Answer:
There are 4 possible solutions:
Step-by-step explanation:
Note that this is a System of Equations with only 4 equations, and 5 unknowns:
[tex]wy=A\\wz=24\\xy=18\\xz=72[/tex]
In general, when there are more unknowns than equations, it is likely that it is not possible to solve for a unique solution, and that there are infinitely many solutions. In this particular case, the values of w, x, y, z, and A, are limited to whole numbers. There ends up being a limited-finite number of solutions.
Based on the dependency of the table, the product of x and z is 72. So, finding all pairs of factors that multiply to 72 will give all of the possibilities for x and z, from which we could solve for the corresponding values of w, y, and A.
Factor pairs of 72:
1 & 72
2 & 36
3 & 24
4 & 18
5 (doesn't divide out evenly)
6 & 12
7 (doesn't divide out evenly)
8 & 9
To find the possible solutions, we can examine each pair of factors sequentially:
Option 1: x=1 & z=72
Return to equation wz=24. When z=72, w=1/3, not a whole number, so there will be no solution where x=1 & z=72.
Option 2: x=2 & z=36
Return to equation wz=24. When z=36, w=2/3, not a whole number, so there will be no solution where x=2 & z=36.
Option 3: x=3 & z=24
Return to equation wz=24. When z=24, w=1.
Return to equation xy=18. When x=3, y=8.
Return to equation wy=A. When w=1 & y=8, A=8.
So, there is a solution with w=1, x=3, y=8, z=24, and A=8.
Option 4: x=4 & z=18
Return to equation wz=24. When z=18, w=4/3, not a whole number, so there will be no solution where x=4 & z=18.
Option 5: x=6 & z=12
Return to equation wz=24. When z=12, w=2.
Return to equation xy=18. When x=6, y=3.
Return to equation wy=A. When w=2 & y=3, A=6.
So, there is a solution with w=2, x=6, y=3, z=12, and A=6.
Option 6: x=8 & z=9
Return to equation wz=24. When z=9, w=8/3, not a whole number, so there will be no solution where x=8 & z=9.
Don't forget, the multiplication is commutative, so the factor pairs could be reversed for 6 more options that we need to check.
Option 7: x=9 & z=8
Return to equation wz=24. When z=8, w=3.
Return to equation xy=18. When x=9, y=2.
Return to equation wy=A. When w=3 & y=2, A=6.
So, there is a solution with w=3, x=9, y=2, z=8, and A=6.
Option 8: x=12 & z=6
Return to equation xy=18. When x=12, y=3/2, not a whole number, so there will be no solution where x=12 & z=6.
Option 9: x=18 & z=4
Return to equation xy=18. When x=18, y=1.
Return to equation wz=24. When z=4, w=6.
Return to equation wy=A. When w=6 & y=1, A=6.
So, there is a solution with w=6, x=18, y=1, z=4, and A=6.
Options 10, 11, & 12: x=24, 36, or 72
Return to equation xy=18. For each of these cases, note that x is greater than 18, and so y will be less than 1, not a whole number, so there will be no solution for any of those options.
There are 4 possible solutions where the values are all whole numbers. The four solutions are as follows:
w=1, x=3, y=8, z=24, and A=8;
w=2, x=6, y=3, z=12, and A=6;
w=3, x=9, y=2, z=8, and A=6;
w=6, x=18, y=1, z=4, and A=6.
