Question

2x+3y=7and (p+q)x+(2p-q)y=21 find value of p and q and have infinitely many soutin

Answer 1

infinitely many solutions
a1/a2. = b1/b2. =c1/c2
2/p+q. =3/(2p-q). = 7/21
2/p+q. = 1/3
p+q=6. ...........(1)
p=6-q

3/(2p-q) =1/3
9=2p-q. ..........(2)
9=2(6-q)-q
9=12-2q-q
9-12=-3q
q=1
substituting value in (1)
p=5
values of p and q are 5 and 1 respectively.
cheers!!!!

Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

__________________________________________________

Given :- 2x + 3y = 7 and  (p+q)x + (2p-q)y = 21 .

             Equation will have infinite number of solutions.

To Find :- The  value of p and q.

Solution :-

2/p+q = 3/2p-q = 7/21

2/p+q = 3/2p-q = 1/3

2/p+q =  1/3 and 3/2p-q = 1/3

p+q = 6 and 2p - q = 9

( p + q ) + ( 2p - q ) = 6 + 9

3p = 15

p = 5

Put p = 5 in p + q = 6 or 2p - q = 9 , for getting the value of q.

q = 1.

Given system of equations will have infinitely many solutions, if p = 5 and q =  1.

__________________________________________________