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How do you solve #1/(x-1) + 2/(x+3) + (2x+2)/(3-2x-x^2)#?

Answer 1

Savannah Anderson

Maybe you see that the third fraction can be factorized.

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

Which leaves the whole thing as: #1/(x-1)+2/(x+3)-(2x+2)/((x-1)(x+3))# (mark the #-# sign)

Next step: make the bottom parts equal: #1/(x-1)xx(x+3)/(x+3)+2/(x+3)xx(x-1)/(x-1)-(2x+2)/((x-1)(x+3))=#

#(1xx(x+3))/((x-1)(x+3))+(2xx(x-1))/((x-1)(x+3))-(2x+2)/((x-1)(x+3))=#

#((x+3)+(2x-2)-(2x+2))/((x-1)(x+3))=#

Put the numbers and the #x#'s together: #((x+cancel(2x)-cancel(2x))+(3-2-2))/((x-1)(x+3))=((x-1))/((x-1)(x+3))=#

Now cancel: #cancel((x-1))/(cancel((x-1))(x+3))=1/(x+3)#

NOTE: Forbidden solutions are #x=1and x=-3# as the numerator of the fraction would be #=0# graph{1/(x+3) [-16.02, 16.02, -8, 8.02]}

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Answer 2

Avery Alexander

To solve the expression 1/(x-1) + 2/(x+3) + (2x+2)/(3-2x-x^2), we need to find a common denominator and combine the fractions. The common denominator is (x-1)(x+3)(3-2x-x^2). After obtaining the common denominator, we can simplify the expression and combine like terms.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.