I have no idea what I'm doing as I haven't touched this topic in months. If anyone can help me, that would be great.
Edit: For some reason, all of the fractions are being messed up so I'm going to post a picture of how they should look.
I have no idea what I'm doing as I haven't touched this topic in months. If anyone can help me, that would be great.
Edit: For some reason, all of the fractions are being messed up so I'm going to post a picture of how they should look.
For 1)
$\displaystyle \frac{3x + 1}{x^2 - 1} - \frac{1}{x + 1}$ notice that $\displaystyle x^2 - 1 = (x - 1)(x + 1)$, so in order to subtract the two fractions you need the same denominator which means multiplying the second fraction by $\displaystyle \frac{x - 1}{x - 1}$ to get the difference$\displaystyle \frac{3x + 1}{x^2 - 1} - \frac{x - 1}{x^2 - 1}$ and then computing the difference as $\displaystyle \frac{4x}{x^2 - 1}$.
to #4:
Keep in mind that
$\displaystyle \frac14-y^2 = -\left(y^2-\frac14\right)=-\left(y-\frac12\right) \left(y+\frac12\right)$
Now your fraction becomes:
$\displaystyle \frac{y-\frac12}{-\left(y-\frac12\right) \left(y+\frac12\right)} = \frac1{-\left(y+\frac12\right)} = -\frac1{y+\frac12}$
to #3:
$\displaystyle \frac4{x-1} = \frac5{2x-2} + \frac{3x}4~,~x \neq 1$
The common denominator of all fractions is $\displaystyle 4(x-1)$
Multiply both sides of the equation by the denominator. You should get:
$\displaystyle 16 = 10 + 3x(x-1)~\implies~3x^2-3x-6=0$
For confirmation only: x = -1 or x = 2