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How do you subtract # b/(b-5) - 2/(b+3) #?

Answer 1

Isaiah Anthony

#(b^2+b+10)/((b-5)(b+3))#

Given -

#b/(b-5)-2/(b+3)#

Find the common denominator. it is -

#(b-5)(b+3)#

Then

#((b+3)b-(b-5)2)/((b-5)(b+3))#

#((b^2+3b)-(2b-10))/((b-5)(b+3))# #(b^2+3b-2b+10)/((b-5)(b+3))#

#(b^2+b+10)/((b-5)(b+3))#

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Answer 2

Nathan Axel

To subtract b/(b-5) - 2/(b+3), you need to find a common denominator. The common denominator in this case is (b-5)(b+3).

To get the first fraction with the common denominator, multiply the numerator and denominator by (b+3). This gives you (b(b+3))/((b-5)(b+3)).

To get the second fraction with the common denominator, multiply the numerator and denominator by (b-5). This gives you (2(b-5))/((b-5)(b+3)).

Now that both fractions have the same denominator, you can subtract the numerators.

The subtraction becomes (b(b+3) - 2(b-5))/((b-5)(b+3)).

Simplifying the numerator gives you (b^2 + 3b - 2b + 10)/((b-5)(b+3)).

Combining like terms in the numerator gives you (b^2 + b + 10)/((b-5)(b+3)).

Therefore, the simplified expression is (b^2 + b + 10)/((b-5)(b+3)).

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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.