Ch 4 Notes
4-1 Inequalities and their Graphs
The solution of an inequality is any number that makes the inequality true. Generally, we will have solutions that look like x > 6 which would mean that makes the inequality true. Generally, we will have solutions that look like x > 6 which would mean that any number bigger than 6 is a solution. 7 is a solution, 8 is a solution, 9, 10, 11, 12, and so on are all solutions, as would be 6.1, 6.01, 6.001 and so on, as long as it is greater than 6. Notice, with an inequality, there is more than one answer; in fact, there are an infinite number of answers in most cases.
Example: Determine if 6 is a solution to 3x + 5 > 26
Simply put 6 in place of the “x” and see if the result is true. If it is, then 6 is a solution, if it isn’t true, then 6 is not a solution.
3(6) + 5 > 26
18 + 5 > 26
23 > 26 is NOT true, therefore 6 is NOT a solution.
Generally we graph the solution to an inequality since it would be impossible to list all the numbers. The graph has three parts:
The Location (What number is x greater or less than?)
The Mark (Is it open for <, > or closed for <, > ?)
The Arrow (Which direction should the answers fall?)
Example: Graph the Inequality -5 > x
It’s always easier to put the variable on the left. If we flip the inequality, we also have to flip the inequality sign, so it becomes x < -5
The Location is -5, the mark is open, and the arrow is pointing to the left.
4-2 Solving Inequalities using Addition and Subtraction
If you have two inequalities which have the same solution, then we call them equivalent inequalities. Similar to equivalent equations, equivalent inequalities will simplify to the same thing.
Example: x + 2 > 3 and x + 7 > 8 both simplify to the same thing
x > 1 and x > 1
Notice, to simplify, we follow the same process as if we had an equal sign. We
added or subtracted the same thing to both sides of the inequality.
As with equations, you can always check your solution to an inequality by placing the value back into the ORIGINAL inequality and checking to see that it works.
4-3 Solving Inequalities using Multiplication and Division
Inequalities can also be solved using multiplication and division, with one additional rule.
Anytime you multiply or divide by a negative, you must change the inequality sign.
Example: Solve: – ½ x > 6
Multiply both sides by the reciprocal:
(-2/1) (– ½ x) > (-2/1) (6)
Two negatives make a positive and the reciprocals cancel out
X < (-2) (6)
Notice the inequality sign changes to less than instead of greater than
X < -12
You do not need to change the inequality sign simply if there is a negative in the inequality, only if you multiply or divide by a negative.
Example: 5x < -25
Divide by +5 to get x alone
X < -5
The sign stays the same because I divided by a positive number
4-4 Solving Multi-Step Inequalities
Solving Multi-Step Inequalities is done the same as solving multi-step equations, simply follow this easy process:
1. Combine any like terms on each side separately
2. If you have a fraction, multiply everything in the equation by the denominator to get rid of the fraction
3. Distribute if needed (if you have parenthesis)
4. Combine like terms again
5. Follow SADMEG to undo the operations on the side of the variable, remember the rule if you multiply or divide by a negative number (You have to switch the inequality sign)
Example:
2x + ½ (4x + 8) – x < 32
X + ½ (4x + 8) < 32 Combine like terms
2x + (4x + 8) < 32 Multiplied by 2 to get rid of fraction
6x + 8 < 32 Combined like terms
6x < 24 Subtracted 8
x < 4 Divided by 6
Be sure to tell me what you did at each step if you don’t show your work!!! (Your answer should look like that above)
Let’s try another:
4(3x – 1) > 2(x + 3)
12x -4 > 2x + 6 Distribute both sides
12x > 2x + 10 Added 4 to both sides to undo the “-4” on the left
10x > 10 Subtracted 2x from both sides to undo the “2x”
x > 1 Divided both sides by 10
Check it to make sure it works: use something bigger than 1, such as 2
4(3*2 – 1) > 2(2 + 3)
4(6 – 1) > 2 (5)
4 (5) > 10
20 > 10 That is true! So we did it correctly!
4-5 Compound Inequalities
A Compound Inequality is made up of two or more inequalities joined by the word “and” or the word “or”. A compound inequality with the word “and” is a Conjunction, while the inequality with the word “or” is a disjunction.
The graph of a conjunction has arrows that meet in the middle, coming toward each other.
