Suppose x, y, and z are positive integers such that xy + yz = 29 and x

How can one solve in under 3 min?

Official answer below:
Start by factoring the first equation, which gives y∗(x+z)=29. Since x,y, and z are positive integers and 29 is prime, we must have either y=1 and (x+z)=29, or y=29 and (x+z)=1.

But since x and z are both positive integers, (x+z)=1 is impossible: hence we must have y=1 and (x+z)=29, so y has a unique solution (y=1, and nothing else).

Now we have x+z=29 and the second equation, which is really z∗(x+1)=81. (Remember that y=1, so you can plug y=1 into the second equation.) Since z=29−x, we can plug that into the second equation and get (29−x)∗(x+1)=81.

That simplifies to x2−28x+52=0, which factors as (x−2)(x−26)=0. Hence we have two values of x. Since z=29−x, we also have two values of z (either z=29−2 or z=29−26). y is thus the only variable with a unique solution, and and the answer is B.