IIT JEE , AIEEE Forum - Mathematics , Algebra

ruhi

hi....here is my ques....
find the number of integral solutions of the equation 2x+y+z=20 where x,y,x >= 0

need a quick reply :0

ruhi

no replies :(

cbg2007

what z is not greater than the 0
check the question.

manvan8580

Here consider the equation as :
y+z=20-2x
Here x can be 1,2,......9
let xbe a parameter that is r.
Number of positive solutions are
=20-2r
Therefore required number=Summatin from r=1to r=9;20-2r
=20*9-2*(9*10/2)=90.
Answer is 90.
Don't forget to rate my answer.............

ruhi

hi manvan.........ur ans is wrong......it is 121......nd m not gettin ur method......sooo can u plz explain it 2 me

ankurgupta91

nw
y+z=20-2x
x can take values 0,1,2........9
if x=0
y+z=20
then we get y=0,z=20 nd z=0,y=20 -------2cases
y=1 ,z=19 ---------2cases
....................y=10,z=10 only one case
at last we get 21 solutions when y+z=20
if x=1
y+z=18
nw we get 19 solutions
this goes uptil when x=9 nd y+z=2
only one solution in that case
so this form an decreasing A.P
21,19,17,..........,1
total sol. =sum of this the terms of this A.P.= 11/2 [21+1] = 121
hope u can understood this .
its very easy .......
thanks................

ruhi

okk...........thanx ankur although it was easy but i wasnt gettin it
thanxxxxxxxxxx

manvan8580

How is it that ankur has taken the value zero,it is given that x,y,z cannot take values of zero as x,y,z>0.Do reply ruhi....................

ruhi

hi manvan.....it is >= 0.........thats y 10 values..........by the way ur method is correct so i ve rated u........soooooooo got that :)

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