Answer to Question #124494 in Algebra for Daryl D. Ligutan

Question

Find all integer solutions to the equation 4x + 3y – 9z = 5 for values of x, y, and z between 0 and 100.

Answer 1

4x+3y-9z=5

=> 4x= 5+9z-3y

=> x = $\frac{(5+9z-3y)}{4}$ = $\frac{5+3(3z-y)}{4}$

For integral values of x , 3z-y will have the values 1,5,9,13.....129 [ an AP with first term 1 and common difference 4] and corresponding values of x are 2,5,8,11......98 [ an AP with first term 2 and common difference 3]

When x = 2, 3y-9z = -3=> y= 3z-1

When x= 5, 3y-9z = -15=> y = 3z-5

When x= 8, 3y-9z= -27=> y = 3z-9

and continuing so on

When x= 98, y= 3z-129

Let z = k , an integer

So integral solutions of the given equation are

(x,y,z) = (2, 3k-1,k) where 0<3k-1<100

or (x,y,z) = (5, 3k-5,k) where 0<3k-5<100

or (x,y,z) = (8, 3k-9,k) where 0<3k-9<100

continuing so on......

or (x,y,z) = (98, 3k-129,k) where 0<3k-129 <100

Therefore the integral solutions are given by (x,y,z) =

(2, 3k-1,k) where 1/3<k<101/3, k$\in$ Z

or (5, 3k-5,k) where 5/3<k<35, k$\in$ Z

or (8, 3k-9,k) where 3<k<109/3, k$\in$ Z

continuing so on.......

or (98, 3k-129,k) where 43<k<229/3, k$\in$ Z, Z being a set of integers