Lehman PTS-3 Summer 2020

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.