Question 574555
{{{(x + y)(z + 1) = 24}}}
{{{(x + y)(y + 1) = 24}}}, and
{{{(y + z)(x + 1) = 24}}}
Combining the first two equations
{{{(x + y)(z + 1) = (x + y)(y + 1)=24}}}
So {{{x+y}}} cannot be zero and {{{z=y}}}
If {{{z=y}}}, the third equation turns into 
{{{(y + y)(x + 1) = 24}}} ---> {{{2y(x+1)=24}}} --> {{{y(x+1)=12}}} --> {{{yx+y)=12}}} which can be substituted into one of the other equations
{{{(x + y)(y + 1) = 24}}} --> {{{xy+x+y^2+y=24}}} --> {{{(xy+y)+y^2+x=24}}} --> {{{12+y^2+x=24}}} --> {{{y^2+x=12}}} --> {{{x=12-y^2}}}
Substituting {{{x=12-y^2}}} into {{{y(x+1)=12}}} we get
{{{y(13-y^2)=12}}} --> {{{13y-y^3=12}}} --> {{{y^3-13y+12=0}}}
That can be solve by factoring, starting by factoring out {{{y-1}}}, because it's obvious that {{{y=1}}} is a solution
{{{y^3-13y+12=0}}} --> {{{(y-1)(y^2+y-12)=0}}} --> {{{(y-1)(y+4)(y-3)=0}}}
SO the solutions are
{{{z=y=1}}} --> {{{x=12-1^2=11}}}
{{{z=y=3}}} --> {{{x=12-3^2=12-9=3}}}
{{{z=y=-4}}} --> {{{x=12-(-4)^2=12-16=-4}}}
The solutions would be (11,1,1), (3,3,3) and (-4,-4,-4).