Math 051 Intermediate Algebra
Lecture Notes for chapter 9.1,9.3-9.5 (audio dated 8/19/07 at 6:46 pm, 30 minutes long) Compound Inequalities
Absolute Value Equations/Inequalities Linear Inequalities in Two Variables Compound Inequalities
To solve compound inequalities using AND, we set up the intersection of the solution sets of the inequalities. To solve the inequalities using OR, we combine the solution sets of the inequalities.
Ex 1. 7x + 2x ≥ –27 and –8x + 5x ≥ –24
We solve each inequality first. 7x + 2x ≥ –27 9x ≥ –27 x ≥ –3 –8x + 5x ≥ –24 –3x ≥ –24 x ≤ 8
The intersection of x ≥ –3 and x ≤ 8 is –3 ≤ x ≤ 8.
The interval notation is [–3, 8].
Ex 2. 6x < 12 or 3x > 3
We solve each inequality first. 6x < 12
x < 2
3x > 3 x > 1
Solving Absolute Value Equations
Let us look at an example of an absolute value equation, | w | = 4. Absolute value equation means the distance between the number w and 0 is 4 units. So, we draw a number line and mark a point 4 units to the left and a point 4 units to the right of 0 each with a closed circle.
We have two solutions for w, –4 or 4.
In general, the absolute value equation | w | = a, a > 0 has two solutions w = a or w = –a. Solve the absolute value equation.
Ex 1. |4x + 3| – 5 = 7
First we need to isolate the absolute value term on the left side. |4x + 3| = 12
Let w be 4x + 3 and a = 12. So, we have two solutions for w. w = –12 or w = 12.
4x + 3 = –12 or 4x + 3 = 12 Next we will solve equation for x. 4x + 3 = –12 or 4x + 3 = 12 4x = –15 or 4x = 9
x = –15/4 = –3.75 or x = 9/4 = 2.25 The solution set is {–3.75, 2.25}
Ex 2. 2 1 1 4
3 t +
− =
We need to isolate the absolute value term.
2 1 5 3 t + = If we let 2 1 3 t
w= + and a = 5, we will write the two solutions for w.
5 or 5 2 1 2 1 5 or 5 3 3 w w t t = − = + + = − =
Next we will multiply 3 to both sides of each equation and solve for t. 2 1 15 or 2 1 15 2 16 or 2 14 8 or 7 t t t t t t + = − + = = − = = − =
The solution set is {–8, 7}. Ex 3. 6 |3 – 4x| + 20 = 14
We will isolate the absolute value term first. 6 |3 – 4x| = – 6
|3 – 4x | = –1
Since absolute value is always positive or equal to 0, there is no solution. Ex 4. |9y + 5| = |6y – 4|
Since we have two square roots in the equation and we understand the roles of w and a in the formula, we will simply write the equation as
9y + 5 = (6y – 4) or 9y + 5 = –(6y – 4) 3y = –9 or 9y + 5 = –6y + 4 y = –3 or 15y = –1
y = –3 or y = –1/15 The solution set is {–3, –1/15}
Solving Absolute Value Inequalities
Let us look at an example of an absolute value inequality, | w | < 3. The absolute value inequality means the distance between the number w and 0 is less than 3 units. So, we draw a number line, mark a point 3 units to the left and a point 3 units to the right of 0 each with an open circle, and shade the part of the line between –3 and 3.
The solution to the inequality is –3 < x < 3.
In general, the absolute value inequality | w | < a, a > 0 has solution of –a < w < a. The absolute value inequality | w | ≤ a, a > 0 has solution of –a ≤ w ≤ a.
Solve the absolute value inequality. Write the solution in interval notation. Ex 1. |2x – 7| + 3 < 16
First we need to isolate the absolute value term. |2x – 7| < 13
Let w = 2x – 7 and a = 13. We will write the solution as –13 < w < 13. –13 < w < 13
–13 < 2x – 7 < 13 Now we solve for x. –6 < 2x < 20
–3 < x < 10
The solution in interval notation is (–3, 10). Ex 2. –3|6 – x| + 1 ≥ –5
First we isolate the absolute value term. –3|6 – x| ≥ –6
|6 – x| ≤ 2
Since we understand the roles of w and a, we will write the solution as –a ≤ w ≤ a. –2 ≤ 6 – x ≤ 2
–8 ≤ –x ≤ –4 8 ≥ x ≥ 4
Or, we can write it backward as 4 ≤ x ≤ 8. The solution in interval notation is [4, 8].
Next we will look at an example of an absolute value inequality, | w | ≥ 5. The absolute value
inequality means the distance between the number w and 0 is greater than or equal to 5 units. So, we draw a number line, mark a point 5 units to the left and a point 5 units to the right of 0 each with a closed circle, and shade the line to the left of –5 and right of 5.
The solution to the inequality is w ≤ –5 or w ≥ 5.
In general, the absolute value inequality | w | ≥ a, a > 0 has solutions of w ≤ –a or w ≥ a. The absolute value inequality | w | > a, a > 0 has solutions, w < –a or w > a.
Solve the absolute value inequality. Write the solution in interval notation. Ex 3. 2|x – 6| – 1 > 9
First we will isolate the absolute value term. 2|x – 6| > 10
|x – 6| > 5
We let w = x – 6 and a = 5. The solutions are w < –5 or w > 5
x – 6 < –5 or x – 6 > 5 x < 1 or x > 11
The solution in interval notation is (–∞, 1) U (11, ∞). Ex 4. 3|1 – 2x| + 5 ≥ 17
We will isolate the absolute value term. 3|1 – 2x| ≥ 12
|1 – 2x| ≥ 4
We understand the roles of w and a and write the solutions. 1 – 2x ≤ –4 or 1 – 2x ≥ 4
–2x ≤ –5 or –2x ≥ 3 x ≥ 2.5 or x ≤ –1.5
The solution in interval notation is (–∞,–1.5] U [2.5, ∞). Ex 6. |3x – 1| + 8 < 2
First we need to isolate the absolute value term. |3x – 1| < –6
Since absolute value is always positive or equal to 0, the inequality is false. So, there is no solution for this problem.
Ex 7. |x – 9| + 5 > 1
We need to isolate the absolute value term first. |x – 9| > –4
Since absolute value is always positive, the inequality is always true. So, the solution is the set of all real numbers, i.e. (–∞, ∞).
Linear Inequalities in Two Variables
Ex 1. Graph 2x – 3y > 6.
Method 1 - First we change the inequality to an equation and find the x-and y-intercepts. 2x – 3y = 6
x y 0 –2 3 0
Since the inequality is a strict inequality, we will graph the line with a dotted or broken line.
Next we select a point, say (0, 0) as a test point. We substitute 0 for x and 0 for y into the original inequality and determine if it is true or false.
2(0) – 3(0) > 6 0 > 6 false!
We will shade the half of the plane not including the test point, i.e. below the dotted line.
Method 2 – First we change the inequality to an equation and find the x-and y-intercepts. 2x – 3y = 6
x y 0 –2 3 0
2x – 3y > 6 3 2 6 2 2 3 y x y x − > − + < −
We will draw the line as a dotted line because it is a strict inequality. The inequality 2 2
3
y< x− means
the y-coordinates of the points will less than the y-coordinates of the points on the line. So, we shade the half plane below the dotted line.
Ex 2. Solve 2 2
5
y≥ x+
First we will find the x- and y-intercepts of the corresponding equation 2 2
5
y= x+ .
x y
0 2
–5 0
