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

2. Do Now Simplify each expression. 1. 2x + 5 – 3x 2. –(w – 2)
3. 6(2 – 3g)
Graph on a number line.
4. t > –2
5. Is 2 a solution of the inequality –2x < –6? Explain.

3. Objectives TSW solve linear equations using a variety of methods.
TSW solve linear inequalities.

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

5. 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.

6. 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.

8. 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.

9. 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?

10. Example 2
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?

11. Example 3: Solving Equations with the Distributive Property
Solve 4(m + 12) = –36
Method 2
Method 1

12. Example 4
Solve 3(2 –3p) = 42.
Method 1
Method 2

13. Example 5
Solve –3(5 – 4r) = –9.
Method 1
Method 2

14. 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.

15. Example 6: Solving Equations with Variables on Both Sides
Solve 3k– 14k + 25 = 2 – 6k – 12.

16. Example 7
Solve 3(w + 7) – 5w = w + 12.

17. 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.

18. Example 8: Identifying Identities and Contractions
Solve 3v – 9 – 4v = –(5 + v).

19. Example 9: Identifying Identities and Contractions
Solve 2(x – 6) = –5x – x.

20. Example 10
Solve 5(x – 6) = 3x – x.

21. Example 11
Solve 3(2 –3x) = –7x – 2(x –3).

22. 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 of an inequality by a negative number, you must reverse the inequality symbol.

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

24. To check an inequality, test
the value being compared with x
a value less than that, and
a value greater than that.
Helpful Hint

25. Example 12: Solving Inequalities
Solve and graph 8a –2 ≥ 13a + 8.

26. Example 13
Solve and graph x + 8 ≥ 4x + 17.

27. 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?

28. 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)

29. Lesson Quiz: Part III
5. Solve and graph.
12 + 3q > 9q – 18