Help with problem

moosie2323 said:
Can anyone help me out with this particular problem.

1/x^2-5x+6=2/2x^2-6x
Click to expand...

You don't say what you need to do.

Please share with us your work/thoughts so that we know where to begin to help you.
 
Please use proper grouping symbols. the way you have it written it means:

\(\displaystyle \frac{1}{x^{2}}-5x+6=x^{2}-6x\)

I am assuming you mean:

\(\displaystyle \frac{1}{x^{2}-5x+6}=\frac{2}{2x^{2}-6x}\)

If the latter, start by factoring the denominators.

\(\displaystyle \frac{1}{(x-3)(x-2)}=\frac{2}{2x(x-3)}\)

Now, find the LCD.

You can also cross multiply. Then, you have a quadratic on each side you can simplify and solve.

Remember, just because you may find a solution it may not be viable. Always plug back in to see if it works.

i.e. if you get 2 or 3 as a solution, then they are extraneous.
 
moosie2323 said:
The LCd would be (x-2)(x-3) that's where i get lost after this point.
Click to expand...

I would doit the following way:

\(\displaystyle \frac{1}{(x-3)(x-2)} \, = \, \frac{1}{x(x-3)}\)

\(\displaystyle \frac{1}{(x-3)(x-2)} \, - \, \frac{1}{x(x-3)} \, = 0\)

the LCD is x(x-3)(x-2)

then

\(\displaystyle \frac{x}{x(x-3)(x-2)} \, - \, \frac{x-2}{x(x-3)(x-2)} \, = 0\)

Now simplify....
 
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