Download presentation

Presentation is loading. Please wait.

Published byKenneth Walton Modified over 4 years ago

1
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities Section 2.1 Solving Linear Equations and Inequalities

2
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities Homework Pg 94 #22 - 39

3
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities Linear Equations in One variable Nonlinear Equations 4x = 8 3x – = –9 2x – 5 = 0.1x +2 Notice that the variable in a linear equation is not under a radical sign and is not raised to a power other than 1. The variable is also not an exponent and is not in a denominator. Solving a linear equation requires isolating the variable on one side of the equation by using the properties of equality. + 1 = 32 + 1 = 41 3 – 2 x = –5

4
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities

5
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities The local phone company charges $12.95 a month for the first 200 of air time, plus $0.07 for each additional minute. If Nina’s bill for the month was $14.56, how many additional minutes did she use? Consumer Application

6
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities Solving Equations with the Distributive Property Solve 4(m + 12) = –36 Divide both sides by 4. The quantity (m + 12) is multiplied by 4, so divide by 4 first. 4(m + 12) = –36 4 m + 12 = –9 m = –21 –12 –12 Subtract 12 from both sides.

7
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities Divide both sides by 3. The quantity (2 – 3p) is multiplied by 3, so divide by 3 first. 3(2 – 3p) = 42 3 Solve 3(2 –3p) = 42. Subtract 2 from both sides. –3p = 12 2 – 3p = 14 –2 –2 –3 Divide both sides by –3. p = –4

8
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities Solving Equations with Variables on Both Sides Simplify each side by combining like terms. –11k + 25 = –6k – 10 Collect variables on the right side. Add. Collect constants on the left side. Isolate the variable. +11k 25 = 5k – 10 35 = 5k 5 7 = k +10 + 10 Solve 3k– 14k + 25 = 2 – 6k – 12.

9
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities You have solved equations that have a single solution. Equations may also have infinitely many solutions or no solution. An equation that is true for all values of the variable, such as x = x, is an identity. An equation that has no solutions, such as 3 = 5, is a contradiction because there are no values that make it true.

10
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities Solve 3v – 9 – 4v = –(5 + v). Identifying Identities and Contractions 3v – 9 – 4v = –(5 + v) Simplify. –9 – v = –5 – v + v + v –9 ≠ –5x Contradiction The equation has no solution. The solution set is the empty set, which is represented by the symbol.

11
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities Solve 2(x – 6) = –5x – 12 + 7x. : Identifying Identities and Contractions 2(x – 6) = –5x – 12 + 7x Simplify. 2x – 12 = 2x – 12 –2x –12 = –12 Identity The solutions set is all real number, or .

12
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities These properties also apply to inequalities expressed with >, ≥, and ≤.

13
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities Solve and graph 8a –2 ≥ 13a + 8. Solving Inequalities Subtract 13a from both sides. 8a – 2 ≥ 13a + 8 –13a –5a – 2 ≥ 8 Add 2 to both sides. +2 +2 –5a ≥ 10 Divide both sides by –5 and reverse the inequality. –5 –5 –5a ≤ 10 a ≤ –2

14
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities Check Test values in the original inequality. –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 Test x = –4Test x = –2Test x = –1 8(–4) – 2 ≥ 13(–4) + 88(–2) – 2 ≥ 13(–2) + 88(–1) – 2 ≥ 13(–1) + 8 –34 ≥ –44 So –4 is a solution. So –1 is not a solution. So –2 is a solution. –18 ≥ –18–10 ≥ –5 x Solve and graph 8a – 2 ≥ 13a + 8.

15
Holt Algebra 2 2-1 Solving Linear Equations and Inequalities Solve and graph x + 8 ≥ 4x + 17. Subtract x from both sides. x + 8 ≥ 4x + 17 –x 8 ≥ 3x +17 Subtract 17 from both sides. –17 –9 ≥ 3x Divide both sides by 3. 3 –9 ≥ 3x –3 ≥ x or x ≤ –3

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google