+306 Four positive integers w, x, y, and z satisfy the following equations:
\(\begin{align*} wx+2w+x&=348 \\ xy+4x+3y&=373 \\ yz+8y+6z&=544 \end{align*}\)
What are w, x, y, and z?
I was able to factor the equations:
\(\begin{align*} (w+1)(x+2)&=350,\\ (x+3)(y+4)&=385,\\ (y+6)(z+8)&=592. \end{align*}\)
and find each of the variables in terms of another variable using those factored equations but I'm not sure what to do from here. Thanks in advance for your help!
See answer #2 by heureka to a very similar question here: https://web2.0calc.com/questions/pls-help-on-part-b
P.S. I can give you the values of w, x, y and z if that helps in any way.
+306 In this problem, the value of the product wxyz is not given, so I don't think I can solve it in the same way. I'm also more interested in how to solve the problem than just the answers. Thanks for linking the other question and for your help - I enjoyed reading heureka's solution!
Here, unlike Heureka's question where there were four equations, there are only three equations in the four unknowns, so it's unlikely that there is a unique solution.
The usual method for this sort of problem, where the number of unknowns exceeds the number of equations by one, is to assign some arbitrary value to one of the unknowns and to then find the other unknowns in terms of that value.
With this problem then, start by letting x = t, (t an arbitrary integer).
Then,
\(\displaystyle w(t+2)+t=348,\text{ so, }w=\frac{348-t}{t+2}.\)
\(\displaystyle w = \frac{350 -(t+2)}{t+2}=\frac{350}{t+2}-1.\)
w is to be an integer, so it's necessary that t + 2 should divide into 350 without remander.
350 has factors 2, 5, 5, 7 so t + 2 can take one of the values, 2, 5, 7, 10, 14, 25, 35, 50, 70, 175 and 350,
so t can take one of the values, 0, 3, 5, 8, 12, 23, 33, 48, 68, 173, 348.
Similarly,
\(\displaystyle y(t+3)+4t=373, \text{ so, }y=\frac{373-4t}{t+3}=\frac{385-4(t+3)}{t+3}=\frac{385}{t+3}-4.\)
385 has prime factors 5, 7, 11, so t + 3 has to take one of the values 5, 7, 11, 35, 55, 77, 385,
in which case t has to take one of the values 2, 4, 8, 32, 52, 74, 382.
Comparing the two sets of possibles for t, the only one that appears in both is t = 8, leading to the unique solution
w = 34, x = 8, y = 31, z = 8.
Using the same routine with the original problem finds two possible solutions to the three equations in four unknowns problem,
w = 24, x = 20, y = 6, z = 14,
and
w = 74, x = 6, y = 20, z = 4.
The fourth equation then narrows this down to a unique solution.
