|
This Lesson (SOLVING ABSOLUTE VALUE EQUATIONS) was created by by Theo(13342)
  About Theo:
This lesson provides an overview of SOLVING ABSOLUTE VALUE EQUATIONS
Last update was at $$$$ REFERENCES http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut21_abseq.htm http://tutorial.math.lamar.edu/Classes/Alg/SolveAbsValueEqns.aspx http://www.purplemath.com/modules/solveabs.htm ABSOLUTE VALUE SYMBOLS Absolute value of an expression is represented by the expression being enclosed by vertical bars that look like this: || Example: Absolute value of (x^2 + 2x + 3) is represented by enclosing it in vertical bars as shown: |x^2 + 2x + 3|. On your keyboard, the | symbol is usually found on the right hand side second row down from the keyboard numbers and is the SHIFT option of the \ key. If you look at the key, you should see \ underneath and | on top although the | may look like it's separated in the middle which can make it look like 2 vertical lines on top of each other rather than 1. BASIC DEFINITION OF ABSOLUTE VALUE THE ABSOLUTE VALUE OF AN EXPRESSION IS EQUAL TO THE VALUE OF THE EXPRESSION IF THE EXPRESSION IS POSITIVE OR EQUAL TO 0 AND IS EQUAL TO MINUS THE VALUE OF THE EXPRESSION IF THE EXPRESSION IS NEGATIVE. Examples: |5| = 5 |-5| = - (-5) |x| = x when x is >= 0 |x| = -x when x is < 0 |x+7| = (x+7) when (x+7) >= 0 |x+7| = -(x+7) when (x+7) < 0 |3x-3| = (3x-3) when (3x-3) >= 0 |3x-3| = -(3x-3) when (3x-3) < 0 NOTE: It is the value of the EXPRESSION that is positive or negative, not just the variable within the expression. Example: The expression (x+7) is positive when (x+7) is >= 0 and is negative when (x+7) < 0. (x+7) >= 0 when x >= -7. (x+7) < 0 when x < -7. ARITHMETIC SYMBOLS In this lesson, the following logical operators will be used: equal: = greater than: > less than: < greater than or equal: >= less than or equal: <= The following arithmetic / algebraic operators may or may not be used where applicable. x^n: x raised to the nth power root (n,x): nth root of x sqrt(x): square root of x x*y: x multiplied by y x/y: x divided by y x+y: y added to x x-y: y subtracted from x EXAMPLE NUMBER 1 |x| >= 0 We determine that the absolute value expression is (x). We do an up front check to determine if our absolute value equation is valid or not. We determine that it is a valid equation because the absolute value of any expression will always be greater than or equal to 0. We further determine that this particular equation will be true for all real values of x because no matter what the value of x is, the absolute value of that will always be greater than or equal to 0. Our answer is that x can be any real number. EXAMPLE NUMBER 2 |x+5| >= 0 We determine that the absolute value expression is (x+5). We do an up front check to determine if our absolute value equation is valid or not. We determine that it is a valid equation because the absolute value of any expression will always be greater than or equal to 0. We further determine that this equation will be true for all real values of x because no matter what the value of x is, the absolute value of (x+5) will always be greater than or equal to 0. Our answer is that x can be any real number. EXAMPLE NUMBER 3 |x| < 0 We determine that the expression within the absolute value signs is x. We do an up front check to determine if our absolute value equation is valid or not. We determine that it is NOT a valid equation because the absolute value of any expression can never be negative. Our answer is that there is no possible solution to this equation. EXAMPLE NUMBER 4 |x+5| <= 0 We determine that the absolute value expression is (x+5). We do an up front check to determine if our absolute value equation is valid or not. We determine that it is a valid equation because the absolute value of any expression can be equal to 0 even if it can never be negative. We will solve for |x+5| = 0 only. We will not solve for |x+5| < 0 because that part of the equation is invalid because the absolute value of any expression can never be < 0. If (x+5) is positive, then |x+5| = 0 means that (x+5) = 0. Solving for x, we get: x = -5 If (x+5) is negative, then |x+5| = 0 means that (x+5) = -0 Since 0 and -0 are the same number, then the solution is x = -5 again. We have 1 possible solution and that is that x = -5. That value of x is our only solution to this problem and it will only satisfy the part of the equation that states that |x+5| = 0. If x is any value other than -5, the expression within the absolute value sign will be other than 0 which means that the absolute value of that expression will be greater than 0 which will not satisfy the requirements of the equation. Our answer is that |x+5| <= 0 is true when x = -5 only. EXAMPLE NUMBER 5 |5-3x| <= 7 We determine that that the absolute value expression is (5-3x) We do an up front check to determine if our absolute value equation is valid or not. We determine that this absolute value equation is valid since the solution we are looking can be greater than or equal to zero even if it has to be less than or equal to 7. |5-3x| <= 