Transcription of Absolute Value Equations Absolute Value Inequalities
1 - - Value Equations and InequalitiesAbsolute Value EquationsAbsolute Value - - 20 33 Distance is greater than is is less than is is greater than is less than definition, the equation |x|= 3 can be solved by finding real numbers at a distance of three units from 0. Two numbers satisfy this equation, 3 and 3. So the solution set is {}.3, 3 - - 3 Properties of Absolute Value1. For 0, if and only if or .ba bab a b>== = 2. if and only if or .a bab a b=== For any positive number b:3. if and only if .a bbab< <<4. if and only if or .a ba b ab> < > - - 4 Example 1 SOLVING Absolute Value EQUATIONSS olve 312x =SolutionFor the given expression 5 3xto have Absolute Value 12, it must represent either 12 or 12 . This requires applying Property 1, with a= 5 3xand b= - - 5 Example 1 SOLVING Absolute Value EQUATIONSS olve 312x =Solution5 312x =5 312x =or5 312x = Property 137x =or317x = Subtract or173x=Divide by 3.
2 - - 6 Example 1 SOLVING Absolute Value EQUATIONSS olve 312x =Solution73x= or173x=Divide by 3. Check the solutions by substituting them in the original Absolute Value equation. The solution set is{}7 17,.33 - - 7 Example 1 SOLVING Absolute Value EQUATIONSS olve =+Solution43 6xx =+or43(6)xx = +Property 239x=or436xx = 3x=or53x= 35x= 436xx =+{}3 The solutionset is ,3 .5 - - 8 Example 2 SOLVING Absolute Value 2 17x+<Use Property 3, replacing awith 2x+ 1 and bwith 17x+<7 2 17x < +<Property 382 6x < <Subtract 1 from each <<Divide each part by - - 9 Example 2 SOLVING Absolute Value 2 17x+<43x < <Divide each part by final inequality gives the solution set ( 4, 3). - - 10 Example 2 SOLVING Absolute Value 2 17x+>2 17x+>217x+ < Property 4 Subtract 1 from each each part by 17x+>28x< or26x>4x< or3x> - - 11 Example 2 SOLVING Absolute Value 2 17x+>Divide each part by < or3x>The solution set is (, 4) (3, ).
3 - - 12 Example 3 SOLVING AN Absolute Value INEQUALITY REQUIRING A TRANSFORMATIONS olve Solution27 >27 14x >27 5x >Add 1 to each < or27 5x >Property 477x < or73x >1x>or37x< Subtract by 7; reverse the direction of each - - 13 Example 3 SOLVING AN Absolute Value INEQUALITY REQUIRING A TRANSFORMATIONS olve Solution27 >1x>or37x< Divide by 7; reverse the direction of each inequality.( )3 The solution set is ,1,.7 - - 14 Example 4 SOLVING SPECIAL CASES OF Absolute Value Equations AND INEQULAITIESS olve Solution Since the Absolute Value of a number is always nonnegative, the inequality is always true. The solution set includes all real - - 15 Example 4 SOLVING SPECIAL CASES OF Absolute Value Equations AND INEQULAITIESS olve Solution There is no number whose Absolute Value is less than 3 (or less than anynegative number).
4 The solution set is .47 3x < - - 16 Example 4 SOLVING SPECIAL CASES OF Absolute Value Equations AND INEQULAITIESS olve Solution The Absolute Value of a number will be 0 only if that number is 0. Therefore, 5150x+= +=is equivalent to515 0x+=which has solution set { 3}. Check by substituting into the original equation.
