It's not entirely clear to me what you're trying to solve, but it looks like the initial equation is
[tex]\dfrac{a-2}{a+3} -1 = \dfrac3{a+2}[/tex]
First convert each term into a fraction with the same (i.e. the least common) denominator. The first term needs to be multiplied by a + 2; the second term by (a + 3) (a + 2); and the third term by a + 3 :
[tex]\dfrac{a-2}{a+3}\cdot\dfrac{a+2}{a+2} -1\cdot\dfrac{(a+3)(a+2)}{(a+3)(a+2)} = \dfrac3{a+2}\cdot\dfrac{a+3}{a+3} \\\\ \dfrac{(a-2)(a+2)}{(a+3)(a+2)} - \dfrac{(a+3)(a+2)}{(a+3)(a+2)} = \dfrac{3(a+3)}{(a+3)(a+2)}[/tex]
Now that everything has the same denominator, we can combine the fractions into one. Move every term to one side and join the numerators:
[tex]\dfrac{(a-2)(a+2)-(a+3)(a+2)-3(a+3)}{(a+3)(a+2)} = 0[/tex]
Simplify the numerator:
[tex]\dfrac{(a^2-4)-(a^2+5a+6)-(3a+9)}{(a+3)(a+2)} = 0 \\\\ \dfrac{-8a-19}{(a+3)(a+2)} = 0[/tex]
If neither a = -3 nor a = -2, we can ignore the denominator:
[tex]-8a-19 = 0[/tex]
Solve for a :
[tex]-8a = 19 \\\\ \boxed{a = -\dfrac{19}8}[/tex]
