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How do you combine #n/(5-n)+(2n-5)/(n-5)#?

Answer 1

Savannah Anderson

#1#

Using an example. Suppose we had -5

Another way of writing this is: #-(+5)#

So we can 'force' a change in sign so that both denominators are the same:

Write #n/(5-n)" as "n/(-(-5+n))#

Just changing the order we have: #n/(-(-5+n)) -> - n/(n-5)#

So write: #color(red)(n/(5-n))color(green)(+(2n-5)/(n-5))" as "color(green)((2n-5)/(n-5))color(red)( - n/(n-5))#

#=(2n-5-n)/(n-5)" " =" " (n-5)/(n-5)" "=" "1#

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Answer 2

Abigail Allen

To combine the expressions n/(5-n) and (2n-5)/(n-5), we need to find a common denominator. The common denominator in this case is (5-n)(n-5).

Next, we can rewrite the fractions with the common denominator:

n/(5-n) = -n/(n-5) (2n-5)/(n-5) = (2n-5)/(n-5)

Now, we can add the fractions together:

-n/(n-5) + (2n-5)/(n-5) = (-n + 2n - 5)/(n-5) = (n - 5)/(n-5)

Simplifying further, we have:

(n - 5)/(n-5) = 1

Therefore, the combined expression simplifies to 1.

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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.