please help me (: ALgebra 2!!!!!!!!!!!!!!!!!!!!!

Answer:
[tex]\frac{3x+1}{x}, x\neq -1[/tex]
Step-by-step explanation:
We can start by combining the fractions in the numerator. We can multiply them by x/x and x + 1/x + 1 respectively to get a common denominator of x(x + 1).
[tex]\frac{4}{x+1}*\frac{x}{x} +\frac{2}{x} *\frac{x+1}{x+1}\\\\\frac{4x}{x(x+1)}+\frac{2x+2}{x(x+1)}\\\\\frac{6x+2}{x(x+1)}[/tex]
Now we can rewrite an simplify:
[tex]\frac{\frac{6x+2}{x(x+1)} }{\frac{2}{x+1} }\\\\\\\frac{6x+2}{x(x+1)}*\frac{x+1}{2} \\\\\frac{3x+1}{x}[/tex]
The domain would be x ≠ -1 because the starting domain is [tex]x\neq -1; x\neq 0[/tex] and it changed to [tex]x\neq 0[/tex], so for the two fractions to be equal, the domain would be [tex]x\neq -1[/tex].