For what value(s) of x does the equality  `((x+1) (x-2)) / (x-2) = x+1` hold?  A. −1 only B. 2 only C. Any value D. Any value except –1 E. Any value except 2

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steveschoen's profile pic

steveschoen | College Teacher | (Level 1) Associate Educator

Posted on

Since we have x-2 on the top and bottom, they can cancel out.  So, x+1 = x+1, which would mean x can be any number.  However, we need to remember x+1 = x+1 wasn't the original problem.  We did have the x-2's.  Given that, x can't be 2, since if x = 2, we would have division by 0 on the left (2-2 = 0).  So, we would have any number but 2.  So, the answer is E.

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llltkl | College Teacher | (Level 3) Valedictorian

Posted on

For the given equality:

`((x+1)(x-2))/(x-2)=(x+1)` to hold the denominator on the L.H.S. i.e

`(x-2)!=` `0`    (because `0/0` is undefined)

`rArr x!= 2`

Hence,` x` can have any value except `2` .

Therefore, the correct answer is option E).

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Check:

For` x=-1` , L.H.S.=`((-1+1)(-1-2))/((-1-2))=0`     R.H.S.=`-1+1=0`

For `x=3` , L.H.S.=`((3+1)(3-2))/((3-2))=4`       R.H.S.=`3+1=4`

aruv's profile pic

aruv | High School Teacher | (Level 2) Valedictorian

Posted on

`((x+1)(x-2))/(x-2)=x+1`

LHS is defined for all real values of x except at x=2 because

`((2+1)(2-2))/(2-2)=(3xx0)/0=0/0`  , which is indeterminat form. It is easy to understad from this function

`f(x)=((x+1)(x-2))/(x-2)-(x+1)`

limit of f(x) as x->2 exist but f(x) is not defined when x=2.

When we simplify the f(x) then f(x) is defined for a value of x. After simplification f(x) will change its characteristics.

Thus correct answer is E

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