-3 < x < 12 Read as -3 < x AND x < 12 (Notice, it’s true that -3 < 12, x is in the middle)
A disjunction has arrows that go away from each other on the graph.
X < -2 or x > 15 (Notice, you can’t keep the signs right if you put x in the middle, it would look like -2 > x > 15, but -2 is NOT > 15 thus it must be a disjunction)
To solve a compound inequality, you can break it into two separate statements, then put it back together, when possible, at the end.
Example: -1 < 4x + 7 < 11
Break it into -1 < 4x + 7 and 4x + 7 < 11
Solve each:
-8 < 4x and 4x < 4
-2 < x and x < 1
Put it back into one statement:
-2 < x < 1
Sets:
Sets are simply groups of numbers, when we look at the intersection of sets, we want to know what do the two sets have in common.
Set A = { 1, 2, 5, 8, 10}
Set B = { 3, 4, 5, 9, 10, 12, 14}
The intersection of the two sets, A∩B, would be = {5, 10} since they have 5 and 10 in common.
The Union of the two sets AUB would be = {1, 2, 3, 4, 5, 8, 9, 10, 12, 14} since together, they have those elements (numbers) in the set.
If two sets have nothing in common, the intersection would be empty; this is called the null set. It is indicated with a { } or a zero with a back slash through it.
4-6 Absolute Value Equations and Inequalities
When solving Absolute Value problems, isolate the absolute value first. Then split the Absolute Value into two separate equations or inequalities, make one positive and one negative.
Example: -2│x + 3│ - 5 = -7
Add the 5 first -2│x + 3│ = -2
Divide by -2 │x + 3│ = 1
Now split it into two statements, one positive, one negative
(x + 3) = 1 and –(x + 3) = 1
Solve each x = -2 and x = -4
Check each solution in the original:
-2│x + 3│ - 5 = 7
-2│-2 + 3│ - 5 = -7 and -2│-4 + 3│ - 5 = -7
-2│1│- 5 = -7 and -2 │-1│- 5 = -7
(-2) (1) – 5 = -7 and (-2) (1) – 5 = -7
-2 – 5 = -7 and -2 -5 = -7
-7 = -7 -7 = -7
It checks!
The solution of an inequality is any number that makes the inequality true. Generally, we will have solutions that look like x > 6 which would mean that makes the inequality true. Generally, we will have solutions that look like x > 6 which would mean that any number bigger than 6 is a solution. 7 is a solution, 8 is a solution, 9, 10, 11, 12, and so on are all solutions, as would be 6.1, 6.01, 6.001 and so on, as long as it is greater than 6. Notice, with an inequality, there is more than one answer; in fact, there are an infinite number of answers in most cases.
Example: Determine if 6 is a solution to 3x + 5 > 26
Simply put 6 in place of the “x” and see if the result is true. If it is, then 6 is a solution, if it isn’t true, then 6 is not a solution.
3(6) + 5 > 26
18 + 5 > 26
23 > 26 is NOT true, therefore 6 is NOT a solution.
Generally we graph the solution to an inequality since it would be impossible to list all the numbers. The graph has three parts:
The Location (What number is x greater or less than?)
The Mark (Is it open for <, > or closed for <, > ?)
The Arrow (Which direction should the answers fall?)
Example: Graph the Inequality -5 > x
It’s always easier to put the variable on the left. If we flip the inequality, we also have to flip the inequality sign, so it becomes x < -5
The Location is -5, the mark is open, and the arrow is pointing to the left.
4-2 Solving Inequalities using Addition and Subtraction
If you have two inequalities which have the same solution, then we call them equivalent inequalities. Similar to equivalent equations, equivalent inequalities will simplify to the same thing.
Example: x + 2 > 3 and x + 7 > 8 both simplify to the same thing
x > 1 and x > 1
Notice, to simplify, we follow the same process as if we had an equal sign. We
added or subtracted the same thing to both sides of the inequality.
As with equations, you can always check your solution to an inequality by placing the value back into the ORIGINAL inequality and checking to see that it works.
4-3 Solving Inequalities using Multiplication and Division
Inequalities can also be solved using multiplication and division, with one additional rule.
Anytime you multiply or divide by a negative, you must change the inequality sign.
