Q13/004) Solve in positive integers x + y + z = 97 and (4x/5) + (5y/6) + (6z/7) = 82.

We are given

x + y + z = 97 …1

(4x/5) + (5y/6) + (6z/7) = 82. …(2)

Subtract (2) from (1) to get

x/5 + y/6 + z/7 = 15  … (3)

now as 5,6,7 ar pairwise co primes

x is a multiple of 5, y of 6 and z of 7

so x = 5a, y = 6b, z = 7c where a,b,c are integers

so 5a + 6b + 7c = 97 ..(4) from (1)

a+ b+c = 15   … (5) from (2)
multiply (5) by 5 and subtract from (4) to get

b + 2c = 22  … 6

subtract (5) from (6) to get

c = 7+ a

so solution set as c = a + 7 so b= 22-2(a+7) = 8- 2a from (6)

so (a, b = 8-2a, c= a+7) is the solution over integers

for positive integers a can be 1 or 2 or 3

a = 1,  b = 6, c = 8=> (x,y,z) = (5,36,56)

a = 2, b= 4, c= 9 =>(x,y,z) = (10,24,63)

a=3 , b= 2, c= 10=>(x,y,z) = (15,12,70)