Download CA 2.1.3_Enhanced_Instruction

Introduction
Solving inequalities is similar to solving equations. To
find the solution to an inequality, use methods similar to
those used in solving equations. In addition to using
properties of equality, we will also use properties of
inequalities to change inequalities into simpler
equivalent inequalities.
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2.1.3: Solving Linear Inequalities
Key Concepts
• The properties of inequality are the rules that allow
you to balance, manipulate, and solve inequalities. The
properties are summarized in the following table.
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2.1.3: Solving Linear Inequalities
Key Concepts, continued
Properties of Inequality
Property
Example
If a > b and b > c, then a > c.
If 10 > 6 and 6 > 2, then 10 > 2.
If a > b, then b < a.
If 10 > 6, then 6 < 10.
If a > b, then –a < –b.
If 10 > 6, then –10 < –6.
If a > b, then a ± c > b ± c.
If 10 > 6, then 10 ± 2 > 6 ± 2.
If a > b and c > 0, then a • c > b • c.
If 10 > 6 and 2 > 0, then 8 • 2 > b • 2.
If a > b and c < 0, then a • c < b • c.
If 10 > 6 and –1 < 0, then 10 • –1 < 6 • –1.
If a > b and c > 0, then a ÷ c > b ÷ c. If 10 > 6 and 2 > 0, then 8 ÷ 2 > 6 ÷ 2.
If a > b and c < 0, then a ÷ c < b ÷ c. If 10 > 6 and –1 < 0, then 10 ÷ –1 < 6 ÷ –1.
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2.1.3: Solving Linear Inequalities
Key Concepts, continued
• When solving more complicated inequalities, first
simplify the inequality by clearing any parentheses.
Do this by either distributing by the leading number or
dividing both sides of the inequality by the leading
number. Then solve the inequality by following the
steps learned earlier, as outlined on the following
slide.
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2.1.3: Solving Linear Inequalities
Key Concepts, continued
Solving Inequalities
1. If a variable appears on both sides of the inequality, choose
which side of the inequality you would like the variable to
appear on.
2. Add or subtract the other variable from both sides of the
inequality using either the addition or subtraction property
of equality.
3. Simplify both expressions.
4. Continue to solve the inequality as you did in earlier examples
by working to isolate the variable.
5. Check that your answer is correct by substituting a value
included in your solution statement into the original inequality
to ensure a true statement.
2.1.3: Solving Linear Inequalities
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Key Concepts, continued
• It is important to remember that when solving
inequalities, the direction of the inequality symbol
(<, >, ≤, or ≥) changes when you divide or multiply by
a negative number. Here’s an example.
• If we had the simple statement that 4 < 8, we know
that we can multiply both sides of the inequality by a
number, such as 3, and the statement will still be true.
4<8
4•3<8•3
12 < 24
Original inequality
Multiply both expressions of the
inequality by 3.
This is a true statement.
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2.1.3: Solving Linear Inequalities
Key Concepts, continued
• We can also divide both sides of the inequality by
a number, such as 2.
4<8
4÷2<8÷2
equation
2<4
Original inequality
Divide both expressions of the
by 2.
This is a true statement.
• Notice what happens when we multiply the
inequality by –3.
4<8
4 • –3 < 8 • –3
–12 < –24
Original inequality
Multiply both expressions of the
inequality by –3.
This is NOT a true statement.
2.1.3: Solving Linear Inequalities
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Key Concepts, continued
• To make this a true statement, change the direction
of the inequality symbol.
–12 > –24
This is a true statement.
• The same is true when dividing by a negative
number; you must change the direction of the
inequality symbol.
4<8
4 ÷ –2 < 8 ÷ –2
–2 < –4
–2 > –4
Original inequality
Divide both expressions of the equation
by –2.
This is NOT a true statement. Change
the direction of the inequality symbol.
This is a true statement.
