The correct equation to start with here is
$${1\over X}={1\over A}+{1\over B}+{1\over C}$$
Everything else you did looks correct, so see if you can take it from here.
It's worth noting what's wrong in writing $X={1\over A}+{1\over B}+{1\over C}$. Each of $A$, $B$, $C$, and $X$, as you've defined them, have dimensions of time, assumed in hours. This means that $1/A$ etc. have dimensions of "per hour." You yourself called it "speed." Just as you can't (or shouldn't) add apples and oranges, you can't equate "hours" with "per hours." In working problems of this type, it often helps to keep track of the dimensions that various quantities have, and make sure the dimensions match up whenever you're adding or equating them.
Added at OP's request: Since $X$ is what you want to know, it's best here to convert your equation(s) $X=A-6=B-1=C/2$ into $A=X+6$, $B=X+1$, and $C=2X$, and then change the original equation to
$${1\over X}={1\over X+6}+{1\over X+1}+{1\over2X}$$
Bring the $1/2X$ over the left hand side, which gives
$${1\over 2X}={1\over X+6}+{1\over X+1}$$
From this you should wind up with a quadratic in $X$. Can you take it from here?
Let us assume, they take $a, b,$ and $c$ hours to finish the job individually, respectively. Note the following points:
They complete half the job in half an hour $\implies$ full job is completed in a hour. Hence, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$
Also, A takes 1 hour more time than C. So, $a= c+1\implies \frac{1}{a}=\frac{1}{c+1}$
If A works for a hour, and B works for four hours, then $\frac{1}{a}+\frac{4}{b}=1 \implies \frac{1}{b}=\frac{1-\frac{1}{a}}{4}=\frac{1-\frac{1}{c+1}}{4}=\frac{c}{4+4c}$.
Now, solving for $c$ gives us the quadratic: $3c^2-4c-4=0 \implies c=2, c= -\frac{2}{3}$. As, $c >0$, we have, the rate of work C does $=0.5$.
Solving by your method: $a =\frac{100}{t}, b = \frac{100-a}{4}=\frac{25t-25}{t}$. Then, we get, $$a +b + c =100 \implies 100+25t-25 + ct = 100t \implies t = \frac{75}{75-c}$$
Putting this in Eq.(3), we get, $$\frac{100}{c}=\frac{c}{75-c}\equiv 7500-100c = c^2 \equiv c^2+100c-7500=0 \implies c = \frac{1}{2}, -\frac{3}{2}$$
Rejecting the negative solution, we have $c =\frac12$, the same answer as obtained earlier.
No mistake in your approach!
Best Answer
Let the efficiencies of the first man be $a$, the second be $b$ and that of the woman, $c$.
It is given that: $$a = b + c \tag{1}$$ $$\frac{1}{b} = \frac{1}{b+c} + 3 \tag{2}$$ $$\frac{1}{a}=\frac{2}{b}-8 \tag{3}$$
Solving this, can you get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$?