Solving rational equation and adding them

[kpx001]

1.adding

8x^2+3x+1/(x^2-1)+7x-4/x+(4x+x^2)/(x-1)

LCD = x(x-1)(x+1)

x(8x^2+3x+1)+7x-4(x-1)(x+1)+x(4x+x^2)(x+1)/x(x-1)(x+1)

is this correct way to solve it or am i doing something wrong? can someone show steps and answer?
my attempt:

8x^3+3x^2+x+x^2-1(7x-4)+4x^2+x^3(x+1)/x(x-1)(x+1)


Solve for x

1.) 2/(3x+3)=3x/(x-2)
LCD= 3(x+1)(x-2)

2(x-2)=3x(3x+3)
2x-4=9x^2+9x

i got 2 x types?? how do i solve for x and did i do something wrong?

next prob=

2.) x/(x+5)+5(x-5)=x^2/(x^2-25)

LCD= (x+5)(x-5)

x(x-5)+5(x+5)=x^2
x^2-5x+5x+25=x^2
again stumpped cuz x^2 and a x


next prob

3.) 3/(x+1)+3/(x-1)=4
LCD= (x+1)(x-1)

3(x-1)+3(x+1)=4(x+1)(x-1)
more x^2 and x???

 

[stapel]

Has your class not yet covered quadratics and how to solve them?

Thank you.

Eliz.

 

[kpx001]

we have, am i doing something wrong??
please help? Ill understand if i get an example but i wanna learn how to solve these please. sry if im being negative.
by quadratics u mean factoring or the quadratic formula?

 

[stapel]

we have...

Great. Then solve the quadratic you created for (2) and (3), check the solutions in the original equations, and you'll be done with those two.

I'm afraid I can't follow your work in (1), and your formatting is ambiguous, so I hesitate to venture there.

Eliz.

 

[kpx001]

2.

i get 25=0

x^2-5x+5x+25=x^2
25=0

3.

(2x-4)(2x+1)

x= 2; -1/2

 

[kpx001]

can u give me help on what i did wrong?