How do you add or subtract #7/(x-5)-(2+x)/(x-5)#?

Answer 1

#-1#

Here's how I did it:

To add or subtract fractions, we must have the same denominator for all expressions. In this case, both expressions have the same denominator, #x-5#, so we do not need to worry about that.
This means we can combine the two expressions and just worry about the numerator: #(7-(2+x))/(x-5)#
Now we distribute the negative to everything in the parenthesis: #(7-2-x)/(x-5)#
Subtract #7# with #2#: #(5-x)/(x-5)#
Rewrite that as: #(-x+5)/(x-5)#
Factor out the negative: #(-(x-5))/(x-5)#
#(-(cancel(x-5)))/(cancel(x-5))#
#-1#
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Answer 2

To add or subtract the given expression, 7/(x-5) - (2+x)/(x-5), we first need to find a common denominator. In this case, the common denominator is (x-5).

To add or subtract fractions with the same denominator, we can simply combine the numerators and keep the common denominator.

So, the expression simplifies to (7 - (2+x))/(x-5).

Further simplifying, we have (7 - 2 - x)/(x-5), which becomes (5 - x)/(x-5).

Therefore, the simplified expression is (5 - x)/(x-5).

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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.

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.

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.

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.

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