How do we solve problems with the rationals expression operations?

How do we solve problems with the rationals expression operations?

Let me show you.

When you get a normal problem with multiplication, you just multiply across.

When you get a division problem you have to use the KCF method. 

You Keep the first fraction, then you Change the sign, and finally you Flip the fraction. 

When you add or subtract fractions you normally just add the numerator and keep the denominator, but that's only if you have the same denominator.


When you don't have the same denominator, you have to convert them.

1	+	1
3		6

One third would be equal to 2/6 and one sixth would still stay the same.

Then you add both fraction and get your answer. 

So when you get a expression like this :

You have to do the KCF method. You keep the first fraction, change the sign to multiply, and flip the 2nd fraction. While you're at it, simplify it. 

You'd get    x    times (x-3)(x+3)

                                                                         x+3                3 * x

That would simplify to   x-3  

                                      x

That's how you do a radical expression operation, that simple.

Sources: 

1. John Schnatterly: johnschnatterly.blogspot.com
2. http://www.mathsisfun.com/algebra/rational-numbers-operations.html