Math Help Forum - View Single Post - Division where one denominator is x-2 and second is 2. Test on monday!

masters · #2 November 15th, 2008, 10:44 AM

Quote:

Originally Posted by vane505

I can solve something like $(y-3/y)=(10y^2/y^2)$ where I can get the Lowest Common Denominator $y^2$ .

What I don't get is how to solve these kinds of problems, where I can't find the Lowest Common Denominator (or at least I don't know how):

$(2/x-2)-(x/2)=(x/x-2)$

I tried getting the LCD by doing like this but I don't get the correct answer:
LCD:
$(x-2): 1*(x-2)2: 1*2LCD: 1*2*(x-2) = 2x-4$

Appreciate Answers

$\frac{2}{x-2}-\frac{x}{2}=\frac{x}{x-2}$

Your LCD is correct, but there is no need to expand it at this point.

The LCD will be the product of all the different factors in the denominator using their largest exponent if they have one.

In your case, you have two different denominators: 2 and x-2.

The LCD is their product: 2(x-2).

You didn't say what answer you got, so I don't know where you might have gone wrong in your solution attempt.

Multiply each term by your LCD.

$2(x-2)\left(\frac{2}{x-2}\right)-2(x-2)\left(\frac{x}{2}\right)=2(x-2)\left(\frac{x}{x-2}\right)$

$2\rlap{///////}(x-2)\left(\frac{2}{\rlap{///////}(x-2)}\right)-\rlap{//}(2)(x-2)\left(\frac{x}{\rlap{//}(2)}\right)=2\rlap{///////}(x-2)\left(\frac{x}{\rlap{///////}(x-2)}\right)$

$4-x(x-2)=2x$

$4-x^2+2x-2x=0$

$x^2-4=0$

$(x-2)(x+2)=0$

$x=2 \ \ or \ \ x=-2$

$x\neq2$ . Therefore $x=-2$ is the only solution.

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