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INEQUALITIES AND ABSOLUTE VALUE EQUATIONS
Unit Overview
An inequality is a mathematical statement that compares algebraic expressions using greater than (>), less than (<), and other inequality symbols. A compound inequality is a pair of inequalities joined by and or or. In this unit, properties of inequalities will be used to solve linear inequalities and compound inequalities in one variable. The unit concludes continues with a study of absolute-value equations and inequalities.
Introduction to Solving Inequalities
Inequality - a mathematical statement that compares algebraic quantities
Inequality Symbols Meaning Keyboard Entry
less than greater than less than or equal greater than or equal not equal
*Solving inequalities is just like solving equations, use opposite operations to isolate the variable.
Example #1 : Solve for x.
3(5 x โ 7) โฅ 54 15 21 54 21 21 15 75 5
x
x x
Inequalities--Bridge Capacity (02:27)
*When multiplying or dividing by a negative number, the inequality sign must be reversed.
Example #2 : Solve for y.
2 y + 9 < 5 y + 15 โ5 y โ5 y โ3 y + 9 < 15 โ9 โ 3 3
โ y โ
y > โ2 *Notice that the inequality sign is flipped because of the division by โ3.
Example #3 : Solve for x. (^3) ( 7) 3 4 x โ โค x โ
( ) 3 ( 7) ( 3) *Multiply both sides by 4. 4 3( 7) 4( 3) *Distribute. 3 21 4 12
21 12 9 *Rewrite with on left side, 9 inequality sign is rev
r
e sed.
x x x x x x
x x
x x x x
Solving Inequalities: Two Operations (01:25)
Example #5 : Graph the solution of each inequality. x < 4 y โฅ โ
Compound Inequalities compound inequalities: a pair of inequalities joined by โandโ or โorโ.
To solve a compound inequality joined with โ and โ, find the values of the variable that satisfy both inequalities.
*โ and โ means the intersection of the solutions
Example #6 : 2 x + 3 > 1 and 5 x โ 9 < 6 2 x > โ2 and 5 x < 15 x > โ1 and x < 3
The solution is written { x | โ1 < x < 3} (set notation) โall numbers x , such that โ1 is less than x is less than 3โ.
To solve a compound inequality joined with โ or โ, find the values of the variable that satisfy at least one inequality.
โ or โ means the union of the solutions
Example #7 : Find all solutions for b.
3 b + 7 โค 1 or 2 b โ 3 โฅ 1 3 b โค โ6 or 2 b โฅ 4 b โค โ2 or b โฅ 2
The solution is written { b | b โค โ2 or b โฅ 2} (set notation) โall numbers b such that b is less than or equal to โ2 or b is greater than or equal to 2โ.
Example #8 : Find all solutions for y when 4 y โ 9 โค 15 and 4 y โ 9 โฅ โ 1.
Since the two inequalities have a common statement (4 y โ 9) and they are an intersection ( and ) of the two solutions, the inequalities can be written so that the common statement is "sandwiched" between the two inequalities.
1 4 9 4 9 15 Make sure the inequality symbols are both pointing the same direction. 1 15 Combine the two inequalities into one inequality statement. 1 4 9 15 9 9 9 Isolate 4 by adding 9
t l
o al
y y
y y
y
parts. 8 4 24 (^8 4 24) Isolate by dividing all parts by 4.
2 6
y y (^) y
y
This solution can be interpreted as y โฅ 2 and y โค 6
The solution is written { y | 2 < y < 6} (set notation) โall numbers y, such that 2 is less than x is less than 6โ. In other words, y can be any number in between and including 2 and 6.
Solving Compound Inequalities (02:25)
4 y โ 9 โฅ โ1 is the same as โ 1 โค 4 y โ 9
Solving Absolute Value Equations and Inequalities
absolute value - the distance a number is from zero (always positive).
*Two bars around the number denote absolute value.
โ 5 = 5 6 = 6
Why do absolute value equations have two solutions?
In a simple absolute value equation such as | x | = 5, let's examine what two values of x make the expression true?
Since the | 5 | = 5, then x = 5. Also, since the |โ5 | = 5, then x = โ5.
