Question 942215
Three equations to solve, too many for one posting, but the skills to use are very much the same.


The second equation:
(x+1)/(2-x)+(x+2)/(x+3)-(4x+7)/(6-x-x^2)=0, in pure text, appearing as rendered to
{{{(x+1)/(2-x)+(x+2)/(x+3)-(4x+7)/(6-x-x^2)=0}}}.


FRACTION SKILLS!  A few simple algebraic tricks, too.
The partial fraction on the right location can be adjusted to an ADDITION if you multiply it by {{{-1}}}.


{{{(x+1)/(2-x)+(x+2)/(x+3)+(4x+7)/(-6+x+x^2)=0}}}


Some factoring:
{{{(x+1)/(2-x)+(x+2)/(x+3)+(4x+7)/((x+3)(x-2))=0}}}


Fix the left position rational expression also...
{{{-(x+1)/(x-2)+(x+2)/(x+3)+(4x+7)/((x+3)(x-2))=0}}}, both those tricks not necessary but can make some of the work a little more comfortable.


LOWEST COMMON DENOMINATOR - Typical Fraction Skill -
Multiply the left member and right member by (x-2)(x+3) and simplify.
The left member is a sum of rational expressions, and the right member is 0.  Not showing this actual multiplication step, but showing just the first result is this:


{{{-(x+1)(x+3)+(x+2)(x-2)+(4x+7)=0}}}


You should be able to handle the rest.  You will get a quadratic equation to finish either by factoring or by general solution to quadratic equation.