maximum value of y satisfying this is y=15 and for that x=0;

now decrease y by 3 and increase x by 4 ,you will have x and y
satisfying the equation.

keep on doing this till you reach minimum value of y i.e 0

this you can do  5 times decreasing y=15 by 3 every time

so there will be 5 solutions .

On 8/28/11, Piyush Grover <[email protected]> wrote:
> 3x+4y = 60
> it's a straight line equation whose x intercept is 20 and y intercept is 15.
> Draw it in first quadrant
> (as x, y are positive integers)
> now x = (60 - 4y)/3 = 4(15-y)/3
> now for y = 1, 2...15 you need to check whether (15-y) is divisible by 3 or
> not. It's simple y = 3, 6, 9, 12
>
> -Piyush
>
> On Sun, Aug 28, 2011 at 6:38 PM, Dave <[email protected]> wrote:
>
>> @Sivaviknesh: The smallest values of x and y are 1. The largest value
>> of y is (60 - 3) / 4 = 14. Solve for x: x = (60 - 4y) / 3. Since x is
>> an integer, 60 - 4y must be a multiple of 3. Since 60 - 3y is a
>> multiple of 3, (60 - 4y) - (60 - 3y) = y must be a multiple of 3.
>> Thus, y = 3, 6, 9, 12. Then you can solve for the corresponding values
>> of x.
>>
>> Dave
>>
>> On Aug 28, 7:46 am, sivaviknesh s <[email protected]> wrote:
>> > *Find the number of solutions for 3x+4y=60, if x and y are positive
>> > integers.*
>> >
>> > Is there any standard method for solving these type of ques ..or only
>> trial
>> > and error ???
>> >
>> > --
>> > Regards,
>> > $iva
>>
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