Therefore, the solution to | x | = 5 is x = 5 or x = โ5.
Let's expand the understanding of an absolute value equation further.
Let's think about the following question: In the absolute value equation, |3 x + 4| = 19, what two values can 3 x + 4 be and why?
The value of 3 x + 4 in |3 x + 4| = 19 can be 19 because why?
Click here to check your answer.
The absolute value of |19| = 19, therefore 3 x + 4 can equal 19.
The value of 3 x + 4 in |3 x + 4| = 19 can be โ19 because why?
Click here to check your answer.
The absolute value of |โ19| = 19, therefore 3 x + 4 can equal โ19.
What two equations can be written to find the values of x that solve |3 x + 4| = 19?
Click here to check your answer.
3 x + 4 = 19 OR 3 x + 4 = โ
Follow these steps to solve absolute value equations, but keep in mind why these steps work by remembering the explanation above.
1.) Rewrite the equation without the absolute value notation. 2.) Rewrite a second time using the opposite of what the original equation was equal to, and connect with the word โ or โ. 3.) Solve both equations and check both answers in the original equation.
Example #1 : Solve 2 x โ 1 = 3 2 x โ 1 = 3 or 2 x โ 1 = โ3 *the โ3 is the opposite of 2 x = 4 or 2 x = โ2 what the original was x = 2 or x = โ1 equal to.
Check : 2(2) โ 1 = 3 or 2( โ1) โ 1 = 3 4 โ 1 = 3 or โ 2 โ 1 = 3 3 = 3 or โ 3 = 3 3 = 3๏ or 3 = 3๏ Therefore, the solution is x = 2 or x = โ1.
Example #2 : Solve 2 x + 1 = x + 5 2 x + 1 = x + 5 or 2 x + 1 = โ x โ 5 *again use the opposite x = 4 or 3 x = โ x = โ
Check : 2(4) + 1 = 4 + 5 or 2( โ2) + 1 = โ2 + 5 8 + 1 = 9 or โ 4 + 1 = 3 9 = 9 or โ 3 = 3 9 = 9๏ or 3 = 3๏
Therefore, the solution is x = 4 or x = โ2.
*Add 1 to each side. 5 11 *Subtract 5 from each side. 4 11
4
(^11) *Divide both sides by 4.
x x
x x
x x x x
x
x
Solve the second equation x โ 1 = โ(5 x +10) for x.
1 (5 10) *Rewrite the right side reversing the 1 5 10 signof each term since the minus sign is in front of the parenthesis. 1 5 10 *Add 1 to each side. 5 9 *Add 5 t
o each side. 6
x x x x
x x
x x x x
x x
(^6 9) *Divide both sides by 6 6
x
x
Check: 1 5 10 OR 1 5 10
11 4 55 40 OR 3 2 15 20
15 OR 5 5
15 15 OR 5 5
(Doesn't Check)
โ = ๏ฃซ^ ๏ฃถ^ + โ = ๏ฃซ^ ๏ฃถ+
extraneous solution : An extraneous is a solution that does not satisfy the original equation. Since โ11/4 does not satisfy the original equation, โ11/4 is an extraneous solution.
*It is important to check all solutions in the original equation to determine whether a solution is extraneous.
Therefore, the solution is x = โ3/2.
absolute value inequalities - an inequality that contains an absolute value.
To solve absolute value inequalities: 1.) Rewrite the inequality without the absolute value notation. 2.) Rewrite a second time, change the inequality sign, and use opposites. 3.) Solve both inequalities and check both answers in the original inequality. 4.) If the inequality is a < or โค , connect with the word โandโ. 5.) If the inequality is a > or โฅ , connect with the word โorโ.
Example #5 : Solve 3 x + 2 > 4 3 x + 2 > 4 or 3 x + 2 < โ4 *flip the sign and use the 3 x > 2 or 3 x < โ6 opposite x > 2 3
or x < โ
Check : 3 x + 2 > 4
tru
10 13 ( e)
x โ + โค
โ + โค
โ + โค
Therefore, the solution to this absolute value inequality is { x | x > 6 5
โ and x < 6}.
Stop! Go to Questions #14-33 to complete this unit.
