As the g.c.d(6,10,15)= 1 which divides the right hand side , solution exists Take any two terms of 6x + 10y + 15z, lets take 10y + 15z now equate it to g.c.d(10,15)u 10y + 15z =5u....(1) Substitute this original equation we get 6x + 5u =-1 this is an equation in two variables 6(-1) + 5(1) =-1 x=-1 +5t u=1-6t Substitute u in (1) and solve 2y+3z=1-6t we can see 2(-1)+ 3(1)= 1 multiply by 1-6t all over 2(6t-1) +3(1-6t)=1-6t y= 6t-1 + 3k z=1-6t -2k x=-1+5t where t and k can take values 0,1,2,...-1,-2,... You can check if we take particular value t0 and k=0 we get x=-1,y=-1 and z=1 substitute in original equation it is satisfied
As the g.c.d(6,10,15)= 1 which divides the right hand side , solution exists Take any two terms of 6x + 10y + 15z, lets take 10y + 15z now equate it to g.c.d(10,15)u 10y + 15z =5u....(1) Substitute this original equation we get 6x + 5u =-1 this is an equation in two variables 6(-1) + 5(1) =-1 x=-1 +5t u=1-6t Substitute u in (1) and solve 2y+3z=1-6t we can see 2(-1)+ 3(1)= 1 multiply by 1-6t all over 2(6t-1) +3(1-6t)=1-6t y= 6t-1 + 3k z=1-6t -2k x=-1+5t where t and k can take values 0,1,2,...-1,-2,... You can check if we take particular value t0 and k=0 we get x=-1,y=-1 and z=1 substitute in original equation it is satisfied
thank you for your great explanation ma'am. you really help me😊❤
Thanks ma'am 🙏🏼
Thank you so much
Nice
Thanks
please mention the mane of the theorem that you have used
I do not think the theorem has a name . It is proved using induction.
Thanks for Diophantine equation.
In ur eqn you were always able to equal two variables with 1 but in my eqn its not working ......
Will u pls solve
6x + 10y + 15z = -1
As the g.c.d(6,10,15)= 1 which divides the right hand side , solution exists
Take any two terms of 6x + 10y + 15z, lets take 10y + 15z now equate it to g.c.d(10,15)u
10y + 15z =5u....(1)
Substitute this original equation we get 6x + 5u =-1 this is an equation in two variables
6(-1) + 5(1) =-1
x=-1 +5t
u=1-6t
Substitute u in (1) and solve
2y+3z=1-6t
we can see 2(-1)+ 3(1)= 1 multiply by 1-6t all over
2(6t-1) +3(1-6t)=1-6t
y= 6t-1 + 3k
z=1-6t -2k
x=-1+5t where t and k can take values 0,1,2,...-1,-2,...
You can check if we take particular value t0 and k=0 we get x=-1,y=-1 and z=1
substitute in original equation it is satisfied
As the g.c.d(6,10,15)= 1 which divides the right hand side , solution exists
Take any two terms of 6x + 10y + 15z, lets take 10y + 15z now equate it to g.c.d(10,15)u
10y + 15z =5u....(1)
Substitute this original equation we get 6x + 5u =-1 this is an equation in two variables
6(-1) + 5(1) =-1
x=-1 +5t
u=1-6t
Substitute u in (1) and solve
2y+3z=1-6t
we can see 2(-1)+ 3(1)= 1 multiply by 1-6t all over
2(6t-1) +3(1-6t)=1-6t
y= 6t-1 + 3k
z=1-6t -2k
x=-1+5t where t and k can take values 0,1,2,...-1,-2,...
You can check if we take particular value t0 and k=0 we get x=-1,y=-1 and z=1
substitute in original equation it is satisfied
Need proof of the theorem
a|(bx +cy), therefore a|(bx +cy +fz +.....+nn) if it divides gcd. this is one of the first rule of divisibility, which is bezouts theorem
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Thank you so much