Transcript Mod1Lesson3Notes

Solving Linear
Inequalities
Included in this presentation:
1. Solving Linear Inequalities
2. Solving Compound Inequalities
3. Linear Inequalities Applications
Steps to Solving Linear Inequalities
1. Simplify both sides of the inequality. (Get rid of parenthesis and put “like
terms” together.
2. Isolate the variable.
When isolating the variable in an inequality you can do the following to both
sides of the inequality to keep equivalent statements.
•
•
•
Add or subtract a number
Multiply or divide by a positive number.
Multiply or divide by a negative number AND reverse the direction of the
inequality sign.
The only difference in solving for a variable in a linear inequality and solving for
a variable in a linear equation is….
WHEN YOU DIVIDE OR MULTIPLY BY A NEGATIVE NUMBER IN AN
INEQUALITY, YOU HAVE TO CHANGE THE INEQUALITY SIGN.
Graphing Solutions
 When you are graphing solutions to inequalities if the
sign is < or > then you put an open circle at the
endpoints.
 If the sign is £ or then
³ you put a closed circle at the
endpoints.
Example 1: Variables on 1 side
Solve the inequality for x, then graph the solution.
3x + 2(x - 5) ³ 4
3x + 2x - 10 ³ 4
5x - 10 ³ 4
5x ³ 14
14
x³
5
Step 1: Distribute the 2.
Step 2: Combine 3x and 2x.
Step 3: Add the 10 to the right
side.
Step 4: Divide by 5 to isolate the
variable.
The graph is on the next slide.
Example 1 continued
Solve the inequality for x, then graph the solution.
3x + 2(x - 5) ³ 4
3x + 2x - 10 ³ 4
5x - 10 ³ 4
5x ³ 14
14
x³
5
This means that any number substituted in for x
in the original statement, that is larger than or
equal to 14/5, will make a true statement.
Example 2: Variables on 1 side
Solve the inequality for x, then graph the solution.
5 - 7(2x + 1) < -2
5 - 14x - 7 < -2
-14 x - 2 < -2
-14 x < 0
0
x>
-14
x>0
Step 1: Distribute the -7.
Step 2: Combine “like” terms
Step 3: Add 2 to both sides.
Step 4: Divide by -14
(WHEN YOU DIVIDE BY A NEGATIVE
NUMBER, YOU MUST CHANGE THE
INEQUALITY SIGN.
The graph is on the next slide.
Example 2 continued
Solve the inequality for x, then graph the solution.
5 - 7(2x + 1) < -2
5 - 14x - 7 < -2
-14 x - 2 < -2
-14 x < 0
0
x>
-14
This means that any number substituted in for x in
the original statement, that is larger than 0, will
x>0
make a true statement.
Example 3: Variables on both sides
Solve the inequality for x, then graph the solution.
2x + 5(x - 3) £ 10x + 3
2x + 5x -15 £ 10x + 3
7x -15 £ 10x + 3
-3x -15 £ 3
-3x £ 18
x ³ -6
Step 1: Distribute the 5.
Step 2: Combine “like” terms
Step 3: Subtract 10x from both sides.
Step 4: Add 15 to both sides
Step 5: Divide by -3.
(WHEN YOU DIVIDE BY A NEGATIVE
NUMBER, YOU MUST CHANGE THE
INEQUALITY SIGN.
The graph is on the next slide.
Example 3: Variables on both sides
Solve the inequality for x, then graph the solution.
2x + 5(x - 3) £ 10x + 3
2x + 5x -15 £ 10x + 3
7x -15 £ 10x + 3
-3x -15 £ 3
-3x £ 18
x ³ -6
This means that any number substituted in for x in
the original statement, that is larger than or equal
to -6, will make a true statement.
Example 4: Variables on both sides
Solve the inequality for x, then graph the solution.
-5x
+ 1 < -2(3x - 4)
3
Step 1: Distribute the -2.
-5x
+ 1 < -6x + 8
3
Step 2: Multiply by 3 to get rid of the fraction.
-5x + 3 < -18x + 24
Step 3: Add 18x to both sides.
13x + 3 < 24
Step 4: Subtract 3 from both sides.
13x < 21
Step 5: Divide by 13.
x<
21
13
The graph is on the next slide.
Example 4: Variables on both sides
Solve the inequality for x, then graph the solution.
-5x
+ 1 < -2(3x - 4)
3
-5x
+ 1 < -6x + 8
3
-5x + 3 < -18x + 24
13x + 3 < 24
13x < 21
x<
21
13
This means that any number substituted in for x in
the original statement, that is smaller than 21/13
will make a true statement.
Compound Inequalities
Compound inequalities are inequalities that are
paired together with “and” or “or.”
The solutions of an “And” compound inequality are bounded by endpoints
and include all the numbers in between the endpoints.
This is an “and” compound inequality
-3 < x < 5
This means that the values of x have to be greater than -3 AND less than
5. The graph of the solutions looks like…..
Solving “AND” Inequalities
Solve the inequality for x, then graph the solution.
-4 £ x - 6 £ 2
+6
2£
+6 +6
x
£8
This means that the values of x have to
between 2 and 8 or equal to 2 and 8 to
make the “AND” inequality true.
When solving AND
inequalities, you need to
isolate the variable in the
middle by adding,
subtracting, multiplying, and
dividing numbers to each
section.
