Question

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

Answer 1

Domain:

$x-1\neq0\\\\\boxed{x\neq1}$

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

1. $(3x+1)^2>0$

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

 $x-1\ \textgreater \ 0\\\\\boxed{x\ \textgreater \ 1}$

2. $(3x+1)^2=0$

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

$(3x+1)^2=0\\\\3x+1=0\\\\3x=-1\qquad|:3\\\\\\\boxed{x=-\dfrac{1}{3}}$

So the solution to this inequality is:

$\boxed{x=-\dfrac{1}{3} \qquad\vee\qquad x\ \textgreater \ 1}$

Answer 2

Alright! As you can see we got

(3x + 1)^2/x-1 0

So to do that we have to divide one or two numbers,

(3x+1)^2=0-3x+1=0

An also the first one is going to be positive.

Finally, the inequality is

x= -1/3

x > 1