1. ## will someone please verify the solution to my problems

I have done the work, but i need someone to verify if they are correct.

Simplify.

a)4x^2y^5/x^4y^2 = 4x^2y^3

b)3xy/9x^2-12x = 3y/3x-4

c) 4x^2-20x/x+3 * 4x^2-1/2x^2-9x-5 = 4x^2(2x-1)/(x+3)

d) 4x+1/3x-5 + 2x/12x-20 = (9x+2)/(6x-10)

e) how would i simplify 3x-1/x^2+4x+3 - x+6/2x^2+7x+1

2. a)missing a (-)

$\frac{{4x^2y^5}}{{x^4y^2}}=4x^2y^5x^{-4}y^{-2}=4x^{-2}y^3$

b) err...

$\frac{{3xy}}{{9x^2-12x}}=\frac{{3xy}}{{3x(3x-4)}}=\frac{{y}}{{3x-4}}$

c) this is what i did

$\frac{{4x^2-20x}}{{x+3}}*\frac{{4x^2-1}}{{2x^2-9x-5)}}=\frac{{4x(x-5)}}{{x+3}}*\frac{{4x^2-1}}{{(2x+1)(x-5)}}=\frac{{4x}}{{x+3}}*\frac{{4x^2-1}}{{(2x+1)}}$

continue

d) something is wrong...

$\frac{{4x+1}}{{3x-5}}+\frac{{2x}}{{12x-20)}}=\frac{{4x+1}}{{3x-5}}+\frac{{x}}{{6x-10)}}=\frac{{4x+1}}{{3x-5}}+\frac{{x}}{{2(3x-5)}}$= $\frac{{4x+1+2x}}{{3x-5}}=\frac{{6x+1}}{{3x-5}}$

e) jeez too long

you should do the algebraic sum as usual

$\frac{{a(x)}}{{b(x)}}-\frac{{c(x)}}{{d(x)}}=\frac{{a(x)d(x)-b(x)c(x)}}{{b(x)d(x)}}$

3. Originally Posted by Black Kawairothlite
a)missing a (-)

$\frac{{4x^2y^5}}{{x^4y^2}}=4x^2y^5x^{-4}y^{-2}=4x^{-2}y^3$
Oh, i see!

4. Originally Posted by Black Kawairothlite
a)missing a (-)

$\frac{{4x^2y^5}}{{x^4y^2}}=4x^2y^5x^{-4}y^{-2}=4x^{-2}y^3$

b) err...

$\frac{{3xy}}{{9x^2-12x}}=\frac{{3xy}}{{3x(3x-4)}}=\frac{{y}}{{3x-4}}$
Absolutely, I forgot to cancel that 3 out by accident

5. Originally Posted by Black Kawairothlite
a)missing a (-)

$\frac{{4x^2y^5}}{{x^4y^2}}=4x^2y^5x^{-4}y^{-2}=4x^{-2}y^3$

b) err...

$\frac{{3xy}}{{9x^2-12x}}=\frac{{3xy}}{{3x(3x-4)}}=\frac{{y}}{{3x-4}}$

c) this is what i did

$\frac{{4x^2-20x}}{{x+3}}*\frac{{4x^2-1}}{{2x^2-9x-5)}}=\frac{{4x(x-5)}}{{x+3}}*\frac{{4x^2-1}}{{(2x+1)(x-5)}}=\frac{{4x}}{{x+3}}*\frac{{4x^2-1}}{{(2x+1)}}$

continue
I appreciate the help.