up Solve. 1. 3x = = z – 100 = w = 98.6 x = 34 y = 225 z = 121 w = 19.5 y 15.
Systems of Equations: Substitution
Solve Linear Systems by Substitution January 28, 2014 Pages
© by S-Squared, Inc. All Rights Reserved.
Linear Equations and Inequalities in One Variable What is an equation? =
Solve Equations With Variables on Both Sides. Steps to Solve Equations with Variables on Both Sides  1) Do distributive property  2) Combine like terms.
* Collect the like terms 1. 2a = 2a x -2x + 9 = 6x z – – 5z = 2z - 6.
Chapter 2 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 2-1 Solving Linear Equations and Inequalities.
6-2 Solving Systems Using Substitution Hubarth Algebra.
Solving Equations with Variables on Both Sides. Review O Suppose you want to solve -4m m = -3 What would you do as your first step? Explain.
Algebra 1 EOC Summer School Lesson 10: Solve Linear Equations and Inequalities.
3.5 Solving Equations with Variables on Both Sides.
My Equations Booklet.
Solving Multistep Equations
Solving Equations with the Variable on Both Sides
Lesson 3.5 Solving Equations with the Variable on Both Sides
Solving Linear Equations and Inequalities
Solving Linear Inequalities in One Unknown
Solving Multi-Step Equations
Objective 3.6 solve multi-step inequalities.
SOLVING EQUATIONS, INEQUALITIES, AND ALGEBRAIC PROPORTIONS
2 Understanding Variables and Solving Equations.
Algebraic Inequalities
Variables on Both Sides with Equations
6-2 Solving Systems Using Substitution
Example 2 4 m 8 m 5m 12 m x y.
Solving Multi-Step Equations
Example 2 4 m 8 m 5m 12 m x y.
Lesson 3.1 How do you solve one-step equations using subtraction, addition, division, and multiplication? Solve one-step equations by using inverse operations.
Solving Multi-Step Equations
OBJECTIVE: Students will solve multistep equations.
Equations: Multi-Step Examples ..
1.3 Solving Linear Equations
Solving Two-Step Equations Lesson 2-2 Learning goal.
Multi-Step Equations TeacherTwins©2014.
Solving Multi-Step Equations
Solving Multi-Step Equations
Learn to solve equations with integers.
Multi-Step Equations TeacherTwins©2014.
Using the Properties Together
Solving Linear Equations and Inequalities
Solving Multi-Step Equations
Section Solving Linear Systems Algebraically
Solving Multi-Step Equations
2-3 Equations With Variables on Both Sides
Lesson 7-6 Multiplying a Polynomial by a Monomial
Unit 2B/3A Solving Equations
Solving basic equations
Bellwork x – 2(x + 10) = 12.
1. How do I Solve Linear Equations
If an equation contains fractions, it may help to multiply both sides of the equation by the least common denominator (LCD) to clear the fractions before.
By: Savana Bixler Solving Equations.
Presentation transcript:

Solving Inequalities with variables on both sides of the Sign Chapter 6.3

Solving Inequalities with variables on both sides of the Sign Lesson Objective: NCSCOS 4.01- Use linear functions to model and solve problems Students will know how to solve equations with variables on both sides. Students will know how to use the distributive property to solve problems. Students will know how to combine like terms to solve problems

Solving Inequalities with variables on both sides of the Sign 5x 2x - 8 x 10 < 20 3x + 12 Example 1: First we must get all the variables on one side of the equation. Subtract 3x from both sides Second we have to get all the numbers on the other side Add 8 to both sides Divide both sides by 2 -3x -3x 2 2 + 8 +8

Solving Inequalities with variables on both sides of the Sign 6 – x > 5x + 30 5y – 4y ≥ 3y + 2 4m – 8 ≤ 10 – 2m 8 + 4k < -10 + k (2/5)x – 8 ≤ 9 – (3/5)x Practice

Solving Inequalities with variables on both sides of the Sign Example 2: Distribute on both sides Add 10x to both sides Add 4 to both sides Divide by 18 4(2x – 1) ≥ -10(x – 5) 18x – 4 ≥ 50 8x – 4 ≥ -10x + 50 18x ≥ 54 x ≥ 3 +10x +10x +4 +4 18 18

Solving Inequalities with variables on both sides of the Sign -3(x + 5) < 3(x – 1) 5(x + 2) ≤ 2(3 – x) 4(m + 3) > 36 4(y – 2) < 2(5 – y) 3(x – 8) ≥ 3x More Practice

Solving Inequalities with variables on both sides of the Sign Example 3: Distribute on both sides Combine like terms on either side of the equals sign Subtract 3x from both sides Subtract 24 from both sides Divide by 2 5(x + 3) + 9 > 3(x – 4) + 6 5x + 15 + 9 > 3x – 12 + 6 5x + 24 > 3x - 6 2x + 24 > -6 __ 2x > -30 x > -15 __ -3x -3x -24 -24 2 2

Solving Inequalities with variables on both sides of the Sign 3(x – 8) – 5 ≤ 9(x + 2) + 1 2(x + 2) + x ≥ 5 + 4(x – 2) 5x – (3x – 3) > 2x + 3(x + 4) Last Practice!

Solving Inequalities with variables on both sides of the Sign Quiz 6.3 2x + 3 ≤ 3x – 5 7x – 3 > 3 – 2x 2(x + 3) ≥ 3(x – 4) 5(2x – 4) < -6(2 – x) 3(x + 2) – 2x > 4(x -3) + 6