Question 969763
<pre>
Although the other tutor got the right answer, she is incorrect
for multiplying both sides of the inequality by the variable x.
That's because a variable x might not be positive, and if you
multiply through by a negative, that would reverse the > to <.
Also she did not use test values on the intervals.  

Here is a correct solution:
</pre>
Solve the following inequality 63x-32>63/x
<pre>
{{{63x-32}}}{{{"">""}}}{{{63/x}}}

Get 0 on the right side by subtracting {{{63/x}}} from both sides:

{{{63x-32-63/x}}}{{{"">""}}}{{{"0"}}}

Get the LCD = x on the left side:

{{{63x*expr(x/x)-32*expr(x/x)-63/x}}}{{{"">""}}}{{{"0"}}}

{{{(63x^2-32x-63)/x}}}{{{"">""}}}{{{"0"}}}

Factor the numerator on the left:

{{{((7x-9)(9x+7))/x{{{"">""}}}{{{"0"}}}

We find all the critical numbers.  They are the values
obtained by setting the numerator=0 and the denominator=0

The numerator (7x-9)(9x+7) when set equal to 0 gives 
critical numbers {{{9/7}}} and {{{-7/9}}}
The denominator x when set equal to 0 gives the critical
number 0.

We draw a number line and mark the critical numbers on it,
in order from smallest to largest.

-----o------o--------------------o--------------
   -7/9     0                   9/7    

We have 4 intervals to test by substituting a test point in the interval:
1. Left of {{{-7/9}}}, which is {{{(matrix(1,3,-infinity,",",-7/9))}}}
2. Between {{{-7/9}}} and {{{"0"}}}, which is {{{(matrix(1,3,-7/9,",","0"))}}}
3. Between {{{"0"}}} and {{{9/7}}}, which is {{{(matrix(1,3,0,",",9/7))}}}
4. Right of {{{9/7}}}, which is {{{(matrix(1,3,9/7,",",infinity))}}}

1. The easiest number in {{{(matrix(1,3,-infinity,",",-7/9))}}} to test in the original is x = -1

{{{63x-32}}}{{{"">""}}}{{{63/x}}}
{{{63(-1)-32}}}{{{"">""}}}{{{63/(-1)}}}
{{{-63-32}}}{{{"">""}}}{{{-63}}} 
{{{-95}}}{{{"">""}}}{{{-63}}}
That's false so the solution set does not include {{{(matrix(1,3,-infinity,",",-7/9))}}}

2. The easiest number in {{{(matrix(1,3,-7/9,",",0))}}} to test in the original is x = -.1

{{{63x-32}}}{{{"">""}}}{{{63/x}}}
{{{63(-.1)-32}}}{{{"">""}}}{{{63/(-.1)}}}
{{{-6.3-32}}}{{{"">""}}}{{{-630}}} 
{{{-38.3}}}{{{"">""}}}{{{-630}}}
That's true so the solution set does include {{{(matrix(1,3,-7/9,",",0))}}}

3. The easiest number in {{{(matrix(1,3,0,",",9/7))}}} to test in the original is x = 1

{{{63x-32}}}{{{"">""}}}{{{63/x}}}
{{{63(1)-32}}}{{{"">""}}}{{{63/(-.1)}}}
{{{63-32}}}{{{"">""}}}{{{-630}}} 
{{{31}}}{{{"">""}}}{{{-630}}}
That's true so the solution set does include {{{(matrix(1,3,-7/9,",",0))}}}

3. The easiest number in {{{(matrix(1,3,9/7,",",infinity))}}} to test in the original is x = 2

{{{63x-32}}}{{{"">""}}}{{{63/x}}}
{{{63(2)-32}}}{{{"">""}}}{{{63/(2)}}}
{{{126-32}}}{{{"">""}}}{{{31.5}}} 
{{{94}}}{{{"">""}}}{{{31.5}}}
That's true so the solution set does include {{{(matrix(1,3,9/7,",",infinity))}}}

None of the critical numbers are part of the solution since the
symbol is > and not &#8805;.

So the solution set is:

{{{matrix(1,3,
(matrix(1,3,-7/9,",",0)),U,(matrix(1,3,9/7,",",infinity)))}}}

</pre>
a.) Is the point x=0 included in the solution set of the inequality?
<pre>
No because if we substitute x=0,

{{{63(0)-32}}}{{{"">""}}}{{{cross(63/0)}}}

It is meaningless to divide by zero.

</pre>
b.) are the other finite end points of the interval included in the solution set?
<pre>
No because the symbol is > and not &#8805;
</pre>
c.) what is the solution set? (in interval notation)
<pre>
{{{matrix(1,3,
(matrix(1,3,-7/9,",",0)),U,(matrix(1,3,9/7,",",infinity)))}}}


Edwin</pre>