math final simplifying math final review

Question

Simplify

(x²-2x)/(x²-4x)   ÷    (x2+x-6)/(x²+x-20)

 

Thanks

Answer 1

[(x²-2x)/(x²-4x)]/[(x²+x-6)/(x²+x-20)]

 

First off.. realize that in dividing fractions we invert and multiply

So, invert the [(x²+x-6)/(x²+x-20)] to  (x²+x-20)/(x²+x-6) and then multiply instead of divide

 

[(x²-2x)/(x²-4x)][(x²+x-20)/(x²+x-6)]

 

Now, let's factor what can be factored

 

(x²-2x) = x(x-2)

(x²-4x) = x(x-4)

 

Thus our first fraction is (x(x-2))/(x(x-4)) = (x-2)/(x-4)

 

For the 2nd fraction:  (x²+x-20) = (x+5)(x-4)

                                (x²+x-6) = (x+3)(x-2)

 

Putting it back together we have:

 

[(x-2)/(x-4)][((x+5)(x-4))/((x+3)(x-2))]

 

The first part has (x-4) in the denominator which cancels with (x-4) from numerator of 2nd part

Similarly, the (x-2) in numerator of the 1st part cancels with (x-2) from denominator of 2nd part

This leaves us with:

 

(x+5)/(x+3)