Solution:
Let the no. of students in each row be x
let the no. of rows be y
therefore the total no. of students are students in a row*no. of rows
therefore the no. of students are xy.
According to 1 condition
no. of students in a row if increased by 3 then the rows decrease by 1.
therefore the equation satisfying the condition is (X+3)(Y-1)=XY ---------------as the no.of students is xy let it be 1 eqn
similarly according to 2 condition
(x-3)(y+2)=xy --------------------as the no. of students is xy let it be 2 eqn
therefore solving the 1 eqn we get
xy-1x+3y-3=xy
here xy onLHS and RHS gets cancelled
therefore the equation is -x+3y-3=0 ------------------let it be 3 eqn
similarly by solving 2 eqn we get
xy+2x-3y-6=xy
here also xy on LHS and RHS gets cancelled.
therefore the eqn is 2x-3y-6=0 --------------------let it be 4 eqn
multiplying 3 eqn by 2 we get
-2x+6y-6=0 --------------------let it be 5 eqn
adding eqn 4 and 5 we get
2x-3y-6=0
-2x+6y-6=0
3y-12=0 -----------------2x and -2x gets cancelled
therefore 3y=12 -------------taking 12 on RHS
y=12/3
y=3
substituting the value of y in 3 eqn
-x+3y-3=0
-x+3*4-3=0 ------------as y=4
-x+12-3=0
-x+9=0
therefore x=9
total no. of students is xy=9*4
=36
therefore the number of students are 36
