Violagirl said:
On this problem: x/(x-2)=1+1/(X-9)
I know I have to find the common denominator on both sides so it'd be (x-2) and (x-9. On the right side for the single digit 1, would it mulitply into both quantities or just (x-2)?
I'm not sure what you're saying here...?
I will guess that the equation is as follows:
. . . . .\(\displaystyle \frac{x}{x\, -\, 2}\, =\, 1\, +\, \frac{1}{x\, -\, 9}\)
In particular, I will guess that you mean "x" and "X" to be, contrary to standard practice, actually the same variable.
I
think you are multiplying through by the common denominator, (x - 2)(x - 9), as:
. . . . .\(\displaystyle \left(\frac{(x\, -\, 2)(x\, -\, 9)}{1}\right)\, \left(\frac{x}{x\, -\, 2}\right)\, =\, \left(\frac{(x\, -\, 2)(x\, -\, 9)}{1}\right)\, \left(\frac{1}{1}\right)\, +\, \left(\frac{(x\, -\, 2)(x\, -\, 9)}{1}\right)\, \left(\frac{1}{x\, -\, 9}\right)\)
...leading to:
. . . . .\(\displaystyle \left(\frac{(x\, -\, 9)}{1}\right)\, \left(\frac{x}{1}\right)\, =\, (x\, -\, 2)(x\, -\, 9)\, +\, \left(\frac{(x\, -\, 2)}{1}\right)\, \left(\frac{1}{1}\right)\)
. . . . .\(\displaystyle x(x\, -\, 9)\, =\, (x\, -\, 2)(x\, -\, 9)\, +\, (x\, -\, 2)\)
But I'm afraid I don't understand what you're saying after that...?
Please reply with clarification,
showing your work and reasoning. Thank you!
Eliz.