When typing questions like this on this webpage, you MUST use brackets
to determine the correct order of operation.
I will interpret your 1/x^2+3x+2 + 2/x^2-1 as
1/(x^2+3x+2) + 2/(x^2-1)
not even looking at what you did, you can find your errors later
= 1/( (x+2)(x+1) ) + 2/( (x-1)(x+1) )
so the LCD is (x-1)(x+1)(x+2)
= (x-1)/( (x-1)(x+1)(x+2) ) + 2(x+2)/( (x-1)(x+1)(x+2) )
= (3x + 3)/( (x-1)(x+1)(x+2) )
= 3(x+1)/( (x-1)(x+1)(x+2) )
= 3/( (x-1)(x+2) ) , x ≠ -1
Your answer is correct, because you wrote it out
on paper in proper fraction form. Unfortunately we
cannot type fractions here, so as I said before, brackets are needed.
1/x^2+3x+2 + 2/x^2-1
I first did 1/x^2+2x+x+2 +2/(x-1)(x+1)
1/x(x+2)x+2 +2/(x-1)(x+1)
1/x+2(x+1) +2/(x-1)(x+1)
x-1+2(x+2)/(x+2)(x-1)(x+1)
x-1+2x+4/(x+2)(x-1)(x+1)
3x+3/(x+2)(x-1)(x+1)
3(x+1)/(x+2)(x-1)(x+1) the x+1 cancel out so
3/(x+2)(x-1)
3/x^2-x+2x-2
answer 3/x^2+x-2
Is this correct? thank you for checking my work
2 answers
Thank you