**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**