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# Equation problem

Find the number of integral solutions of 2x + 3y = 763 such that x and y are both positive....

Note by Kislay Raj
4 years, 8 months ago

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We can rewrite $$x=\frac{763-3y}{2}$$.

Since $$x$$ and $$y$$ are positive integers$$3y$$ should not exceed $$763$$. Now we see that $$x$$ can only be positive when $$3y$$ is odd. Also, since $$y$$ is positive it can only take values up to $$\frac{762}{3}=254$$ ($$3y$$ will never be 763 due to the positive constraint).

Now since $$y$$ has to be odd there are $$127$$ odd numbers(between 1 and 254), so no of possible positive integer solution is $$127$$.

- 4 years, 8 months ago

Good solution!

Staff - 4 years, 8 months ago

we have 2x + 3y = 763 divide throughout by 2, the smaller coefficient; thus x + y + y/2 = 381 + 1/2 => x + y + (y-1)/2 = 381 .................................1.) since x and y are to be integers , therefore (y-1)/2 = integer we multiply (y-1)/2 by 5 in order to make the coefficient of y differ by unity from a multiple of 2.

=> (5y-5)/2 = integer => 2y-2 + (y-1)/2 = integer and therefore (y-1)/2 = integer =p (suppose) therefore y-1 =2p => y= 2p+1..................................................................2.) substitute this value of y in the first equation => x + 3p + 1 = 381 => x =380 -3p................................................................3.) but if p>126 , we get negative value of x since we need only positive values of x and y, hence p can take values between 0 and 126 only

hence corresponding values of x and y we will get.... so to get positive integral values for x and y , p= 0,1,2,3, ........126 therefore we will have 127 solutions altogether

- 4 years, 8 months ago

y=(763-2x)/3 Minimum value of x = 2 and maximum value of x=380 such that it makes the above equation divisible by 3. x=2,5,8........,380. i.e. it forms an A.P. So, 380= 2+(n-1)3 378/3=n-1 127=n.

- 3 years ago