Question 127382
Assume it's
:
{{{3 - ((a-1))/((a+2)) = ((a^2-1))/((a+2))}}}
:
{{{3 = ((a-1))/((a+2)) + ((a^2-1))/((a+2))}}}; Added {{{((a-1))/((a+2))}}} to both sides
:
They have the same denominators so we can just write it:
{{{3 = (( a - 1 + a^2 - 1))/((a+2))}}} = {{{3 = ((a^2 + a - 2))/((a+2))}}}; combine like terms, we have quadratic now
That will factor to:
{{{3 = ((a-1)(a+2))/((a+2))}}}
Cancel out the (a+2)'s and we have a simple equation
a - 1 = 3
:
a = 3 + 1
:
a = 4
:
:
Check solution in the original equation, substitute 4 for a  
{{{3 - ((4-1))/((4+2)) = ((4^2-1))/((4+2))}}}
{{{3 - (3)/(6) = ((16-1))/(6)}}}
{{{3 - (3)/(6) = (15)/(6)}}}
3 is equal to 18/6:
{{{18/6 - (3)/(6) = (15)/(6)}}}; confirms our solution of a = 4

:
Could you follow what we did here? Any questions?