Question

combine the following rational expressions
(2x+2)/(3x^2+6x+3)+x-7/3x^2-3

Answer 1

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

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

LCD = 3(x + 1)^2(x - 1)

Take it from here.

Answer 2

After many reminders for past posts, you must know by now that brackets are critical for
these type of problems, so

(2x+2)/(3x^2+6x+3)+x-7/(3x^2-3)
= 2(x+1)(3)(x^2 + 2x + 1)/(3(x^2 - 1))
= 6(x+1)(x+1)(x+1)/(3(x+1)(x-1))
= 2(x+1)^2 /(x-1) , x ≠ ±1

Answer 3

reiny
this 2(x+1)(x+1)/x-1 the answer.

Answer 4

oops , let's try that again, should write these out first.
assuming you meant:
(2x+2)/(3x^2+6x+3)+ (x-7)/(3x^2-3)
= 2(x+1)/((3)(x+1)(x+1)) + (x-7)/(3(x+1)(x-1))
LCD is 3(x+1)(x+1)(x-1)

= ( 2(x+1) + (x-7)(x+1) )/( (3(x+1)(x+1)(x-1) )
= (2x+2 + x^2 - 6x - 7)/( 3(x+1)(x+1)(x-1) )
= x^2 - 4x - 5)/ (3(x+1)(x+1)(x-1) )
= (x+1)(x-5) / (3(x+1)(x+1)(x-1) )
= (x-5) / (3(x+1)(x-1) ) , x ± 1