Solving binary linear equations: {X / 3 - (y + 1) / 6 = 3,2 (Y-X / 2) = 3 (Y-X / 18)}

Solving binary linear equations: {X / 3 - (y + 1) / 6 = 3,2 (Y-X / 2) = 3 (Y-X / 18)}

1 formula * 6 is 2x - (y + 1) = 18, and reduced to 2x-y = 19 （4）
Formula 2 is reduced to Y-X = (Y-X) / 6
3 formula * 6 is 6y-6x = Y-X, and 5y-5x = 0 （5）
(4) If * 5 is 10x-5y = 95 （6）
(5) + (6) is 5x = 95, x = 19
y=19

How to solve the equation of three unknowns and two equivalent relations, for example: x + y + Z = 10,10x + 5Y + 2Z = 120

If it is an integer solution, the elimination method can be used
Z = 10-x-y, substitute into 2 Formula: 10x + 5Y + 20-2x-2y = 120
8x+3y=100
y=(100-8x)/3=33-3x+(1+x)/3
We get: 1 + x = 3T, t is any integer
so
x=3t-1,y=33-3(3t-1)+t=36-8t
z=10-3t+1-36+8t=5t-25
If it is a real number solution, then there are also countless groups of solutions, the above t can be any real number