|6-5x| = 3x -7

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- Jun 28th 2006, 09:17 PMbobby77please help
|6-5x| = 3x -7

- Jun 28th 2006, 10:49 PMIskarius
|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. - Jun 29th 2006, 12:40 AMRich B.
bobby77|6-5x| = 3x -7

Hi Bobby:

As you are perhaps aware, the meaning of. 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.*absolute value is the distance to the origin (i.e., zero) on the real [number]line*

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,"! 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.**CHECK YOUR "SOLUTIONS**

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. - Jun 29th 2006, 03:27 PMQuickQuote:

Originally Posted by**Iskarius**

rather, $\displaystyle \left|6-5x\right|$ can be solved for $\displaystyle \pm\left(6-5x\right)$ - Jun 29th 2006, 03:27 PMThePerfectHacker
Given,

$\displaystyle

|6-5x|=3x-7$

If, $\displaystyle 6-5x\geq 0 \rightarrow x\leq 1.2$ then $\displaystyle |6-5x|=6-5x$ thus,

$\displaystyle 6-5x=3x-7$

Thus,

$\displaystyle -8x=-13$

Thus,

$\displaystyle x=1.624 \leq 1.2$--->a solution

If $\displaystyle 6-5x<0 \rightarrow x>1.2$ then, $\displaystyle |6-5x|=5x-6$

Thus,

$\displaystyle 5x-6=3x-7$ thus,

$\displaystyle 2x=-1$ thus, $\displaystyle x=-.5\not > 1.2$--->no solution. - Jul 8th 2006, 07:06 AMRich B.
Given,

http://www.mathhelpforum.com/math-he...940dc619-1.gif

If, http://www.mathhelpforum.com/math-he...541b4315-1.gif then http://www.mathhelpforum.com/math-he...ff5047fc-1.gif thus,

http://www.mathhelpforum.com/math-he...98079dcd-1.gif

Thus,

http://www.mathhelpforum.com/math-he...4727cb79-1.gif

Thus,

http://www.mathhelpforum.com/math-he...210e6511-1.gif--->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.

- Jul 8th 2006, 06:03 PMThePerfectHackerQuote:

Originally Posted by**Rich B.**

- Jul 8th 2006, 09:01 PMRich 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 - Jul 9th 2006, 05:09 AMSoroban
Hello, bobby77!

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

Quote:

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

We have two graphs: .$\displaystyle y \,= \,|6 - 5x|$ and $\displaystyle y \,= \,3x - 7$

. . and we are asked to find their intersection(s).

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

The graph of $\displaystyle y = |x|$ is a "V" with its vertex at the origin.

This graph has been moved $\displaystyle \frac{6}{5}$ 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: .$\displaystyle y\:=\:3x-7$

This is a straight line with intercepts $\displaystyle \left(\frac{7}{3},0\right),\;(0,\text{-}7)$

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.

- Jul 9th 2006, 05:33 AMThePerfectHackerQuote:

Originally Posted by**Rich B.**

$\displaystyle 1.625\leq 1.2$

Thus, there are no solutions are Soroban said. - Jul 9th 2006, 05:41 AMThePerfectHacker
In fact you can turn and absolute value equation into a quadradic.

Using the identity, $\displaystyle |x|=\sqrt{x^2}$.

Thus, the equation becomes,

$\displaystyle \sqrt{(6-5x)^2}=3x-7$

*Assume*that a solution exists.

Then,

$\displaystyle (6-5x)^2=(3x-7)^2$

Then,

$\displaystyle (6-5x)^2-(3x-7)^2=0$

Thus,

$\displaystyle (6-5x+3x-7)(6-5x-3x+7)=0$

Thus,

$\displaystyle (-2x-1)(-8x+13)=0$

From here we have,

$\displaystyle x=-.5,1.625$

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,

$\displaystyle 6-5x\geq 0$ and $\displaystyle 3x-7\geq 0$