Assistance needed with solving this equation

[Ravingsofthesane]

Hello everyone, Need a little assistance with this one. The instructions state solve the equation and check your solution.

X-5 24
-------- = ---------
X X+5


From my understanding I have to make this a quadratic equation and solve


X-5 24
------- - ------- = 0
X X+5


So this is the point where I think I am making my mistake

5x - 25 24
---------- - -------
X+5 X+5

so if I made a mistake with that one I am now pretty lost


5(x-5) - 24
------------
X+5


and i am not sure where to go next with this or even if I did part of this wrong. Can anyone offer a little assistance?

thanks in advance[/code]

 

[stapel]

I think you mean the equation to be as follows:

. . . . .(x - 5)/x = 24/(x + 5)

I'm not sure why you're moving everything over to one side...?

This is of the form "(fraction) = (fraction)", so just cross-multiply:

. . . . .(x - 5)(x + 5) = 24x

Then solve the resulting quadratic. Remember to check your solutions, since x = -5 and x = 0 won't work in the original equation.

Hope that helps a bit.

Eliz.

 

[Ravingsofthesane]

I think you mean the equation to be as follows:

. . . . .(x - 5)/x = 24/(x + 5)

I'm not sure why you're moving everything over to one side...?

This is of the form "(fraction) = (fraction)", so just cross-multiply:

. . . . .(x - 5)(x + 5) = 24x

Then solve the resulting quadratic. Remember to check your solutions, since x = -5 and x = 0 won't work in the original equation.

Hope that helps a bit.

Eliz.

thanks .. actually it was not ment to be moved over to one side , it looks like somehow when it was posted the spaces that seperated the values were removed


thanks for the assistance .. this helps me out alot

 

[soroban]

Hello, Ravings!

\(\displaystyle \L\frac{x\,-\,5}{x}\:=\:\frac{24}{x\,+\,5}\)

Note that: \(\displaystyle x\,\neq\,0,\:x\,\neq\,-5\)

Multiply through by the LCD, \(\displaystyle x(x + 5):\;\;(x\,-\,5)(x\,+\,5)\:=\:24x\)

We have: .\(\displaystyle x^2\,-\,25\:=\:24x\;\;\Rightarrow\;\;x^2\,-\,24x\,-\,25\:=\:0\)

. . which factors: .\(\displaystyle (x\,-\,25)(x\,+\,1)\:=\:0\)

. . and has roots: .\(\displaystyle x\,=\,24,\,-1\)

 

[Ravingsofthesane]

so I see the resulting quadratic is

(x - 5) (X+5) - 24x = 0

so X can not = 5 , -5 or 0


so Positive 5 is the only solution that works in the origional equation.

the issue now is the correct anwser X = 5 , x = -1
or

x=5 , x = 1


and where did the 1 in this come from?

 

[stapel]

the resulting quadratic is (x - 5) (X+5) - 24x = 0

Yes, although you do need to stick with only one variable, since "x" and "X" are not the same thing.

so X can not = 5 , -5 or 0

The denominators of the original equation had only two zeroes. How are you getting three proscribed values?

so Positive 5 is the only solution that works in the origional equation.

How are you getting this? And didn't you just say that positive 5 could not be the solution?

is the correct anwser X = 5 , x = -1 or x=5 , x = 1

Neither.

Have you solved the quadratic yet? You have:

. . . . .(x - 5)(x + 5) - 24x = 0

. . . . .x² - 25 - 24x = 0

. . . . .x² - 24x - 25 = 0

Then what?

where did the 1 in this come from?

Solve the quadratic to find out.

Eliz.

 

[Denis]

so I see the resulting quadratic is
(x - 5) (X+5) - 24x = 0
...blah...blah...

...and that becomes x^2 - 24x + 25 = 0

I don't know why so much rhetoric has been used here, when you're obviously
learning to solve quadratics; factor that equation:
x^2 - 24x + 25 = 0
(x - 25)(x + 1) = 0
x - 25 = 0 or x + 1 = 0
x = 25 or x = -1

That's it.

If you're asked to use the quadratic formula to solve it, fine:
substitute a=1, b=-24 and c=25 in the formula; you'll get same answer.