B.4 Solving Inequalities Algebraically and Graphically, Exams of Algebra

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B.4 Solving Inequalities Algebraically and

Graphically

Properties of Inequalities

The inequality symbols <, ≤, >, and ≥ are used to compare two numbers and to denote subsets of real numbers. For instance, the simple inequality x ≥ 3 denotes all real numbers x that are greater than or equal to 3 As with an equation, you solve an inequality in the variable x by finding all values of x for which the inequality is true. These values are solutions of the inequality and are said to satisfy the inequality. For example, the number 9 is a solution to 5x - 7 > 3x + 9 because when you substitute x = 9, 5(9) - 7 > 3(9) + 9 Substitute x = 9 45 - 7 > 27 + 9 38 > 36 is a true statement.

1. When each side of an inequalities is multiplied or divided by a negative number, the direction of the inequality symbol must be reversed in order to maintain a true statement. -2 < 5 (-3)(-2) > (-3)(5) Reverse sign, Multiply by - 6 > -
2. Two inequalities that have the same solution set are equivalent inequalities. x + 2 < 5 and x < 3 x + 2 - 2 < 5 - 2 Subtract 2 from both sides x < 3

Properties of Inequalities

Properties of Inequalities

Let a, b, c, and d be real numbers.

1. Transitive Property if a < b and b < c then a < c
2. Addition of Inequalities if a < b and c < d then a + c < b + d
3. Addition of a Constant if a < b then a + c < b + c
4. Multiplying by a Constant for c > 0, if a < b then ac < bc for c < 0, if a < b then ac > bc Each of the properties above is true if the symbol < is replaced by ≤ and > is replaced by ≥. 5

Solving a Linear Equality

Algebraic Solution: 5x - 7 > 3x + 9 -3x -3x Subtract -3x from both sides 2x - 7 > 9 +7 +7 Add 7 to both sides 2x > 16 2 2 Divide both sides by 2 x > 8

So, the solution set is all real numbers that are greater than 8. The interval notation for this solution set is (8, ∞)

Solving an Inequality

Example 2 Solve 1 - (3/2)x ≥ x - 4

Solving an Inequality

Graphical Solution 1 - (3/2)x ≥ x - 4 Let y 1 = 1 - (3/2)x and y 2 = x - 4 You can see that the point of intersection is (2, -2). The graph of y 1 lies above the graph of y (^2) to the left of their point of intersection, which implies y 1 ≥ y 2 for all x ≤ 2

Solving a Double Inequality

Example 3 Solve -3 ≤ 6x - 1 and 6x - 1 < 3

What would the interval notation be?

Solving a Double Inequality

Graphical Solution Let y 1 = 6x - 1 y 2 = - y 3 = 3 Use the intersect feature to find that the points of intersection are (-⅓, -3) and (⅔, 3).

The graph of y 1 lies above the graph of y 2 to the right of (-⅓ , -3) AND the graph of y 1 lies below the graph of y 3 to the left of (⅔ , 3). This implies that y 2 ≤ y 1 < y 3 when -⅓ ≤ x < ⅔

Inequalities Involving Absolute Value

Solving an Absolute Value Inequality Let x be a variable or an algebraic expression and let a be a real number such that a ≥ 0.

1. The solutions of |x|< a are all values of x that lie between -a and a |x|< a if and only if -a < x < a Double inequality
2. The solutions of |x|> a are all values of x that are less than -a or greater than a. |x|> a if and only if x < -a or x > a Compound inequality

These rules are also valid if < is replaced by ≤ and > is replaced by ≥.

● What would each of these look like on a number line?

Solving a Double Inequality

Algebraic Solution a. |x - 5|< 2 -2 < x - 5 < 2 Write the double inequality 3 < x < 7 Add 5 to each part

The interval notation for this solution set is (3, 7).

Solving a Double Inequality

Graphical Solution a. |x - 5|< 2 Let y 1 = |x - 5| and y 2 = 2

Use the intersect feature on your graphing calculator.

The points of intersection are (3, 2) and (7, 2). The graph of y 1 lies below the graph of y 2 when 3 < x < 7.

Solving a Double Inequality

Graphical Solution b. |x - 5|> 2 Let y 1 = |x - 5| and y 2 = 2

The points of intersection are (3, 2) and (7, 2). The graph of y 1 lies above the graph of y 2 when x < 3 or when x > 7

Polynomial Inequalities

To solve a polynomial inequality such as x 2 - 2x - 3 > 0, use the fact that a polynomial can change signs only at its zeros (the x-values that make the polynomial equal to zero). These zeros are the critical numbers of the inequality, and the resulting open interval are the test intervals for the inequality. For example, x 2 - 2x - 3 = (x + 1)(x - 3) and has two zeros, x = -1 and x = 3, which divide the real number line into three test intervals: (-∞, -1) , (-1, 3) , and (3, ∞). To solve the inequality x 2 - 2x -3 > 0, you need to test only one value from each test interval.