Remember when you divide
by a negative or multiply by a
negative number you will still
need to change the inequality
sign.
Example 5: “AND” Compound Inequality
Solve the inequality for x, then graph the solution.
 10  3 x  4  6
 14  3 x  2
 14
2
x
3
3
Step 1: Subtract 4 from each section.
Step 2: Divide each section by 3.
This means that the values of x have to
between -14/3 and 2/3 to make the “AND”
inequality true.
Example 6: “AND” Compound Inequality
Solve the inequality for x, then graph the solution.
10 £ -5(x + 8) £ 15
Step 1: Distribute the -5 through the middle
section.
10 £ -5x - 40 £ 15
Step 2: Add 40 to each section.
50 £ -5x £ 55
-10 ³ x ³ -11
Step 3: Divide each section by -5.
Remember when you divide by a negative
number you change the inequality signs.
-11 £ x £ -10
Its not necessary to rewrite the inequality but
it may help interpret the answer.
This means that the values of x have to between -11 and -10 or equal to 11 and -10 to make the “AND” inequality true.
Compound Inequalities
Compound inequalities are inequalities that are
paired together with “and” or “or.”
The solutions of an “OR” compound inequality are not bounded by
endpoints and the solutions go in opposite ways.
This is an “or” compound inequality
x < -4 or x>1
This means that the values of x have to be less than or equal to -4 or
greater than or equal to 1. The graph of the solutions looks like…..
Solving “OR” Inequalities
Solve the inequality for x, then graph the solution.
When solving “OR” compound inequalities solve each inequality
separately and then graph the solutions on the same number line.
Step 1:
Subtract 3
Step 2:
Divide by 5
5x + 3 < 8 or
2x + 9 > 23
5x < 5
or
2x > 14
x <1
or
x>7
Step 1:
Subtract 9
Step 2:
Divide by 2
This means that the values of x have to be less than 1 or greater than 7 to make
the “OR” inequality true.
Example 7: Solving “OR” Inequality
Solve the inequality for x, then graph the solution.
Steps:
-3(x - 1) + 7 < -2 or
-2(x + 1) + 8 > 10
1. Distribute -3
2. Combine like terms.
3. Subtract 10
4. Divide by -3
-3x + 3 + 7 < -2
-3x + 10 < -2
or
or
-2x - 2 + 8 > 10
- 2x + 6 > 10
Flip sign of inequality
when divide by a
negative number
-3x < -12
or
- 2x > 4
x>4
or
x < -2
Steps:
1. Distribute -2
2. Combine like terms.
3. Subtract 6
4. Divide by -2
Flip sign of inequality
when divide by a
negative number
This means that the values of x have to be greater than 4 or less than -2 to
make the “OR” inequality true.
Linear Inequalities Applications
When given word problems that can be represented with
linear inequalities, follow the following steps.
1. Define the variable.
2. Write the inequality the represents the situation.
3. Solve the inequality to state the answer.
Example 8: Linear Inequalities Applications
The product of 5 and a number, x, decreased by 9 is at
most 7. What are the possible solutions of x?
Let “x” represent the number we don’t know.
Product means multiply, decreased by 9 means subtract 9 and at most means
that it can be equal to 7 or smaller than 7….so
5x-9 £ 7
5x - 9 £ 7
5x £ 16
16
x£
5
The possible x values that will be solutions
of this inequality are any numbers that are
smaller than 16/5 or equal to 16/5.
Example 9: Linear Inequalities Applications
The product of -2 and the quantity of the sum of a number,
x, and 4 is either less than -9 or more than 5 . What are the
possible solutions of x?
Let “x” represent the number we don’t know.
Product means multiply, quantity means that I have to have the sum of x and 4
and then multiply that to the -2.….so
-2(x + 4) < -9 or - 2(x + 4) > 5
-2x - 8 < -9 or -2x-8 > 5
-2x < -1
1
x>
2
or
or
-2x >13
-13
x<
2
The possible x values that will be solutions
of this inequality are any numbers that are
smaller than -13/2 or greater than ½.
Example 10: Linear Inequalities Applications
The quantity of twice a number, x, increased by 7 is
between 2 and 28 . What are the possible solutions of x?
Let “x” represent the number we don’t know.
Twice a number means multiply by 2, increased by 7 means add 7, between
means the quantity is in between those 2 numbers but can’t equal those
numbers….so
2 < 2x + 7 < 28
-5 < 2x < 21
-5
21
<x<
2
2
The possible x values that will be solutions
of this inequality are any numbers that are
larger than -5/2 AND less than 21/2.
Example 10: Linear Inequalities Applications
You have $130 and go shopping. You pay $7.25 for lunch
and the rest of the money you have to buy shirts. If shirts
cost $15 a piece, what is the inequality that would
represent the number of shirts you could buy? How many
shirts could you buy?
Let “x” represent the number of shirts that we buy.
15x £130 - 7.25
15x £122.75
x £ 8.1833333
15 times the quantity of shirts you buy has to be
less than or equal to the amount of money you
have after lunch. So it says to spend less than
or equal to the amount of money you have left
you would have to buy 8.183333 shirts….which
means you can buy 8 shirts or less. You can not
buy part of a shirt so your solutions would be
limited to 8, 7, 6, 5, 4, 3, 2, 1, or 0 shirts.