How do you solve #(3x+6)/(x^2-4)=(x+1)/(x-2)#?

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Answer 1 · 2016-08-24T14:17:51

#x =2#

Well there is the long way. Cross multiply
#(3x+6)(x-2)=(x^2-4)(x +1)#
Multiply out solve

Or there is the efficient way. Factorise each term first.
#[3(x +2)]/[(x+2)(x-2)]=(x +1)/(x-2)#

Cancel and we are left with 3=#x#+1

Answer 2 · 2016-08-24T14:34:51

This equation has no solution.

#((3(x + 2))/((x + 2)(x - 2)))/((x + 1)/(x - 2)) = 1#

#(3(x + 2)(x - 2))/((x + 2)(x - 2)(x + 1)) = 1#

Cancelling using the property #a/a = 1#, we are left with:

#3/(x + 1) = 1#

#3 = 1(x + 1)#

#3 = x + 1#

#x = 2#

However, this value of #x# is extraneous, since it renders the denominator #0# (which in turn makes the equation undefined).

Hence, this equation has no solution #{O/}#.

Hopefully this helps!