Solve ((3)/(x^2-6x+9))+((x-2)/(3x-9))=(x)/(2x-6)

If you plug 10 back into the equation, you'll see that x = 10 is not a solution.
 
\(\displaystyle \L \frac{3}{x^2-6x+9} + \frac{x-2}{3x-9} = \frac{x}{2x-6}\)

Combine the two fractions on the left hand side into one:

\(\displaystyle \L \frac{3(3x-9) + (x-2)(x^2-6x+9)}{(x^2-6x+9)(3x-9)} = \frac{x}{2x-6}\)

Factor even more

\(\displaystyle \L \frac{9(x-3) + (x-2)(x-3)^3}{(x^2-3)^3 9(x-3)} = \frac{x}{2(x-3)}\)

WoW! Look at all the things that can be canceled! Once you cancel everything possible, solve for x
 
let's try this ...

3/(x<sup>2</sup>-6x+9) + (x-2)/(3x-9) = x/(2x-6)

factor all the denominators ...

3/(x-3)<sup>2</sup> + (x-2)/[3(x-3)] = x/[2(x-3)]

the common denominator is 6(x-3)<sup>2</sup> ...

18/[6(x-3)<sup>2</sup>] + 2(x-3)(x-2)/[6(x-3)<sup>2</sup>] = 3x(x-3)/[6(x-3)<sup>2</sup>]

since we have a common denominator, the numerators form the equation ...

18 + 2(x-3)(x-2) = 3x(x-3)

now, solve this equation for x ... remember that x cannot equal 3.
 
So, x=14? that's what i got
however, when i plug it into the equation, it doesn't work.
18+2(x-3)(x-2)=3x(x-3)
18+2(x-2)=3x
18+2x-4=3x
2x+14=3x
2x=3x-14
-x=-14
x=14

right????
 
evan399 said:
18+2(x-3)(x-2)=3x(x-3)
18+2(x-2)=3x
Click to expand...
Where did the "(x - 3)" factors go? They disappeared from two of the three terms...? :shock:

Eliz.
 
evan399 said:
Don't they cancel each other out?
Click to expand...
How does what "cancel each other out"? I'm sorry, but I don't understand your reasoning...? :oops:

Eliz.
 
There's an (x-3) on each side of the equation, so when one moves over, it cancels them out right?
 
evan399 said:
There's an (x-3) on each side of the equation, so when one moves over, it cancels them out right?
Click to expand...

if you have:

A*n + B*n = C*n

Then you can "cancel" n out (with some restriction). In mathematical terms, you have 'n' as a common factor. You can, in that case, 'divide' the common factor out.

However, in your case, you do not have the '(x-3)' in the first term - so you do not have a common factor. Thus you cannot "cancel" out.
 
So if I have 18+2(x-3)(x-2)=3x(x-3)
then
18+2(x^2-5x-6)=3x^2-9x
18+(2x^2-10x-12)=3x^2-9x
2x^2-10x+6=3x^2-9x
-x+6=x^2

now what?
 
evan399 said:
So if I have 18+2(x-3)(x-2)=3x(x-3) then

18+2(x^2-5x-6)=3x^2-9x : NO ; -3 times -2 = +6

18+(2x^2-10x-12)=3x^2-9x

2x^2-10x+6=3x^2-9x

-x+6=x^2 : are you saying you don't know what to do if that was correct equation?

now what?
Click to expand...

To see where you're at, can you solve this:
x^2 - 6x - 7 = 0
 
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