Solving Linear Equations and 2 1 Inequalities Solving

Solving Linear Equations and 2 -1 Inequalities Solving Linear Equations and Inequalities Warm Up Lesson Presentation Lesson Quiz Holt Algebra 22

2 -1 Solving Linear Equations and Inequalities Warm Up Simplify each expression. 2. –(w – 2) –w + 2 1. 2 x + 5 – 3 x –x + 5 3. 6(2 – 3 g) 12 – 18 g Graph on a number line. 4. t > – 2 – 4 – 3 – 2 – 1 0 1 2 3 4 5 5. Is 2 a solution of the inequality – 2 x < – 6? Explain. No; when 2 is substituted for x, the inequality is false: – 4 < – 6 Holt Algebra 2

2 -1 Solving Linear Equations and Inequalities Objectives Solve linear equations using a variety of methods. Solve linear inequalities. Holt Algebra 2

2 -1 Solving Linear Equations and Inequalities Vocabulary equation solution set of an equation linear equation in one variable identify contradiction inequality Holt Algebra 2

2 -1 Solving Linear Equations and Inequalities Linear Equations in One variable 4 x = 8 3 x – = – 9 2 x – 5 = 0. 1 x +2 Nonlinear Equations + 1 = 32 + 1 = 41 3 – 2 x = – 5 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. Holt Algebra 2

2 -1 Solving Linear Equations and Inequalities Check It Out! Example 1 Stacked cups are to be placed in a pantry. One cup is 3. 25 in. high and each additional cup raises the stack 0. 25 in. How many cups fit between two shelves 14 in. apart? Holt Algebra 2

2 -1 Solving Linear Equations and Inequalities Check It Out! Example 1 Continued Let c represent the number of additional cups needed. Model additional cup one cup plus height 3. 25 Holt Algebra 2 + 0. 25 times * number of total additional = height cups c = 14. 00

2 -1 Solving Linear Equations and Inequalities Check It Out! Example 1 Continued Solve. 3. 25 + 0. 25 c = 14. 00 – 3. 25 0. 25 c = 10. 75 0. 25 Subtract 3. 25 from both sides. Divide both sides by 0. 25. c = 43 44 cups fit between the 14 in. shelves. Holt Algebra 2

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

2 -1 Solving Linear Equations and Inequalities Example 2 Continued Check 4(m + 12) = – 36 4(– 21 + 12) 4(– 9) – 36 Holt Algebra 2 – 36

2 -1 Solving Linear Equations and Inequalities Example 2 Continued Solve 4(m + 12) = – 36 Method 2 Distribute before solving. 4 m + 48 = – 36 – 48 Distribute 4. Subtract 48 from both sides. 4 m = – 84 4 m – 84 = 4 4 m = – 21 Holt Algebra 2 Divide both sides by 4.

2 -1 Solving Linear Equations and Inequalities If there are variables on both sides of the equation, (1) simplify each side. (2) collect all variable terms on one side and all constants terms on the other side. (3) isolate the variables as you did in the previous problems. Holt Algebra 2

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

2 -1 Solving Linear Equations and Inequalities Check It Out! Example 3 Solve 3(w + 7) – 5 w = w + 12. Simplify each side by combining – 2 w + 21 = w + 12 like terms. +2 w Collect variables on the right side. 21 = – 12 9 3 Holt Algebra 2 3 w + 12 – 12 = 3 w 3 3=w Add. Collect constants on the left side. Isolate the variable.

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 Example 4 A: Identifying Identities and Contractions Solve 3 v – 9 – 4 v = –(5 + v) – 9 – v = – 5 – v +v +v – 9 ≠ – 5 x Simplify. 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 Example 4 B: Identifying Identities and Contractions Solve 2(x – 6) = – 5 x – 12 + 7 x Simplify. 2 x – 12 = 2 x – 12 – 2 x – 12 = – 12 Identity The solutions set is all real number, or . Holt Algebra 2

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

2 -1 Solving Linear Equations and Inequalities Check It Out! Example 5 Solve and graph x + 8 ≥ 4 x + 17 –x –x 8 ≥ 3 x +17 – 9 ≥ 3 x 3 3 – 3 ≥ x or x ≤ – 3 Holt Algebra 2 Subtract x from both sides. Subtract 17 from both sides. Divide both sides by 3.

2 -1 Solving Linear Equations and Inequalities Lesson Quiz: Part I 1. Alex pays $19. 99 for cable service each month. He also pays $2. 50 for each movie he orders through the cable company’s pay-per-view service. If his bill last month was $32. 49, how many movies did Alex order? 5 movies Holt Algebra 2

2 -1 Solving Linear Equations and Inequalities Lesson Quiz: Part II Solve. x=6 2. 2(3 x – 1) = 34 3. 4 y – 9 – 6 y = 2(y + 5) – 3 y = – 4 4. r + 8 – 5 r = 2(4 – 2 r) all real numbers, or 5. – 4(2 m + 7) = no solution, or Holt Algebra 2 (6 – 16 m)

2 -1 Solving Linear Equations and Inequalities Lesson Quiz: Part III 5. Solve and graph. 12 + 3 q > 9 q – 18 q<5 ° – 2 – 1 Holt Algebra 2 0 1 2 3 4 5 6 7