Certainly! Let's solve the equation step by step:
[tex]\[ \frac{1}{x+1} - \frac{1}{x-3} = 0 \][/tex]
1. Combine the fractions on the left-hand side:
To combine the fractions, we need a common denominator. The common denominator for [tex]\( \frac{1}{x+1} \)[/tex] and [tex]\( \frac{1}{x-3} \)[/tex] is [tex]\((x+1)(x-3)\)[/tex].
[tex]\[ \frac{1}{x+1} - \frac{1}{x-3} = \frac{(x-3) - (x+1)}{(x+1)(x-3)} \][/tex]
2. Simplify the numerator:
[tex]\[ (x-3) - (x+1) = x - 3 - x - 1 = -4 \][/tex]
So, we have:
[tex]\[ \frac{1}{x+1} - \frac{1}{x-3} = \frac{-4}{(x+1)(x-3)} \][/tex]
3. Set the simplified equation equal to zero:
[tex]\[ \frac{-4}{(x+1)(x-3)} = 0 \][/tex]
A fraction is equal to zero only if its numerator is zero. Therefore,
[tex]\[ -4 = 0 \][/tex]
This is a contradiction because -4 will never equal 0. Therefore, there are no solutions for [tex]\( x \)[/tex] that satisfy the equation.
As we derived, the equation has no solutions. Thus, there are no values of [tex]\( x \)[/tex] that make the initial equation true.
