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

* Warm Up Simplify each expression. 1. 2x + 5 – 3x 2. –(w – 2) 3. 6(2 – 3g) 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.

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

An equation is a mathematical statement that two expressions are equivalent. The solution set of an equation is the value or values of the variable that make the equation true. A linear equation in one variable can be written in the form ax = b, where a and b are constants and a ≠ 0.

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.

To isolate the variable, perform the inverse or opposite of every operation in the equation on both sides of the equation Do inverse operations in the reverse order of operations.

Example 1: Consumer Application 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?

Let m represent the number of additional minutes that Nina used. Example 1 Continued Let m represent the number of additional minutes that Nina used. Model additional minute charge number of additional minutes monthly charge total charge plus times = = 14.56 12.95 + 0.07 * m

Example 1 Continued Solve. 12.95 + 0.07m = 14.56 Subtract 12.95 from both sides. –12.95 –12.95 0.07m = 1.61 Divide both sides by 0.07. 0.07 0.07 m = 23 Nina used 23 additional minutes.

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. You fill in the rest of the information. Model additional cup height number of additional cups total height plus times one cup = = + * 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 * Example 2: Solving Equations with the Distributive Property Solve 4(m + 12) = –36 Method 1 Divide both sides by 4 first, then solve.

* Example 2 Continued Solve 4(m + 12) = –36 Method 2 Distribute the 4 first, then solve.

*  Example 2 Continued Check your answer 4(m + 12) = –36 4(____+ 12) –36 4(___) –36 ___ –36 

* Check It Out! Example 2a Solve 3(2 –3p) = 42. Method 1 Divide by 3 first.

Check It Out! Example 2a Continued * Check It Out! Example 2a Continued Solve 3(2 – 3p) = 42 . Method 2 Distribute the 3 first.

* Check It Out! Example 2b Solve –3(5 – 4r) = –9. Choose your method; either divide by -3 or distribute -3.

If there are variables on both sides of the equation: simplify each side collect all variable terms on one side and all constants terms on the other side 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.

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

Example 4B: Identifying Identities and Contractions * Example 4B: Identifying Identities and Contractions Solve 2(x – 6) = –5x – 12 + 7x.

* Check It Out! Example 4a Solve 5(x – 6) = 3x – 18 + 2x. The equation has no solution. The solution set is the empty set, which is represented by the symbol .

* Check It Out! Example 4b Solve 3(2 –3x) = –7x – 2(x –3).

An inequality is a statement that compares two expressions by using the symbols <, >, ≤, ≥, or ≠. The graph of an inequality is the solution set, the set of all points on the number line that satisfy the inequality. The properties of equality are true for inequalities, with one important difference. If you multiply or divide both sides by a negative number, you must reverse the inequality symbol.

These properties also apply to inequalities expressed with >, ≥, and ≤.

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 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 a ≤ –2

* Check It Out! Example 5 Solve and graph x + 8 ≥ 4x + 17.