How do you simplify a fraction? The answer is, you divide the top and bottom by the same thing.
4
6
=
4
÷
2
6
÷
2
=
2
3
4
6
=
4
÷
2
6
÷
2
=
2
3
size 12{ { {4} over {6} } = { {4 div 2} over {6 div 2} } = { {2} over {3} } } {}
(1)
So
4646 size 12{ { {4} over {6} } } {} and
2323 size 12{ { {2} over {3} } } {} are two different ways of writing the same number.
Table 1

On the left, a pizza divided into six equal slices: the four shadedin regions represent
4646 size 12{ { {4} over {6} } } {} of a pizza. On the right, a pizza divided into three equal slices: the two shadedin regions represent
2323 size 12{ { {2} over {3} } } {} of a pizza. The two areas are identical:
4646 size 12{ { {4} over {6} } } {} and
2323 size 12{ { {2} over {3} } } {} are two different ways of expressing the same amount of pizza. 

In some cases, you have to repeat this process more than once before the fraction is fully simplified.
40
48
=
40
÷
4
48
÷
4
=
10
12
=
10
÷
2
12
÷
2
=
5
6
40
48
=
40
÷
4
48
÷
4
=
10
12
=
10
÷
2
12
÷
2
=
5
6
size 12{ { {"40"} over {"48"} } = { {"40" div 4} over {"48" div 4} } = { {"10"} over {"12"} } = { {"10" div 2} over {"12" div 2} } = { {5} over {6} } } {}
(2)
It is vital to remember that we have not divided this fraction by 4, or by 2, or by 8. We have rewritten the fraction in another form:
40484048 size 12{ { {"40"} over {"48"} } } {} is the same number as
5656 size 12{ { {5} over {6} } } {}. In strictly practical terms, if you are given the choice between
40484048 size 12{ { {"40"} over {"48"} } } {} of a pizza or
5656 size 12{ { {5} over {6} } } {} of a pizza, it does not matter which one you choose, because they are the same amount of pizza.
You can divide the top and bottom of a fraction by the same number, but you cannot subtract the same number from the top and bottom of a fraction!
4048=40−3948−39=194048=40−3948−39=19 size 12{ { {"40"} over {"48"} } = { {"40"  "39"} over {"48"  "39"} } = { {1} over {9} } } {}
✗✗
Wrong!
Given the choice, a hungry person would be wise to choose
40484048 size 12{ { {"40"} over {"48"} } } {} of a pizza instead of
1919 size 12{ { {1} over {9} } } {}.
Dividing the top and bottom of a fraction by the same number leaves the fraction unchanged, and that is how you simplify fractions. Subtracting the same number from the top and bottom changes the value of the fraction, and is therefore an illegal simplification.
All this is review. But if you understand these basic fraction concepts, you are ahead of many Algebra II students! And if you can apply these same concepts when variables are involved, then you are ready to simplify rational expressions, because there are no new concepts involved.
As an example, consider the following:
x
2
−
9
x
2
+
6x
+
9
x
2
−
9
x
2
+
6x
+
9
size 12{ { {x rSup { size 8{2} }  9} over {x rSup { size 8{2} } +6x+9} } } {}
(3)
You might at first be tempted to cancel the common
x2x2 size 12{x rSup { size 8{2} } } {} terms on the top and bottom. But this would be, mathematically, subtracting x2x2 size 12{x rSup { size 8{2} } } {} from both the top and the bottom; which, as we have seen, is an illegal fraction operation.
Table 2
x
2
−
9
x
2
+
6x
+
9
x
2
−
9
x
2
+
6x
+
9
size 12{ { {x rSup { size 8{2} }  9} over {x rSup { size 8{2} } +6x+9} } } {}


=
−
9
6x
+
9
=
−
9
6x
+
9
size 12{ {}= { {  9} over {6x+9} } } {}

✗✗
Wrong!

To properly simplify this expression, begin by factoring both the top and the bottom, and then see if anything cancels.
Table 3
x
2
−
9
x
2
+
6x
+
9
x
2
−
9
x
2
+
6x
+
9
size 12{ { {x rSup { size 8{2} }  9} over {x rSup { size 8{2} } +6x+9} } } {}

The problem

=
(
x
+
3
)
(
x
−
3
)
(
x
+
3
)
2
=
(
x
+
3
)
(
x
−
3
)
(
x
+
3
)
2
size 12{ {}= { { \( x+3 \) \( x  3 \) } over { \( x+3 \) rSup { size 8{2} } } } } {}

Always begin rational expression problems by factoring! This factors easily, thanks to(x+a)(x−a)=x2−ax(x+a)(x−a)=x2−ax size 12{ \( x+a \) \( x  a \) =x rSup { size 8{2} }  a rSup { size 8{x} } } {} and
(x+a)2=x2+2ax+a2(x+a)2=x2+2ax+a2 size 12{ \( x+a \) rSup { size 8{2} } =x rSup { size 8{2} } +2 ital "ax"+a rSup { size 8{2} } } {} 
=
x
−
3
x
+
3
=
x
−
3
x
+
3
size 12{ {}= { {x  3} over {x+3} } } {}

Cancel a common (x+3)(x+3) size 12{ \( x+3 \) } {} term on both the top and the bottom. This is legal because this term was multiplied on both top and bottom; so we are effectively dividing the top and bottom by (x+3)(x+3) size 12{ \( x+3 \) } {}, which leaves the fraction unchanged. 
What we have created, of course, is an algebraic generalization:
x
2
−
9
x
2
+
6x
+
9
=
x
−
3
x
+
3
x
2
−
9
x
2
+
6x
+
9
=
x
−
3
x
+
3
size 12{ { {x rSup { size 8{2} }  9} over {x rSup { size 8{2} } +6x+9} } = { {x  3} over {x+3} } } {}
(4)
For any x value, the complicated expression on the left will give the same answer as the much simpler expression on the right. You may want to try one or two values, just to confirm that it works.
As you can see, the skills of factoring and simplifying fractions come together in this exercise. No new skills are required.
"DAISY and BRF versions of this collection are available."