please help with absolute value equation

**bobby77** · June 28th 2006, 09:17 PM

|6-5x| = 3x -7

**Iskarius** · June 28th 2006, 10:49 PM

|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.

**Rich B.** · June 29th 2006, 12:40 AM

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.

**Quick** · June 29th 2006, 03:27 PM

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 $\pm\left(6-5x\right)$

**ThePerfectHacker** · June 29th 2006, 03:27 PM

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.

**Rich B.** · July 8th 2006, 07:06 AM

Given,

If,

then

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.

**ThePerfectHacker** · July 8th 2006, 06:03 PM

Originally Posted by Rich B.

Given,

If,

then

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. $x=1.625$

**Rich B.** · July 8th 2006, 09:01 PM

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

**Soroban** · July 9th 2006, 05:09 AM

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.

**ThePerfectHacker** · July 9th 2006, 05:33 AM

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.

**ThePerfectHacker** · July 9th 2006, 05:41 AM

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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