Notice that for 2 numbers a and b, \((a+1)(b+1)=ab+a+b+1\) This is very similar to the equations above, so we can rewrite the equations like this:

\((x+1)(y+1)-1=19 \rightarrow (x+1)(y+1)=20\\(y+1)(z+1)-1=29\rightarrow(y+1)(z+1)=30\\(z+1)(x+1)-1=5\rightarrow(z+1)(x+1)=6\)

Let us replace (x+1) with a, (y+1) with b, and (z+1) with c, so it is easier to solve:

\(ab=20\\bc=30\\ca=6\)

Multiply all the equations and then take the square root:

\(a^2b^2c^2=3600\\ abc=60\)

Now, to solve for a, b, and c, just divide abc by one of the equations of ab, bc, or ca:

\(a = \frac{abc}{bc}=2\\ b = \frac{abc}{ac}=10\\ c = \frac{abc}{ab}=3\)

Now, subtract each of a, b, and c by 1 to solve for x, y and z, respectively:

\(x =1\\ y=9\\ z=2\)

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