1. ## Solve for x

How would you solve this for $\displaystyle x$??

$\displaystyle \frac{\frac{1}{x+1}-\frac{1}{4}}{x-3} = 0$ (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??

2. Originally Posted by kdl00
How would you solve this for $\displaystyle x$??

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

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)

3. 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.

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

5. 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.

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

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

same procedure as I described before for simplification ...

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

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

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

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

next time, be straightforward about what you want.

7. Originally Posted by skeeter
ok ... I saw your limit post in the calculus section.

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

same procedure as I described before for simplification ...

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

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

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

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

next time, be straightforward about what you want.
kk got it. thanks for the help