Question

how to use Lcd in this problem? [a-2. -_1_ =_3_]find Lcd [ a+3 1 a-2]. (a+3)(a+2). multiply all neumerator to (LCD) a-2 - 1= 3 ___ _ a/+3. a+/2 (a-6 )(a+6) - 1 (a+2)(a+3)(a+2/)/*) negative a +positive2: distribuet: (a(a+3)++2. = (a+3)​

Answer 1

It's not entirely clear to me what you're trying to solve, but it looks like the initial equation is

$\dfrac{a-2}{a+3} -1 = \dfrac3{a+2}$

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 :

$\dfrac{a-2}{a+3}\cdot\dfrac{a+2}{a+2} -1\cdot\dfrac{(a+3)(a+2)}{(a+3)(a+2)} = \dfrac3{a+2}\cdot\dfrac{a+3}{a+3} \\\\ \dfrac{(a-2)(a+2)}{(a+3)(a+2)} - \dfrac{(a+3)(a+2)}{(a+3)(a+2)} = \dfrac{3(a+3)}{(a+3)(a+2)}$

Now that everything has the same denominator, we can combine the fractions into one. Move every term to one side and join the numerators:

$\dfrac{(a-2)(a+2)-(a+3)(a+2)-3(a+3)}{(a+3)(a+2)} = 0$

Simplify the numerator:

$\dfrac{(a^2-4)-(a^2+5a+6)-(3a+9)}{(a+3)(a+2)} = 0 \\\\ \dfrac{-8a-19}{(a+3)(a+2)} = 0$

If neither a = -3 nor a = -2, we can ignore the denominator:

$-8a-19 = 0$

Solve for a :

$-8a = 19 \\\\ \boxed{a = -\dfrac{19}8}$