Aman Bansal
Last Activity: 13 Years ago
Dear Manoj,
If x> 1, then x may be 2 or 3 ... it cannot be 4 or 5... and so on because if x= 4 you have x^2 + 7x = 16 + 28 and it is greater than 37.
a) if x = 2 you have 4 + 14 + 5y + z + w =37 ==> 5y + z+ w = 19
and here y must be 1 or 2 or 3
If y= 1 you have z=1 and w=12 or z=2 and w =11... go on (12 solutions)
If y = 2 you have z= 1 and w= 8 or z=2 and w = 7 ... go on (8 solutions)
If y = 3 you have z= 1 and w=3 or z=2 and w=2 or w=4 and z = 1 (2 solutions)
b) if x = 3 you have 9+21 + 5y + z + w = 37 ==> 5y +z + w = 7
The only solution is y = 1 and z=1 and w = 1 (1 solution)
Number of solutions: 23
Best Of luck
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Aman Bansal
Askiitian Expert