7 means that: (5-3x) <= 7 or (5-3x) >= -7 If the expression within the absolute value signs is <= 7 or >= -7, it will satisfy the inequality. The first possible solution set of (5-3x) <= 7 is solved as follows: Start with: (5-3x) <= 7 Subtract 5 from both sides of the equation to get: -3x <= 2 Divide both sides of the equation by -3 to get: x >= -2/3 Note that multiplying or dividing both sides of an inequality by a negative number reverses the inequality. The second possible solution set of (5-3x) >= -7 is solved as follows: Start with: (5-3x) >= -7 Subtract 5 from both sides of the equation to get: -3x >= -12 Divide both sides of the equation by -3 to get: x <= -12/-3 which becomes: x <= 4 Note, once again, that multiplying or dividing both sides of an inequality by a negative number reverses the inequality. Our answer is: x >= -2/3 or x <= 4 This can also be written as: -2/3 <= x <= 4 To confirm the solution, we need to check values for x that are: < -2/3 (this solution should not be valid)./ = -2/3 > -2/3 and < 4 = 4 > 4 (this solution should not be valid). When x = -1 which is smaller than -2/3, then |5-3x| = |8| which is not <= 7 (invalid as expected). When x = -2/3 which is equal to -2/3, then |5-3x| = |7| which is <= 7 (valid as expected). When x = 0 which is > -2/3 and < 4, then |5-3x| = |5| which is <= 7 (valid as expected). When x = 4 which is equal to 4, then |5-3x| = |5-12| = |-7| = 7 which is <= 7 (valid as expected). When x = 5 which is > 4, then |5-3x| = |5-15| = |-10| = 10 which is not <= 7 (invalid as expected). All conditions confirm that the solution is valid, so the solution is: -2/3 <= x <= 4 EXAMPLE NUMBER 6 |x^2 + x - 9| = 3 We determine that the absolute value expression is (x^2 + x - 9). We do an up front check to determine if our absolute value equation is valid or not. We determine that this absolute value equation is valid since the solution we are looking will be 3 which is greater than or equal to 0. |x^2 + x - 9| = 3 means that: (x^2 + x - 9) = 3 if (x^2 + x - 9) >= 0 or: (x^2 + x - 9) = -3 if (x^2 + x - 9) < 0 These 2 equations are our possible solution sets. We solve for the first possible solution set as follows: Start with: (x^2 + x - 9) = 3 Subtract 3 from both sides of the equation to get: x^2 + x - 12 = 0 Factor this quadratic equation to get: (x+4) * (x-3) = 0 Solve for x to get: x = -4 or x = 3 We solve for the second possible solution set as follows: Start with: (x^2 + x - 9) = -3 Add 3 to both sides of the equation to get: x^2 + x - 6 = 0 Factor this quadratic equation to get: (x+3) * (x-2) = 0 Solve for x to get: x = -3 or x = 2 The 4 possible solutions to this absolute value equation are: x = -4 or x = 3 or x = -3 or x = 2 To confirm, we have to plug these values into the original absolute value equation to see that the equation holds true. When x = -4, then |x^2 + x - 9| = 3 becomes |(-4)^2 + (-4) - 9| = 3 which becomes |16-4-9| = 3 which becomes |3| = 3 which becomes 3 = 3 confirming that x = -4 is a valid solution. When x = 3, then |x^2 + x - 9| = 3 becomes |3^2 + 3 - 9| = 3 which becomes |9+3-9| = 3 which becomes |3| = 3 which becomes 3 = 3 confirming that x = 3 is a valid solution. When x = -3, then |x^2 + x - 9| = 3 becomes |(-3)^2 + (-3) - 9| = 3 which becomes |9-3-9| = 3 which becomes |-3| = 3 which becomes 3 = 3 confirming that x = -3 is a valid solution. x = 2, then |x^2 + x - 9| = 3 becomes |2^2 + 2 - 9| = 3 which becomes |4+2-9| = 3 which becomes |-3| = 3 which becomes 3 = 3 confirming that x = 2 is a valid solution. All the possible solutions check out ok. The possible solutions will not always be valid. Confirming that they are valid is a necessary step that needs to be taken before you can be assured that the possible solutions are actual solutions. EXAMPLE NUMBER 7 |x–3| = |3x+2|–1 This equation has 2 absolute value expressions rather than 1. This is a complication but it is not unmanageable. We will use the same procedure as before to solve this problem. We determine that the absolute value expressions are (x-3) and (3x+2). We do an up front check to determine if our absolute value equation is valid or not. The equation is: |x–3| = |3x+2|–1 |x-3| will always be greater than or equal to 0 for any value of x. This means that the right side of this equation must always be greater than or equal to 0. |3x+2|–1 will always be greater than or equal to 0 if |3x+2| >= 1 Since this is possible for at least some values of x, this equation is considered to be valid and can possibly be solved. Based on the definitions of absolute values, we have 4 possible conditions. x-3 = 3x+2-1 -(x-3) = 3x+2-1 x-3 = -(3x+2)-1 -(x-3) = -(3x+2)-1 These simplify to: x-3 = 3x+1 -x+3 = 3x+1 x-3 = -3x-3 -x+3 = -3x-3 We need to solve for x in each one of these combinations to see which ones will give us a valid solution. solve for x in x-3 = 3x+1 to get x = -2 solve for x in -x+3 = 3x+1 to get x = 1/2 solve for x in x-3 = -3x-3 to get x = 0 solve for x in -x+3 = -3x-3 to get x = -3 Of these possibilities, the 2 that are valid are x = 1/2 and x = -3 The following