Example: Solve: – ½ x > 6
Multiply both sides by the reciprocal:
(-2/1) (– ½ x) > (-2/1) (6)
Two negatives make a positive and the reciprocals cancel out
X < (-2) (6)
Notice the inequality sign changes to less than instead of greater than
X < -12
You do not need to change the inequality sign simply if there is a negative in the inequality, only if you multiply or divide by a negative.
Example: 5x < -25
Divide by +5 to get x alone
X < -5
The sign stays the same because I divided by a positive number
4-4 Solving Multi-Step Inequalities
Solving Multi-Step Inequalities is done the same as solving multi-step equations, simply follow this easy process:
1. Combine any like terms on each side separately
2. If you have a fraction, multiply everything in the equation by the denominator to get rid of the fraction
3. Distribute if needed (if you have parenthesis)
4. Combine like terms again
5. Follow SADMEG to undo the operations on the side of the variable, remember the rule if you multiply or divide by a negative number (You have to switch the inequality sign)
Example:
2x + ½ (4x + 8) – x < 32
X + ½ (4x + 8) < 32 Combine like terms
2x + (4x + 8) < 32 Multiplied by 2 to get rid of fraction
6x + 8 < 32 Combined like terms
6x < 24 Subtracted 8
x < 4 Divided by 6
Be sure to tell me what you did at each step if you don’t show your work!!! (Your answer should look like that above)
Let’s try another:
4(3x – 1) > 2(x + 3)
12x -4 > 2x + 6 Distribute both sides
12x > 2x + 10 Added 4 to both sides to undo the “-4” on the left
10x > 10 Subtracted 2x from both sides to undo the “2x”
x > 1 Divided both sides by 10
Check it to make sure it works: use something bigger than 1, such as 2
4(3*2 – 1) > 2(2 + 3)
4(6 – 1) > 2 (5)
4 (5) > 10
20 > 10 That is true! So we did it correctly!
4-5 Compound Inequalities
A Compound Inequality is made up of two or more inequalities joined by the word “and” or the word “or”. A compound inequality with the word “and” is a Conjunction, while the inequality with the word “or” is a disjunction.
The graph of a conjunction has arrows that meet in the middle, coming toward each other.
-3 < x < 12 Read as -3 < x AND x < 12 (Notice, it’s true that -3 < 12, x is in the middle)
A disjunction has arrows that go away from each other on the graph.
X < -2 or x > 15 (Notice, you can’t keep the signs right if you put x in the middle, it would look like -2 > x > 15, but -2 is NOT > 15 thus it must be a disjunction)
To solve a compound inequality, you can break it into two separate statements, then put it back together, when possible, at the end.
Example: -1 < 4x + 7 < 11
Break it into -1 < 4x + 7 and 4x + 7 < 11
Solve each:
-8 < 4x and 4x < 4
-2 < x and x < 1
Put it back into one statement:
-2 < x < 1
Sets:
Sets are simply groups of numbers, when we look at the intersection of sets, we want to know what do the two sets have in common.
Set A = { 1, 2, 5, 8, 10}
Set B = { 3, 4, 5, 9, 10, 12, 14}
The intersection of the two sets, A∩B, would be = {5, 10} since they have 5 and 10 in common.
The Union of the two sets AUB would be = {1, 2, 3, 4, 5, 8, 9, 10, 12, 14} since together, they have those elements (numbers) in the set.
If two sets have nothing in common, the intersection would be empty; this is called the null set. It is indicated with a { } or a zero with a back slash through it.
4-6 Absolute Value Equations and Inequalities
When solving Absolute Value problems, isolate the absolute value first. Then split the Absolute Value into two separate equations or inequalities, make one positive and one negative.
Example: -2│x + 3│ - 5 = -7
Add the 5 first -2│x + 3│ = -2
Divide by -2 │x + 3│ = 1
Now split it into two statements, one positive, one negative
(x + 3) = 1 and –(x + 3) = 1
Solve each x = -2 and x = -4
Check each solution in the original:
-2│x + 3│ - 5 = 7
-2│-2 + 3│ - 5 = -7 and -2│-4 + 3│ - 5 = -7
-2│1│- 5 = -7 and -2 │-1│- 5 = -7
(-2) (1) – 5 = -7 and (-2) (1) – 5 = -7
-2 – 5 = -7 and -2 -5 = -7
-7 = -7 -7 = -7
It checks!