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

Warm Up Simplify each expression. 1. 2x + 5 – 3x –x + 5 2. –(w – 2) 3. 6(2 – 3g) 12 – 18g 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 –2x < –6? Explain. No; when 2 is substituted for x, the inequality is false: –4 < –6

Objectives Solve linear equations using a variety of methods. Solve linear inequalities.

Vocabulary equation solution set of an equation linear equation in one variable identify contradiction inequality

Linear Equations in One variable Nonlinear Equations 4x = 8 + 1 = 32 3x – = –9 + 1 = 41 2x – 5 = 0.1x +2 3 – 2x = –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.

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?

Check It Out! Example 1 Continued Let c represent the number of additional cups needed. Model additional cup height number of additional cups total height plus times one cup = = 14.00 3.25 + 0.25 * c

Check It Out! Example 1 Continued Solve. 3.25 + 0.25c = 14.00 –3.25 –3.25 Subtract 3.25 from both sides. 0.25c = 10.75 Divide both sides by 0.25. 0.25 c = 43 44 cups fit between the 14 in. shelves.

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 Divide both sides by 4. 4 4 m + 12 = –9 –12 –12 Subtract 12 from both sides. m = –21

 Example 2 Continued Check 4(m + 12) = –36 4(–21 + 12) –36 4(–9) –36 4(–21 + 12) –36 4(–9) –36 –36 –36 

Example 2 Continued Solve 4(m + 12) = –36 Method 2 Distribute before solving. 4m + 48 = –36 Distribute 4. –48 –48 Subtract 48 from both sides. 4m = –84 = 4m –84 4 4 Divide both sides by 4. m = –21

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.

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

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

You have solved equations that have a single solution 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.

Example 4A: Identifying Identities and Contractions Solve 3v – 9 – 4v = –(5 + v). 3v – 9 – 4v = –(5 + v) –9 – v = –5 – v Simplify. + v + v –9 ≠ –5 x Contradiction The equation has no solution. The solution set is the empty set, which is represented by the symbol .

Example 4B: Identifying Identities and Contractions Solve 2(x – 6) = –5x – 12 + 7x. 2(x – 6) = –5x – 12 + 7x Simplify. 2x – 12 = 2x – 12 –2x –2x –12 = –12  Identity The solutions set is all real number, or .

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

Check It Out! Example 5 Solve and graph x + 8 ≥ 4x + 17. x + 8 ≥ 4x + 17 –x –x Subtract x from both sides. 8 ≥ 3x +17 Subtract 17 from both sides. –17 –17 –9 ≥ 3x –9 ≥ 3x Divide both sides by 3. 3 3 –3 ≥ x or x ≤ –3

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

Lesson Quiz: Part II Solve. 2. 2(3x – 1) = 34 3. 4y – 9 – 6y = 2(y + 5) – 3 4. r + 8 – 5r = 2(4 – 2r) 5. –4(2m + 7) = (6 – 16m) x = 6 y = –4 all real numbers, or  no solution, or

Lesson Quiz: Part III 5. Solve and graph. 12 + 3q > 9q – 18 –2 –1 0 1 2 3 4 5 6 7 °