graph shows that these are the intersection points of the equations y = |x-3| and y = |2x+3|-1 $$$$ EXAMPLE NUMBER 9 |x| + |y| <= 1 We determine that the absolute value expressions are (x) and (y). We do an up front check to determine if our absolute value equation is valid or not. Since the answer can be positive then we have possible solutions to this problem. Looking at this equation, it becomes clear that |x| and |y| must both be <= 1. this is because if either |x| or |y| were > 1, that would force the other absolute value to be < 0 which is impossible with absolute values since they are always >= 0. If we solve for |x|, we will see that: |x| <= 1 - |y| If we solve for |y|, we will see that: |y| <= 1 - |x| Because we have one equation in two unknowns, unless one of the values, either x or y, is given, we will not be able to solve this equation. We can, however, find a range of values for x and y in which the solution of this equation can be found. We already have one possible solution set. That is when: |x| <= 1 |x| <= 1 means that: x <= 1 and: -x <= 1 Multiply this second part by -1 to get: x >= -1 Our possible solution set becomes: x <= 1 and: x >= -1 This can also be written as: -1 <= x <= 1 Our second possible solution set is when |y| <= 1 Solving for |y| <= 1, we get: y <= 1 and: y >= -1 This can also be written as: -1 <= y <= 1 Our possible solution sets are: -1 <= x <= 1 and: -1 <= y <= 1 To show that this solution set will work, we can pick any value of x that satisfies the criteria and then solve for y. Example 1: let x = 0 The original equation of |x| + |y| <= 1 becomes |y| <= 1 If y >= 0, this equation becomes y <= 1 If y < 0, this equation becomes -y <= 1 which becomes y >= -1 Solutions for y are good when: -1 <= y <= 1 which is within the range of our possible solution set for y confirming that if x and y are within the range of their possible solution sets, we can find an answer to this problem that is valid. Example 2: let x = -.7 The original equation of |x| + |y| <= 1 becomes: |-.7| + |y| <= 1 This becomes: .7 + |y| <= 1 This becomes: |y| <= .3 If y >= 0, this becomes y <= .3 If y < 0, this becomes -y <= .3 which becomes y >= -.3 Solutions for y are good when: -.3 <= y <= .3 which is within the range of our possible solution set for y confirming that if x and y are within the range of their possible solution sets, we can find an answer to this problem that is valid. To complete the analysis, we will pick a value for x that is NOT within the requirements. Take x = 2. The original equation of |x| + |y| <= 1 becomes: |2| + |y| <= 1 This equation becomes: 2 + |y| <= 1 Subtract 2 from both sides of this equation to get: |y| <= -1 Since absolute value of anything can never be < 0, this equation is invalid and there are no possible solutions for it. EXAMPLE NUMBER 10 We determine that the absolute value expressions is (x+1)/(2x-3). We do an up front check to determine if our absolute value equation is valid or not. Since the answer can be positive then we have possible solutions to this problem. We look a little further to see that the value of x cannot be equal to 3/2 because that would make the denominator of this equation equal to 0 which would result in an undefined value for the equation. We further determine that a division within the absolute value sign will cause some complications that will need to be dealt with. Those complications are: It has to be determine when the expression within the absolute value signs is positive or negative. In order to determine that, it has to be determined when the expression in the numerator will be positive and when the expression in the denominator will be positive. Once that is determined, it can be determined when the whole expression within the absolute value sign will be positive and when it will be negative. You must remember that the expression will be positive when both the numerator and denominator are positive and when both the numerator and denominator are negative (minus divided by a minus is a plus). You will encounter one more complication when you are multiplying both sides of the equation by the denominator in order to solve the equation. That complication is that you have to take into account when the denominator is positive and when the denominator is negative. When the denominator is positive, the multiplication is normal. When the denominator is negative, the multiplication will reverse the sign of the inequality. You will see how that works when the time comes. The complication when multiplying by a negative denominator comes from the fact you can't see that it is negative. You have to know that it will be and when it will be and you have to take that into account. The other things you have to know is that when you do multiply by the negative expression, you change the inequality but nothing else changes, i.e. no other signs are reversed during that operation. Example: Equation is: (x+1)/(2x-3)) > y While (2x-3) is positive, multiply both sides of this equation by (2x-3). Your answer will be: (x+1) > y * (2x-3) Nothing