Our first step is to solve the bottom equation for : 

so we can substitute it into the top equation:

Now we can plug in our y-values into the bottom equation to find our x-values:

Remember we cannot take a square root of a negative number without getting an imaginary number. As such, we'll just focus on the  values.

Our solution is then: 

The first step is to solve the bottom equation for : 

  since our question specifies for  we just focus on 

We now substitute this equation into the top equation:

we can now plug in our x-values into the bottom equation to find our y-values:

The solutions are then: 

We can solve this equation by using substitution since the bottom equation is already solved for . Substituting the bottom equation into the top we get:

 We then solve the equation for our  values:

Finally, we substitute our values into the bottom equation to get our  values:

Our different solutions are then: 

We can substitute the top equation into the bottom:

and solve for  values:

Now that we have our  values we can plug it into the top equation and find our  values

So, our values are 

First, we must solve for the roots of the cubic polynomial equation.

We obtain that the roots are .

Now there are four regions created by these numbers:

• . In this region, the values of the polynomial are negative (i.e.plug in  and you obtain 

• . In this region, the values of the polynomial are positive (when , polynomial evaluates to )

•  . In this region the polynomial switches again to negative.

• . In this region the values of the polynomial are positive

Hence the two regions we want are  and .

First, subtract  from both sides so you get 

.

Then find the common denominator and simplify 

.

Next, factor out the numerator 

and set each of the three factor equal to zero and solve for .

The solutions are 

.

Now plug in values between , , , and  into the inequality and observe if the conditions of the inequality are met.

Note that . They are met in the interval  and .

Thus, the solution to the inequality is  

. 

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

2) Simplify:

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

     For interval notation:

4) Graph:

Graph the rational expression,

1) Because  and a divide by is undefined in the real number system, there is a vertical asymptote where .

2) As   ,   , and as  ,  .

3) As  ,  , and as   ,  .

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

One approach for solving the inequality: 

For 

1) Determine where  over the x-values  or .

2)  for the intervals  or .

3) Then the solution is .

Another approach for solving the inequality:

1) Write   as , then determine the x-values that cause  to be true: 

2)  is true for  or .

3) Then the solution is .

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.

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

 And so the function is undefined at .  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 .