# Lesson 16: More Inequalities

28 Jul 2020
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### Lesson 16: More Inequalities

• 1. 𝐏𝐓𝐒 𝟑 Bridge to Calculus Workshop Summer 2020 Lesson 16 More Inequalities “If I were again beginning my studies, I would follow the advice of Plato and start with mathematics." – Galileo Galilei -
• 2. Lehman College, Department of Mathematics Compound Inequalities (1 of 6) Example 1. Write a compound inequality representing the set of all real numbers greater than or equal to 0 and less than 4. Then graph the inequality. Solution. The set can be written as two inequalities: The inequalities can be combined in a single inequality: The graph of this compound inequality is shown below: 𝑥 < 40 ≤ 𝑥 and 0 ≤ 𝑥 < 4
• 3. Lehman College, Department of Mathematics Compound Inequalities (2 of 6) Solution 1. Solve the two inequalities separately: Solution 2. Solve the compound inequality in one go: The graph of the solution is shown below: −2 − 2 and 𝑥 + 2 ≤ 4 Subtract 2𝑥 + 2 − 2 > 𝑥 + 2 > −2 𝑥 + 2 − 2 ≤ 4 − 2 𝑥 > −4 and 𝑥 ≤ 2 Simplify Combine−4 < 𝑥 ≤ 2 −2 < 𝑥 + 2 ≤ 4 Write original inequality Subtract 2−2 − 2 < 𝑥 + 2 − 2 ≤ 4 − 2 Simplify−4 < 𝑥 ≤ 2
• 4. Lehman College, Department of Mathematics Compound Inequalities (3 of 6) Example 3. Solve −3 ≤ 2𝑥 + 1 ≤ 5. Graph the solution. Solution. Solve the compound inequality in one go: The graph of the solution is shown below: −3 ≤ 2𝑥 + 1 ≤ 5 Write original inequality Subtract 1−3 − 1 ≤ 2𝑥 + 1 − 1 ≤ 5 − 1 Simplify−4 ≤ 2𝑥 ≤ 4 Divide by 2− 4 2 ≤ 2𝑥 2 ≤ 4 2 Simplify−2 ≤ 𝑥 ≤ 2
• 5. Lehman College, Department of Mathematics Compound Inequalities (4 of 6) Example 4. Solve −2 < −2 − 𝑥 < 1. Graph the solution. Solution. Solve the compound inequality in one go: The graph of the solution is shown below: −2 < −2 − 𝑥 < 1 Write original inequality Add 2−2 + 2 < −2 − 𝑥 + 2 < 1 + 2 Simplify0 < −𝑥 < 3 Multiply by −10 > 𝑥 > −3 Rewrite inequality−3 < 𝑥 < 0
• 6. Lehman College, Department of Mathematics Compound Inequalities (5 of 6) Example 5. Write a compound inequality representing the set of all real numbers less than −1 or greater than 2. Then graph the inequality. Solution. The set can be written as two inequalities: The graph of this compound inequality is shown below: You may have realized by now that or means union (∪) of sets, while and means intersection (∩) of sets. Example 2. Solve the following compound inequalities: 3𝑥 + 1 < 4 or 2𝑥 − 5 > 7. Then graph the solution. 𝑥 > 2𝑥 < −1 or
• 7. Lehman College, Department of Mathematics Compound Inequalities (6 of 6) Solution. Solve the two inequalities separately: The solution is all real numbers less than 1 or greater than 6. The graph of the solution is shown below: 4 − 1 or 2𝑥 − 5 > 7 Isolate 𝑥3𝑥 + 1 − 1 < 3𝑥 + 1 < 4 2𝑥 − 5 + 5 > 7 + 5 3𝑥 < 3 or 2𝑥 > 12 Simplify Solve for 𝑥 3𝑥 3 < 3 3 or 2𝑥 2 > 12 2 Simplify𝑥 < 1 𝑥 > 6or
• 8. Lehman College, Department of Mathematics Absolute-Value Equations (1 of 4) We can solve an absolute-value equation of the form 𝑥 = 𝑐, where 𝑐 ≥ 0, by finding all points on the number line whose distance from zero is 𝑐. Example 6. Solve the equation 𝑥 = 2. Solution. The equation 𝑥 = 2 means 𝑥 is 2 units from zero. From the graph below, we see that both −2 and 2 are 2 units from zero. Therefore, if 𝑥 = 2, then the solutions are: 𝑥 = −2 or 𝑥 = 2.
• 9. Lehman College, Department of Mathematics Absolute-Value Equations (2 of 4) Example 7. Solve the equation 𝑥 − 3 = 2. Solution. The equation 𝑥 − 3 = 2 means 𝑥 is 2 units away from 3. Graphical solution: on the number line below, find the points that are 2 units from 3: Therefore, if 𝑥 − 3 = 2, then the solutions are: 𝑥 = 1 or 𝑥 = 5. We will now learn how to solve absolute value equations using algebra methods.
