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

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

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

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

Holt Algebra 2 2-1 Solving Linear Equations and Inequalities

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

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.

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

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.

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.

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.

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 .

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

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

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.

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

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