find the integral value of x+y+z+4w=20
Answers
If y =1 +2(1) = 3 , x= 18-3(1) = 15.
If y= 1 + 2(2) = 5, x= 18 – 3(2) = 12.
If y= 1 + 2(3) = 7, x = 18 – 3(3) = 9 and so on.
We will substitute the variables in the question such that this case would become similar to previous case
. In previous case,
(X1, X2, X3) >= 1. In this case, (X1, X2, X3) >= 0. Therefore, (X1+1, X2+1, X3+1) >= 1. Substitute X1+1=Y1, X2+1=Y2 and X3+1=Y3 in the given equation such that
(X1+1) + (X2+1) +(X3+1) = 11
=> Y1+Y2+Y3=11.
Number of Integral Solution 2: a + b + c + d < 15.
a + b + c + d = 14,13,12,11,10,9,8,7,6,5,4.(Since we need to find positive integral solutions, sum of 4 variables cannot be less than 4)
Total no of positive solution = 13C3 + 12C3+ 11C3 … 3C3
= 286+220+165+120+84+56+35+20+10+4+1
=1001.
let x = 0 and |y| and |z| be at least equal to 1. Therefore, we need the positive integral solution of b + c = 15, where b = |y| and c = |z|. The number of solutions is 14C1 = 14. Each of these solutions will give two values of y and z and there are 3 ways in which we can keep one of the variables equal to 0. Therefore, total number of ways are 14 x 2 x 2 x 3 = 168.
Now let two of the variables be equal to 0. In this case, the total number of solutions is equal to 6.
Therefore, the total number of integral solutions = 728 + 168 + 6 = 902.