• 10. Lehman College, Department of Mathematics Absolute Value Equations (3 of 4) Example 8. Solve the equation 𝑥 = 8. Solution. There are two values of 𝑥 that have absolute value 8. Therefore 𝑥 = 8 or 𝑥 = −8. Example 9. Solve the equation 𝑥 − 2 = 5. Solution. Because 𝑥 − 2 = 5, then the expression 𝑥 − 2 is equal to 5 or −5: The equation has two solutions: 7 and −3. 𝑥 − 2 is positive 𝑥 − 2 = 5 𝑥 − 2 + 2 = 5 + 2 𝑥 = 7 𝑥 − 2 is negative 𝑥 − 2 = −5 𝑥 − 2 + 2 = −5 + 2 𝑥 = −3
• 11. Lehman College, Department of Mathematics Absolute-Value Equations (4 of 4) Example 10. Solve the equation 2𝑥 − 7 − 5 = 4. Solution. First, isolate the absolute-value expression. Because 2𝑥 − 7 = 9, then the expression 2𝑥 − 7 is equal to 9 or −9: 2𝑥 − 7 is positive 2𝑥 − 7 = 9 2𝑥 − 7 + 7 = 9 + 7 2𝑥 = 16 2𝑥 − 7 is negative 2𝑥 − 7 = −9 2𝑥 − 7 + 7 = −9 + 7 2𝑥 = −2 2𝑥 − 7 − 5 = 4 2𝑥 − 7 − 5 + 5 = 4 + 5 2𝑥 − 7 = 9 𝑥 = 8 𝑥 = −1
• 12. Lehman College, Department of Mathematics Absolute-Value Inequalities (1 of 4) Example 11. Solve the inequality 𝑥 > 2. Then graph the solution. Solution. The inequality 𝑥 > 2 means 𝑥 is more than 2 units from zero. From the graph below, we see that both −2 and 2 are 2 units from zero. Therefore, if 𝑥 > 2, then the solutions are: 𝑥 > 2 or 𝑥 < −2. Example 11. Solve the inequality 𝑥 + 5 ≥ 2. Then graph the solution.
• 13. Lehman College, Department of Mathematics Absolute-Value Inequalities (2 of 4) Example 11. Solve the inequality 𝑥 + 5 ≥ 2. Then graph the solution. Solution. The inequality 𝑥 + 5 ≥ 2 means 𝑥 is 2 or more units from −5. The graph of the solution is shown below: 𝑥 + 5 ≥ 2 𝑥 + 5 ≥ 2 𝑥 + 5 − 5 ≥ 2 − 5 𝑥 ≥ −3 𝑥 + 5 ≤ −2 𝑥 + 5 − 5 ≤ −2 − 5 𝑥 ≤ −7 𝑥 + 5 is positive 𝑥 + 5 is negative
• 14. Lehman College, Department of Mathematics Absolute-Value Inequalities (3 of 4) Example 11. Solve the inequality 𝑥 − 4 < 3. Then graph the solution. Solution. The inequality 𝑥 − 4 < 3 means 𝑥 is less that 3 units from 4. That is 𝑥 − 4 < 3 and 𝑥 − 4 > −3. Write this as one inequality: −3 < 𝑥 − 4 < 3 and solve: The graph of the solution is shown below: −3 < 𝑥 − 4 < 3 −3 + 4 < 𝑥 − 4 + 4 < 3 + 4 1 < 𝑥 < 7
• 15. Lehman College, Department of Mathematics Absolute-Value Inequalities (4 of 4) In general, we use the following properties to solve absolute-value inequalities: Inequality Equivalent Form Graph 𝑥 < 𝑐 −𝑐 < 𝑥 < 𝑐 𝑥 ≤ 𝑐 −𝑐 ≤ 𝑥 ≤ 𝑐 𝑥 > 𝑐 𝑥 < −𝑐𝑥 > 𝑐 or 𝑥 ≥ 𝑐 𝑥 ≤ −𝑐𝑥 ≥ 𝑐 or
• 16. Lehman College, Department of Mathematics Inequalities in the Plane (1 of 3) Example 12. Graph the inequality 𝑥 < − 2. Solution. Step 1. Graph the line 𝑥 = −2. This is a vertical line. Since −2 is not included, we use a dashed line. The line 𝑥 = −2 divides the plane into two half-planes. Step 2. Test a point on one side of the line. We usually test the origin (0, 0) 0 < −2 is not true, so the origin does not satisfy the inequality.
• 17. Lehman College, Department of Mathematics Inequalities in the Plane (1 of 3) Example 12. Graph the inequality 𝑥 + 2𝑦 > 8. Solution. Step 1. Graph the line 𝑥 + 2𝑦 = 8. This line has 𝑦-intercept 4 and 𝑥-intercept 8, and passes through the point (2, 3). The line 𝑥 + 2𝑦 = 8 divides the plane in two. Step 2. Test a point on one side of the line. We usually test the origin (0, 0) 0 > 8 is not true, so the origin does not satisfy the inequality.