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.
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
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
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Thank you so much
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