2.1.3: Solving Linear Inequalities
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Common Errors/Misconceptions
• not understanding why the inequality symbol changes
direction when multiplying or dividing by a negative
number
• replacing inequality symbols with equal signs
• forgetting to switch the inequality symbol when
multiplying or dividing by a negative number
• switching the inequality symbol when subtracting a
number
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2.1.3: Solving Linear Inequalities
Guided Practice
Example 1
Solve the inequality
-3x - 4
7
>5.
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2.1.3: Solving Linear Inequalities
Guided Practice: Example 1, continued
1. Isolate the variable by eliminating the
denominator.
In this inequality, the denominator means “divide by
7.” Eliminate it by performing the inverse operation,
multiplication. Multiply both expressions of the
inequality by 7.
7·
-3x - 4
> 7·5
7
-3x - 4 > 35
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2.1.3: Solving Linear Inequalities
Guided Practice: Example 1, continued
2. Isolate the variable.
Perform the inverse operation of adding 4 to both
expressions of the inequality.
-3x - 4 > 35
-3x
+4 +4
> 39
Now solve.
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2.1.3: Solving Linear Inequalities
Guided Practice: Example 1, continued
3. Divide both sides of the inequality by the
coefficient, –3.
-3x
>
39
-3 -3
x < -13
Notice that the direction of the inequality symbol
changed because we divided by –3.
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2.1.3: Solving Linear Inequalities
Guided Practice: Example 1, continued
4. The solution to the original inequality
-3x - 4
7
> 5 is all numbers less than –13.
To check this, choose any number less than –13 to
show a true statement. Let’s try –20. Be sure to
substitute the value into the original inequality.
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2.1.3: Solving Linear Inequalities
Guided Practice: Example 1, continued
-3x - 4
7
>5
-3(-20) - 4
7
60 - 4
7
56
7
Original inequality
>5
>5
>5
8>5
2.1.3: Solving Linear Inequalities
Substitute –20 for x.
Multiply, then subtract.
Simplify the fraction.
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Guided Practice: Example 1, continued
8 > 5 is a true statement; therefore, all numbers less
than –13 will result in a true statement.
✔
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2.1.3: Solving Linear Inequalities
Guided Practice: Example 1, continued
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2.1.3: Solving Linear Inequalities
Guided Practice
Example 2
Solve the inequality 5x + 4 ≥ 11 – 2x.
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2.1.3: Solving Linear Inequalities
Guided Practice: Example 2, continued
1. Move the variable to one side of the
inequality.
Notice the variable x is in both expressions of the
inequality. Begin by choosing which side you want
your variable to appear on. Just like with equations,
this is a choice, but most people choose to have all
variables on the left side of the inequality. Continue
by adding 2x to both expressions of the inequality.
5x + 4 ³ 11- 2x
+2x
+ 2x
7x + 4 ³ 11
2.1.3: Solving Linear Inequalities
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Guided Practice: Example 2, continued
2. Isolate the variable.
Subtract 4 from both expressions.
7x + 4 ³ 11
-4 -4
7x
³7
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2.1.3: Solving Linear Inequalities
Guided Practice: Example 2, continued
3. Finally, divide both sides of the inequality
by the coefficient of x, 7.
7x
³
7
7
7
x ³1
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2.1.3: Solving Linear Inequalities
Guided Practice: Example 2, continued
4. The solution to the original inequality,
5x + 4 ≥ 11 – 2x, is all numbers greater
than or equal to 1.
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2.1.3: Solving Linear Inequalities
Guided Practice: Example 2, continued
5. To check this solution, choose a number
greater than or equal to 1, such as 2, and
substitute it for all instances of x in the
original inequality.
5x + 4 ≥ 11 – 2x
Original inequality
5(2) + 4 ≥ 11 – 2(2)
Substitute 2 for each instance of x.
10 + 4 ≥ 11 – 4
Simplify each expression.
14 ≥ 7
This is a true statement.
Our check proved true, so we can be sure
that our solution of x ≥ 1 is accurate.
2.1.3: Solving Linear Inequalities
✔
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Guided Practice: Example 2, continued
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2.1.3: Solving Linear Inequalities