First, subtract \(\displaystyle 2\) from both sides so you get 

\(\displaystyle \frac{3x^2-5x+4}{x+1}-2\leq0\).

Then find the common denominator and simplify 

\(\displaystyle \frac{3x^2-5x+4-2(x+1)}{x+1}\leq0\rightarrow\frac{3x^2-7x+2}{x+1}\leq0\).

Next, factor out the numerator 

\(\displaystyle \frac{(3x-1)(x-2)}{x+1}\leq0\) 

and set each of the three factor equal to zero and solve for \(\displaystyle x\).

The solutions are 

\(\displaystyle x=-1;\frac{1}{3};2\).

Now plug in values between \(\displaystyle (-\infty ,-1)\), \(\displaystyle \left(-1,\frac{1}{3}\right]\), \(\displaystyle \left[\frac{1}{3},2\right]\), and \(\displaystyle [2,\infty)\) into the inequality and observe if the conditions of the inequality are met.

Note that \(\displaystyle x\neq-1\). They are met in the interval \(\displaystyle (-\infty ,-1)\) and \(\displaystyle \left[\frac{1}{3},2\right]\).

Thus, the solution to the inequality is  

\(\displaystyle (-\infty ,-1)\cup \left[\frac{1}{3},2\right]\). 

1) Multiply both sides of the equation by the common denominator of the fractions:

     \(\displaystyle 4(x + 1)(\frac{4x + 3}{4} - \frac{2x}{x + 1}) < x\cdot4(x + 1)\)

     \(\displaystyle (4x + 3)(x + 1) < 4x(x + 1)\)

2) Simplify:

     \(\displaystyle x^{2} - 7x + 3 - 8x < 4x^{2} + 4\)

     \(\displaystyle -x + 3 < 4x\)

     \(\displaystyle 3 < 4x + x\)

     \(\displaystyle 3 < 5x\)

     \(\displaystyle \frac{3}{5} < x\)

3) For standard notation, and the fact that inequalities can be read backwards:

    \(\displaystyle x > \frac{3}{5}\)

     For interval notation:

     \(\displaystyle (\frac{3}{5}, +\infty)\) 

4) Graph:

Graph the rational expression,

\(\displaystyle y = \frac{3}{x - 2}\)

1) Because \(\displaystyle 2 - 2 = 0\) and a divide by \(\displaystyle 0\) is undefined in the real number system, there is a vertical asymptote where \(\displaystyle x = 2\).

2) As \(\displaystyle x\) \(\displaystyle \rightarrow\) \(\displaystyle 2^{+}\), \(\displaystyle y\) \(\displaystyle \rightarrow\) \(\displaystyle +\infty\), and as \(\displaystyle x\) \(\displaystyle \rightarrow +\infty\), \(\displaystyle y\) \(\displaystyle \rightarrow 0\).

3) As \(\displaystyle x\) \(\displaystyle \rightarrow 2^{-}\), \(\displaystyle y\) \(\displaystyle \rightarrow -\infty\), and as \(\displaystyle x\) \(\displaystyle \rightarrow\) \(\displaystyle -\infty\), \(\displaystyle y\) \(\displaystyle \rightarrow 0\).

4) The funtion y is exists over the allowed x-intervals:

One approach for solving the inequality: 

For \(\displaystyle y = \frac{3}{x - 2}\)

1) Determine where \(\displaystyle y < 1\) over the x-values \(\displaystyle x < 2\) or \(\displaystyle x > 5\).

2) \(\displaystyle y < 1\) for the intervals \(\displaystyle (-\infty, 2)\) or \(\displaystyle (5, +\infty )\).

3) Then the solution is \(\displaystyle (-\infty, 2) \cup (5, +\infty )\).

Another approach for solving the inequality:

1) Write  \(\displaystyle \frac{3}{x - 2} < 1\) as \(\displaystyle y = \frac{3}{x - 2}\), then determine the x-values that cause \(\displaystyle y < 1\) to be true: 

2) \(\displaystyle y < 1\) is true for \(\displaystyle x < 2\) or \(\displaystyle x > 5\).

3) Then the solution is \(\displaystyle (-\infty, 2) \cup (5, +\infty )\).

When solving for a point where the function will be undefined, you set the denominator equal to zero and solve for .  This creates a vertical asymptote because when the denominator equals zero the function is undefined and we are solving for .  Say for example a function is undefined at .  So at all values of where this function is undefined creating a vertical asymptote.

We begin by finding the zeros of the equation using the numerator.

\(\displaystyle x^{2}+10x+25=0\)

\(\displaystyle (x+5)(x+5)=0\)

\(\displaystyle x=-5\)

So we know that the function will equal zero when \(\displaystyle x=-5\).  If we just look at the numerator of the function, then this graph would be a parabola with its point at \(\displaystyle x=-5\).  Now we will solve for the points where the function is undefined by setting the denominator equal to zero and solving for \(\displaystyle x\).

 \(\displaystyle x-2=0\)

\(\displaystyle x=2\)

 And so the function is undefined at \(\displaystyle x=2\).  If we make a table to solve for some of the points of the graph:

 And if we graph these points we see something like below (which is our answer).  Note that the dotted blue line is the vertical asymptote at \(\displaystyle x=2\).

The zeros of the function are the values of \(\displaystyle x\) where the function will be equal to zero.  In order to find these we set the numerator of the function equal to zero.

 \(\displaystyle x^{2}-4x+4=0\)

\(\displaystyle (x-2)(x-2)=0\)

We only need to solve for \(\displaystyle x-2=0\) once,

\(\displaystyle x-2=0\)

\(\displaystyle x=2\)

So the zeros of this function are \(\displaystyle x=2\).

To solve for the points at which this function will be undefined, we set the denominator equal to zero and solve for \(\displaystyle x\).

\(\displaystyle x^{2}-25=0\)

\(\displaystyle x^{2}=25\)

\(\displaystyle x= \pm 5\)

And so the function is undefined at \(\displaystyle x= \pm 5\)

This inequality wants all values where \(\displaystyle x\) is greater than \(\displaystyle -3\).  So everything up until \(\displaystyle -3\) is included and this is represented by having a dotted line on the graph or an open circle on a number line.

We will first begin by solving for the zeros and undefined points of the inequality.  We solve for the zeros by setting the numerator equal to zero.

\(\displaystyle x^{2}+x-2=0\)

\(\displaystyle \left ( x+2 \right )\left ( x-1 \right )=0\)

\(\displaystyle x=1,-2\) 

And so the zeros of this function are at \(\displaystyle x=1,-2\)

Now we will solve for the undefined points by setting the denominator equal to zero.  Since the denominator is \(\displaystyle x\), then whenever \(\displaystyle x=0\), the function is undefined.  Now we need to find for which values of \(\displaystyle x\) is each factor is greater than zero.

For \(\displaystyle x-1> 0\), any value where \(\displaystyle x> 0\) will be positive and we will be able to graph it.  For \(\displaystyle x+2> 0\), any value where \(\displaystyle x> -2\) will be positive.  Now, we can only graph these values up until \(\displaystyle x=0\) because the function is undefined here.  We are able to pick up the graph again once we reach \(\displaystyle x=1\).  The graph will look like the one below.