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Math Help - please help with absolute value equation

  1. #1
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    please help

    |6-5x| = 3x -7
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  2. #2
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    |6 - 5x| = 3x - 7

    6+5x = 3x - 7 (minus 3x from both sides)

    6 + 2x = -7 (minus 6)

    2x = -13 (divide by 2)

    x = -13/2


    6 - 5x = 3x - 7 (minus 3x from both sides)

    6 - 8x = -7 (add 6)

    -8x = -1 (divide by -8)

    x = 1/8


    x = -13/2 and x = 1/8

    I think that's right.
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  3. #3
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    bobby77|6-5x| = 3x -7

    Hi Bobby:

    As you are perhaps aware, the meaning of absolute value is the distance to the origin (i.e., zero) on the real [number]line. So, for instance, |3|=3 because the distance from 3 to zero on the real line is 3 units. Similarly, the distance from -3 to zero is also 3 units and, therefore |-3|=3.

    Now, suppose I tell you there is a number (or numbers) inside the set {}. Further, I will tell you that the distance from any number in the set is 8. Acording to the discussion above, this means that if x is in that set, then |x|=8. Therefore, the set contains, at most, two numbers which can only be 8 and/or -8. Hence x=8 or x=-8.

    The object of the preceding paragraph is to convince you that, given |u|=k, and k>=0, then it must follow that u=k, and/or u=-k.

    In your case, we are given:

    |6-5x| = 3x-7. So,

    6-5x = (3x-7) or 6-5x = -(3x-7)

    Case 1: 6-5x = (3x-7). Adding 7 to both sides gives 13-5x = 3x. Adding 5x to both sides of this gives 13=8x which is equivalent to 8x=13. Finally, dividing by 8 leaves x=13/8 or 1.625.

    Case 2: 6-5x = -(3x-7)
    Distributing the -1 on the right side gives 6-5x = -3x+7. Skipping the details, this equation has solution x=-0.5.

    Now, the all important step in solving equations involving absolute value is, CHECK YOUR "SOLUTIONS"! If substituting a number for the variable in the original equation results in a true statement, then that number is, in fact, a solution. On the other hand, if such substitution results in a false statement, such as, 3=5, then the substituted number is NOT a solution of the equation.

    I leave it for you to test both 1.625 and -0.5 as possible solutions of |6-5x| = 3x-7.

    By the way, has your teacher yet taught you to graph equations involving absolute value? If so, there is a quick test for the existence of solutions that I will gladly share with you.

    Regards,

    Rich B.
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  4. #4
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    Quote Originally Posted by Iskarius
    |6 - 5x| = 3x - 7

    6+5x = 3x - 7 (minus 3x from both sides)

    6 + 2x = -7 (minus 6)

    2x = -13 (divide by 2)

    ----------

    I think that's right.
    I just want to point out that \left|6-5x\right|\neq6+5x
    rather, \left|6-5x\right| can be solved for \pm\left(6-5x\right)
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  5. #5
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    Given,
    <br />
|6-5x|=3x-7
    If, 6-5x\geq 0 \rightarrow x\leq 1.2 then |6-5x|=6-5x thus,
    6-5x=3x-7
    Thus,
    -8x=-13
    Thus,
    x=1.624 \leq 1.2--->a solution

    If 6-5x<0 \rightarrow x>1.2 then, |6-5x|=5x-6
    Thus,
    5x-6=3x-7 thus,
    2x=-1 thus, x=-.5\not > 1.2--->no solution.
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  6. #6
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    Given,

    If, then thus,

    Thus,

    Thus,
    --->a solution

    Dear P.Hckr:

    I am not sure what your final analysis is here? Are you claiming 1.624 to be a solution of the given equation?

    Regards:

    Rich B.

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  7. #7
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    Quote Originally Posted by Rich B.
    Given,

    If, then thus,

    Thus,

    Thus,
    --->a solution

    Dear P.Hckr:

    I am not sure what your final analysis is here? Are you claiming 1.624 to be a solution of the given equation?

    Regards:

    Rich B.

    Forgive me, typed the wrong charchter. x=1.625
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  8. #8
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    Hello Again:

    Perhaps I have lost my way somewhere en route but, the right side of initial equation, when evaluated at x=1.625, is negative; is it not? And, if so, that doesn't jive with left member being an absolute value. Do you agree with this line of reasoning?

    I await your opinion.

    Rich
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  9. #9
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    Hello, bobby77!

    I was getting very strange answers . . . then I found out why.


    |6-5x| \:= \:3x -7

    We have two graphs: . y \,= \,|6 - 5x| and y \,= \,3x - 7
    . . and we are asked to find their intersection(s).


    The first function is: . y\:=\:|6-5x|\;=\;|5x-6|\:=\:5\left|x-\frac{6}{5}\right|

    The graph of y = |x| is a "V" with its vertex at the origin.
    This graph has been moved \frac{6}{5} units to the right
    . . and has a vertical "stretch" by a factor of 5.

    Its graph looks like this:
    Code:
               *|        *
               6*       *
                |*     *
                | *   *
                |  * *
          - - - + - * - - -
                |  6/5

    The second function is: . y\:=\:3x-7
    This is a straight line with intercepts \left(\frac{7}{3},0\right),\;(0,\text{-}7)
    Its graph looks like this:
    Code:
                |               o
                |             o
                |           o
                |         o
          - - - + - - - o- - - - -
                |     o 7/3
                |   o 
                | o
             -7 o
              o |

    Graph the two functions on the same set of axes and we get:
    Code:
               *|        *        o
               6*       *       o
                |*     *      o
                | *   *     o
                |  * *    o
          - - - + - * - o- - - - - -
                |     o
                |   o 
                | o
             -7 o
              o |

    Get it? . . . The graphs don't intersect!

    Therefore, the equation has no solutions.

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  10. #10
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    Quote Originally Posted by Rich B.
    Hello Again:

    Perhaps I have lost my way somewhere en route but, the right side of initial equation, when evaluated at x=1.625, is negative; is it not? And, if so, that doesn't jive with left member being an absolute value. Do you agree with this line of reasoning?

    I await your opinion.

    Rich
    I made a second mistake,
    1.625\leq 1.2
    Thus, there are no solutions are Soroban said.
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  11. #11
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    In fact you can turn and absolute value equation into a quadradic.
    Using the identity, |x|=\sqrt{x^2}.
    Thus, the equation becomes,
    \sqrt{(6-5x)^2}=3x-7
    Assume that a solution exists.

    Then,
    (6-5x)^2=(3x-7)^2
    Then,
    (6-5x)^2-(3x-7)^2=0
    Thus,
    (6-5x+3x-7)(6-5x-3x+7)=0
    Thus,
    (-2x-1)(-8x+13)=0
    From here we have,
    x=-.5,1.625
    Now we check cuz we assumed and see that they do no work. Thus no solutions.

    You can instead of assuming the solution exists find the conditions for a solution to exist which is when,
    6-5x\geq 0 and 3x-7\geq 0
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