|6-5x| = 3x -7

Printable View

- June 28th 2006, 10:17 PMbobby77please help
|6-5x| = 3x -7

- June 28th 2006, 11:49 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. - June 29th 2006, 01:40 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. - June 29th 2006, 04:27 PMQuickQuote:

Originally Posted by**Iskarius**

rather, can be solved for - June 29th 2006, 04:27 PMThePerfectHacker
Given,

If, then thus,

Thus,

Thus,

--->a solution

If then,

Thus,

thus,

thus, --->no solution. - July 8th 2006, 08:06 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.

- July 8th 2006, 07:03 PMThePerfectHackerQuote:

Originally Posted by**Rich B.**

- July 8th 2006, 10:01 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 - July 9th 2006, 06:09 AMSoroban
Hello, bobby77!

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

Quote:

We have two graphs: . and

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

The first function is: .

The graph of is a "V" with its vertex at the origin.

This graph has been moved 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: .

This is a straight line with intercepts

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.

- July 9th 2006, 06:33 AMThePerfectHackerQuote:

Originally Posted by**Rich B.**

Thus, there are no solutions are Soroban said. - July 9th 2006, 06:41 AMThePerfectHacker
In fact you can turn and absolute value equation into a quadradic.

Using the identity, .

Thus, the equation becomes,

*Assume*that a solution exists.

Then,

Then,

Thus,

Thus,

From here we have,

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,

and