Download Section 2.7 Linear Inequalities in One Variable – Part I

Math 123 – Section 2.7 – Solving Linear Inequalities – Part I − Page 1
Section 2.7
Linear Inequalities in One Variable – Part I
I. Addition Property of Inequality
A.
If a < b, then a + c < b + c. This also is true for >, <, or >.
B.
In words, adding (or subtracting) the same thing from both sides of an inequality
gives us an equivalent inequality.
II. Multiplication Property of Inequality
A.
If a < b and c > 0, then ac < bc. This also is true for >, <, or >.
B.
If a < b and c < 0, then ac > bc. This also is true for >, <, or >.
C.
In words, when you multiply (or divide) both sides of an inequality by the same
positive number, we get an equivalent inequality. However, if we multiply (or divide)
by the same negative number, we have to switch the inequality sign.
III. Solving Linear Inequalities in One Variable
A. Procedure
1.
Multiply each side by the LCD to clear any fractions.
2.
Simplify each side.
3.
Use the Addition Property to get all the variable terms on one side, the
constant terms on the other.
4.
Use the Multiplication Property to isolate the variable.
5.
Graph the solution set.
6.
Write the solution using Set-Builder Notation.
B. Graphing solution sets
1.
For < and >, use a parentheses as a boundary.
2.
For < and >, use a square bracket as a boundary.
3.
When solving, always get the variable terms on the LHS of the inequality.
4.
Once the variable is isolated, the graph will go in the same direction as the
inequality sign.
C. Writing the solution set in interval notation.
1.
We can never reach infinity (positive or negative), so we will always use a
parentheses there.
2.
Whatever our boundary on the graph, that is what we will use on the interval.
3.
The interval always goes from smallest number (on the left) to the largest
number (on the right).
D. Examples – Solve, graph the solution set and write the solution set as an interval.
1.
3<x+8
First, subtract x from each side to get the variable term on the LHS:
−x + 3 < 8
Now subtract 3 from both sides to isolate the variable term.
−x < 5
Next divide both sides by −1 to isolate the variable. What do you have to do
with the inequality sign when you divide by a negative?
x > −5
So the arrow of our graph will go to the right since we have ">", with the
boundary being a “(“.
Answer:
{x| x > −5}
(
−5
(− 5, ∞ )
© Copyright 2009 by John Fetcho. All rights reserved
Math 123 – Section 2.7 – Solving Linear Inequalities – Part I − Page 2
2.
a – 3 < 10
First, add 3 to each side to get:
a < 13
Since we have "<", the arrow of our graph will go to the left, with the boundary
being a “]”.
Answer:
{a | a < 13}
]
13
(− ∞,13]
3.
Now you try one: 11 < x – 2
Answer:
{x | x > 13}
(
13
(13, ∞ )
3
y>6
4
First, multiply both sides by 4 to get rid of the fraction.
−3y > 24
Now divide both sides by –3, remembering to switch the “>” sign.
y < −8
Since we have "<", the arrow of our graph will go to the left.
Answer:
{y | y < −8}
)
−8
(− ∞,−8)
4. −
5
y > −11
6
First, multiply both sides by 6 to get rid of the fraction.
24 – 5y > −66
Now subtract 24 from each side to get:
−5y > −90
Next, divide both sides by –5 to get (don’t forget to change the “>” sign)
y < 18
Since we have "<", the arrow of our graph will go to the left.
Answer:
{y | y < 18}
)
18
(− ∞,18)
5. 4 −
6.
Now you try one: 8 – 6(x – 3) < −4x + 12
Answer:
{x | x > 7}
[
7
[7, ∞ )
© Copyright 2009 by John Fetcho. All rights reserved
Math 123 – Section 2.7 – Solving Linear Inequalities – Part I − Page 3
3x − 5 < 3(x − 2)
Begin by distributing the 3 on the RHS.
3x − 5 < 3x − 6
Subtract 3x from both sides to get the variable terms on the LHS.
−5 < −6
The variable term disappears and we get a false statement.
7.
When this happens, we have no solution and we write this as ∅ .
Answer:
∅
8.
x + 4 < x + 10
We begin by subtracting x from both sides to get the variable terms on the
LHS.
4 < 10
The variable disappears and we get a true statement.
When this happens, any number we substitute in for x will make this a true
statement, so all the real numbers is the solution set, we write as ℝ .
Answer:
{x | x is a real number} or {x | x ∈ ℝ } (− ∞, ∞ )
9.
Now you try one:
Answer:
∅
7x < 7(x − 2)
© Copyright 2009 by John Fetcho. All rights reserved