plaverbach wrote:
How can one solve in under 3 min?
Official ansower above:
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.
The way you have described should get you an answer in less than 2 mins.
To make the calculation a bit faster, recognise that in the second equation \(z * (x+1) = 81\), 1 is already being added to x. So, it might be faster to substitute the value of x in terms of z.
Doing that, the equation becomes:
\(z * (29 - z + 1) = 81\)
\(30z - z^2 = 81\)
\(z^2 - 30z + 81 = 0\)
\((z - 3) (z - 27) = 0\)
So, \(z = 3\) or \(z = 27\)
Now, since z can take two values, without any further calculations, we know that x can take two values as well.
Hence, B is correct.
Hope that makes it clear.
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Psst...If that helped press the thumb on the left