Rational Expression Help

**gurrry** · April 3rd 2011, 04:42 PM

Im having trouble with a problem off a review... Ill try to my best abilities to write it out

Solve.

3 OVER x^2 - 6x + 9

PLUS

x - 2 OVER 3x - 9

SET EQUAL TO

x OVER 2x - 6

First I factor out all the denominators. Get the LCD then multiply it to each individual term to get rid of my fractions..

But this is where I have a problem. I dont understand what the LCD should be. Should it be 6(x-3)(x-3)?

**pickslides** · April 3rd 2011, 04:54 PM

Is this the equation?

$\displaystyle \frac{3}{x^2-6x+9}+\frac{x-2}{3x-9}= \frac{2}{2x-6}$

**Prove It** · April 3rd 2011, 04:55 PM

So you are trying to solve for $\displaystyle x$ in the equation

$\displaystyle \frac{3}{x^2 - 6x + 9} + \frac{x-2}{3x - 9} = \frac{x}{2x-6}$ .

For starters, factor all the denominators

$\displaystyle \frac{3}{(x - 3)^2} + \frac{x - 2}{3(x - 3)} = \frac{x}{2(x - 3)}$ .

Now get a common denominator for both sides, and yes, you are correct that the lowest common denominator is $\displaystyle 6(x - 3)^2$ .

**gurrry** · April 3rd 2011, 05:03 PM

ok.. I multiply 6(x-3)^2 to each section and I get..

18 + 2x-6 = 3x -9

then i try and solve for x and get 21... which isnt the write answer according to this answer sheet

**Prove It** · April 3rd 2011, 05:11 PM

Originally Posted by gurrry

ok.. I multiply 6(x-3)^2 to each section and I get..

18 + 2x-6 = 3x -9

then i try and solve for x and get 21... which isnt the write answer according to this answer sheet

OK, for starters, don't jump to multiplying by the common denominator. I ALWAYS write the fractions with their common denominators first, it is clearer and reduces mistakes.

So $\displaystyle \frac{6\cdot 3}{6(x - 3)^2} + \frac{2(x - 3)(x - 2)}{6(x - 3)^2} = \frac{3x(x - 3)}{6(x - 3)^2}$

$\displaystyle \frac{18}{6(x - 3)^2} + \frac{2x^2 - 10x + 12}{6(x - 3)^2} = \frac{3x^2 - 9x}{6(x- 3)^2}$

$\displaystyle \frac{2x^2 - 10x + 30}{6(x - 3)^2} = \frac{3x^2 - 9x}{6(x - 3)^2}$

$\displaystyle 2x^2 - 10x + 30 = 3x^2 - 9x$ .

Now solve the resulting quadratic.

**gurrry** · April 3rd 2011, 05:46 PM

Originally Posted by Prove It

OK, for starters, don't jump to multiplying by the common denominator. I ALWAYS write the fractions with their common denominators first, it is clearer and reduces mistakes.

So $\displaystyle \frac{6\cdot 3}{6(x - 3)^2} + \frac{2(x - 3)(x - 2)}{6(x - 3)^2} = \frac{3x(x - 3)}{6(x - 3)^2}$

$\displaystyle \frac{18}{6(x - 3)^2} + \frac{2x^2 - 10x + 12}{6(x - 3)^2} = \frac{3x^2 - 9x}{6(x- 3)^2}$

$\displaystyle \frac{2x^2 - 10x + 30}{6(x - 3)^2} = \frac{3x^2 - 9x}{6(x - 3)^2}$

$\displaystyle 2x^2 - 10x + 30 = 3x^2 - 9x$ .

Now solve the resulting quadratic.

So for the 2(x-3)(x-2) part, did you just multiply out (x-3)(x-2) then go back and multiply 2 throughout the entire thing? Or do you distribute the 2 through the x-3 first then use the FOIL method?

**Prove It** · April 3rd 2011, 05:54 PM

Originally Posted by gurrry

So for the 2(x-3)(x-2) part, did you just multiply out (x-3)(x-2) then go back and multiply 2 throughout the entire thing? Or do you distribute the 2 through the x-3 first then use the FOIL method?

It doesn't make any difference, multiplication is associative...

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