Question 1077251
{{{(s+2)/(s^2-s-6) = 3-4/(s-3)}}} can be written as
{{{(s+2)/((s+2)(s-3)) = 3-4/(s-3)}}} factoring a denominator.
For both sides of the equal sign to exist,
it must be {{{s<>-2}}} and {{{s<>3}}} .
STEP 1 - Eliminate denominators:
If {{{s<>-2}}} and {{{s<>3}}} ,
denominators can be eliminated by multiplying times
{{{(s+2)(s-3)}}} to get the equivalent equations
{{{(s+2)(s+2)(s-3)/((s+2)(s-3)) = 3(s+2)(s-3)-4(s+2)(s-3)/(s-3)}}}
{{{s+2 = 3(s+2)(s-3)-4(s+2)}}}
{{{s+2 = 3(s^2-s-6)-4s-8}}}
{{{s+2 = 3s^2-3s-18-4s-8}}}
{{{s+2 = 3s^2-7s-26}}}
{{{s+2-s-2 = 3s^2-7s-26-s-2}}}
{{{0 = 3s^2-8s-28}}}
STEP 2 - Solve the resulting equation:
{{{3s^2-8s-28=0)}}} is a quadratic (degree 2) equation.
Quadratic equations always can be solved by either
"completing the square"
or using the quadratic formula.
Some quadratic equations can also be solved by factoring,
and if they can be solved by factoring,
and you are good at factoring,
it is the easiest way.
Factoring, you get
{{{(3s-14)(s+2)=0)}}} ,
whose solutions are
{{{3s-14=0}}} <---> {{{s=14/3}}}
and {{{s+2=0}}} <---> {{{s=-2}}}
STEP 3 - Make sure to eliminate invalid solutions:
{{{s=-2}}} is not a valid solution,
because it causes denominator {{{s^2-s-6=(s+2)(s-3)}}} to be zero,
and the rational expression
{{{(s+2)/(s^2-s-6)}}} to be undefined.
The only solution to the original equation is
{{{highlight(s=14/3)}}} .


NOTE:
If you meant {{{(s+2)/(s^2-s-6) = 3-4/(s-3)}}} ,
it should be written as (s+2)/(s^2-s-6) = 3-4/(s-3),
because when evaluating that expression for any value of s,
numerators and denominators must be calculated first, before dividing.
According to the order of operation conventions,
s+2/s^2-s-6 = {{{s+2/s^2-s-6}}} .
Most calculators obey those conventions,
and if you do not enter the necessary parentheses,
you get the wrong answer.