Solve for x

**kdl00** · March 3rd 2010, 05:29 PM

How would you solve this for $x$ ??

$ \frac{\frac{1}{x+1}-\frac{1}{4}}{x-3} = 0 $ (fixed it! made it into an equation)

i think you multiply the 1 from the numerator to the denominator, but im not sure if you can do that when there is a subtract sign between the two integers. help??

**skeeter** · March 3rd 2010, 05:38 PM

Originally Posted by kdl00

How would you solve this for $x$ ??

$ \frac{\frac{1}{x+1}-\frac{1}{4}}{x-3} $

i think you multiply the 1 from the numerator to the denominator, but im not sure if you can do that when there is a subtract sign between the two integers. help??

first of all, x cannot be solved for since what you have written is not an equation, but an expression.

the expression, however, can be simplified.

get a common denominator and combine the fractions in the numerator, then divide the resulting fraction by (x-3)

**kdl00** · March 3rd 2010, 05:51 PM

Originally Posted by skeeter

first of all, x cannot be solved for since what you have written is not an equation, but an expression.

the expression, however, can be simplified.

get a common denominator and combine the fractions in the numerator, then divide the resulting fraction by (x-3)

actually, this was a limit question in calc., but i wanted to know how to get rid of the fraction on the numerator, thats why i posted it in algebra.

**skeeter** · March 3rd 2010, 06:16 PM

is the fraction equal to 0 or are you trying to find the limit of the fraction as $x \to 0$ ?

**satx** · March 3rd 2010, 06:18 PM

This expression is indeterminate. The only way a fraction can equal zero is if the numerator equals zero. The only way this numerator could equal zero would be if x was equal to 3, but that would make the denominator equal to zero as well.

**skeeter** · March 3rd 2010, 06:28 PM

ok ... I saw your limit post in the calculus section.

$\lim_{x \to 3} \frac{\frac{1}{x+1} - \frac{1}{4}}{x-3} $

same procedure as I described before for simplification ...

$\lim_{x \to 3} \frac{\frac{4}{4(x+1)} - \frac{1(x+1)}{4(x+1)}}{x-3} $

$\lim_{x \to 3} \frac{\frac{4- (x+1)}{4(x+1)}}{x-3} $

$\lim_{x \to 3} \frac{\frac{3-x}{4(x+1)}}{x-3} $

$\lim_{x \to 3} \frac{-1}{4(x+1)} = -\frac{1}{16} $

next time, be straightforward about what you want.

**kdl00** · March 3rd 2010, 06:34 PM

Originally Posted by skeeter

ok ... I saw your limit post in the calculus section.

$\lim_{x \to 3} \frac{\frac{1}{x+1} - \frac{1}{4}}{x-3} $

same procedure as I described before for simplification ...

$\lim_{x \to 3} \frac{\frac{4}{4(x+1)} - \frac{1(x+1)}{4(x+1)}}{x-3} $

$\lim_{x \to 3} \frac{\frac{4- (x+1)}{4(x+1)}}{x-3} $

$\lim_{x \to 3} \frac{\frac{3-x}{4(x+1)}}{x-3} $

$\lim_{x \to 3} \frac{-1}{4(x+1)} = -\frac{1}{16} $

next time, be straightforward about what you want.

kk got it. thanks for the help

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