happened other than you wound up moving the denominator over to the right hand side of the equation as you would normally do when you multiply both sides of an equation by a denominator on one side of the equation. The inequality is still greater than. While (2x-3) is NEGATIVE, multiply both sides of this equation by (2x-3). Your answer will be: (x+1) < y * (2x-3) The same thing happened EXCEPT now the inequality is reversed. The inequality is now smaller than, when before it was greater than. You will see all this happening as we go through the example. APPLY BASIC DEFINITION OF ABSOLUTE VALUES TO GET THE POSSIBLE SOLUTION SETS by the basic definition of absolute values, |(x+1)/(2x-3)| < 2 becomes: and: When you multiply POSSIBLE SOLUTION SETS You have 2 possible solution sets. set 1 is when the expression within the absolute value signs is positive and is: set 2 is when the expression within the absolute value signs is negative and is: DETERMINE WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGN IS POSITIVE AND WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGN IS NEGATIVE. The numerator is positive when The denominator is positive when Note that x cannot be equal to 3/2 because then the denominator in the equation is equal to 0 which is not a real number and therefore can't be part of any solution to this problem. Numerator is >= 0 when x >= -1 Denominator is >= 0 when x > 3/2 (= 3/2 not allowed). The expression is positive when both the numerator and the denominator are both positive. This happens when The expression is also positive when both the numerator and the denominator are both negative. this happens when The expression is positive when: This means the expression is negative when: This can be written as: remember that x cannot be equal to 3/2. SOLVE FOR WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGNS IS POSITIVE. When the expression within the absolute sign is positive, the possible solution set is: You have two possible scenarios within this solution set. The first scenario is when the denominator is positive. The second scenario is when the denominator is negative. If the denominator is positive, then you would solve this is as follows: Equation to solve is: Multiply both sides of this equation by (2x-3) to get: Remove parentheses to get: subtract 1 from both sides and subtract 3x from both sides to get: divide both sides by 3 and multiply both sides by -1 to get: If the denominator is negative, then you would solve this as a follows: Equation to solve is: Multiply both sides of this equation by (2x-3) to get: Note that the inequality reversed because we were multiplying by a negative number which is represented by the expression (2x-3) in the denominator. Remove parentheses to get: subtract 4x from both sides of the equation and subtract 1 from both sides of the equation to get: multiply both sides of this equation by -1 and divide both sides of this equation by 3 to get: You have 2 possible solutions when the expression within the absolute value sign is positive. Those 2 possible solutions are: When the expression is positive and the denominator is positive, x has to be greater than 3/2, so x > 7/3 does not have to be modified and can stand as is. When the expression is positive and the denominator is negative, x has to be less than -1 so x < 7/3 has to be modified to say x < -1 Your possible solutions for when the absolute value expression is positive are: SOLVE FOR WHEN THE EXPRESSION WITHIN THE ABSOLUTE VALUE SIGNS IS NEGATIVE When the expression within the absolute value signs is negative, the possible solution set is: The boundaries for this occur when: The denominator of 2x-3 is negative throughout this boundary so we only have to solve for when the denominator is negative and do not have to solve for when the denominator is positive. When the denominator is negative you would solve as follows: Equation to solve is: Multiply both sides of this equation by (2x-3) to get: Note that (2x-3) is a negative number so multiplying both sides of this equation by a negative number reverses the inequality. Remove parentheses to get: Subtract 1 from both sides of this equation and add 4x to both sides of this equation to get: divide both sides of this equation by 5 to get: Since the solution set is negative when this can be written as: PUT ALL THE INFORMATION TOGETHER SO YOU CAN ANALYZE IT TO CONFIRM THAT THE ANSWER IS GOOD. When the expression within the absolute value signs is positive, the possible solutions are: When the expression within the absolute value signs is negative, the possible solutions are: These 2 solutions can be combined to show as: We test the intervals to see if they are accurate by taking values within and without those intervals to see if the equation is true or not. Test values for x will be: -2 (should be good) -1 (should be good) 0 (should be good) 1 (should not be good) 7/3 (should not be good) 8/3 (should be good) When x = -2: When x = -1: When x = 0: When x = 1: When x = 7/3: When x = 8/3: This lesson has been accessed 30406 times. |
