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How do you combine #(x+2)/(x-5)+(x-12)/(x-5)#?

Answer 1

Hazel Anglin

#(2x-10)/(x+5)#

#(x+2)/(x-5)##+##(x-12)/(x-5)#

Since both denominators are the same, just combine the fraction, like so,

#((x+2)+(x-12))/(x-5)#

Open up the brackets,

#(x+2+x-12)/(x-5)# #(2x-10)/(x+5)#

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Answer 2

Eva Abramson

#2#

Before we can add/subtract fractions we require them to have a #color(blue)"common denominator"#

These fractions have a common denominator ( x - 5) so we can add the numerators, leaving the denominator as it is.

#rArr(x+2+x-12)/(x-5)#

#=(2x-10)/(x-5)#

The numerator can be simplified by taking out a #color(blue)"common factor"#

#rArr(2x-10)/(x-5)=(2(cancel(x-5))^1)/cancel(x-5)^1#

#color(blue)"cancelling" " a common factor of " (x-5)#

#=2#

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Answer 3

Eva Adamson

#2#

#(x+2)/(x-5)+(x-12)/(x-5)#

#:.=(x+2+x-12)/(x-5)#

#:.=(2x-10)/(x-5)#

#:.=(2cancel((x-5))^color(red)1)/cancel((x-5))^color(red)1#

#:.=2#

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Answer 4

Nathan Axel

To combine the given expressions, (x+2)/(x-5) and (x-12)/(x-5), we can simply add the numerators together and keep the common denominator. This gives us (x+2+x-12)/(x-5). Simplifying the numerator, we have (2x-10)/(x-5). Therefore, the combined expression is (2x-10)/(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.