Can anyone help? There’s 2 solutions. (50 points + brainliest)

Domain:

[tex]x-1\neq0\\\\\boxed{x\neq1}[/tex]

As we see in a numerator we have a square, so [tex](3x+1)^2[/tex] is always 0 or more. Now, there are two possibilities:

1. [tex](3x+1)^2>0[/tex]

We divide two numbers, and first one is positive, so the result will be also positive (it canot be 0 in this case) when [tex]x-1\ \textgreater \ 0[/tex]. So:

 [tex]x-1\ \textgreater \ 0\\\\\boxed{x\ \textgreater \ 1}[/tex]

2. [tex](3x+1)^2=0[/tex]

In this case the result wil be 0 (we have ≥ 0 in our inequality, so it will be a solution).

[tex](3x+1)^2=0\\\\3x+1=0\\\\3x=-1\qquad|:3\\\\\\\boxed{x=-\dfrac{1}{3}}[/tex]

So the solution to this inequality is:

[tex]\boxed{x=-\dfrac{1}{3} \qquad\vee\qquad x\ \textgreater \ 1}[/tex]