Question 888525
<pre>
{{{x^2 - 8x <= -15}}}

Get 0 on the right, everything else on the left:

{{{x^2 - 8x +15 <= 0}}}

{{{(x-3)(x-5)<=0}}}

The critical numbers are the numbers which either make
the left side 0 or undefined.

The critical numbers are 3 and 5.

Place these numbers on a number line:

----------------o-----o------------  
-2 -1  0  1  2  3  4  5  6  7  8  9

we test any value less than 3, say 0, in the inequality

{{{(x-3)(x-5)<=0}}}
{{{(0-3)(0-5)<=0}}}
{{{(-3)(-5)<=0}}}
{{{15<=0}}}

That's false so we do not shade the number line to the
left of 3

----------------o-----o------------  
-2 -1  0  1  2  3  4  5  6  7  8  9

we test any value between 3 and 5, say 4, in the inequality

{{{(x-3)(x-5)<=0}}}
{{{(4-3)(4-5)<=0}}}
{{{(1)(-1)<=0}}}
{{{-1<=0}}}

That's true so we shade the number line between 3 and 5.

----------------o=====o------------  
-2 -1  0  1  2  3  4  5  6  7  8  9

we test any value to the right of 5, say 6, in the inequality

{{{(x-3)(x-5)<=0}}}
{{{(6-3)(6-5)<=0}}}
{{{(3)(1)<=0}}}
{{{3<=0}}}

That's false so we do not shade the number line to the right of 5.

Next we test the critical points themselves:

----------------o=====o------------  
-2 -1  0  1  2  3  4  5  6  7  8  9

Test 3

{{{(x-3)(x-5)<=0}}}
{{{(3-3)(3-5)<=0}}}
{{{(0)(-2)<=0}}}
{{{0<=0}}}

That's true so we darken the circle at 3:

----------------&#9679;=====o------------  
-2 -1  0  1  2  3  4  5  6  7  8  9

Test 3

{{{(x-3)(x-5)<=0}}}
{{{(5-3)(5-5)<=0}}}
{{{(2)(0)<=0}}}
{{{0<=0}}}

----------------&#9679;=====&#9679;------------  
-2 -1  0  1  2  3  4  5  6  7  8  9

That's true so we darken the circle at 5 also.

The interval notation for the solution is [3,5]

The set builder notation is  {{{matrix(1,5,    "{",x,"|",3<=x<=5,"}" )}}} 

Edwin</pre>