What are the excluded values for #(5x+1)/(x^2-1)#?

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Answer 1 · 2017-04-04T12:23:39

See the entire solution process below:

Because we cannot divide by #0# the excluded values are:

#x^2 - 1 != 0#

We can factor #x^2 - 1# using the rule:

#a^2 - b^2 = (a + b)(a - b)#

Letting #a^2 = x^2#, #a =x#, #b^2 = 1# and #b = 1# and substituting gives:

#(x + 1)(x - 1) != 0#

Now, solve each term for #0# to find the excluded values of #x#:

Solution 1)

#x + 1 = 0#

#x + 1 - color(red)(1) = 0 - color(red)(1)#

#x + 0 = -1#

#x = -1#

Solution 2)

#x - 1 = 0#

#x - 1 + color(red)(1) = 0 + color(red)(1)#

#x - 0 = 1#

#x = 1#

The excluded values are: #x = -1# and #x = 1#