If 3xp + y/3q = 7 and 2x + 1 = y + 1 = p = 5 ; find value of q.
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Step-by-step explanation:
Given equations:-
2x+3y=9
(p+q)x+(2p−q)y=3(p+q+1)
Here,
a
2
a
1
=
p+q
2
,
b
2
b
1
=
2p−q
3
,
c
2
c
1
=
3(p+q+1)
9
For a pair of linear equations to have infinitely many solutions:
a
2
a
1
=
b
2
b
1
=
c
2
c
1
So, we need,
p+q
2
=
2p−q
3
=
3(p+q+1)
9
or,
p+q
2
=
2p−q
3
=>2(2p−q)=3(p+q)
=>4p−2q=3p+3q
=>p=5q....(i)
Also,
2p−q
3
=
3(p+q+1)
9
=>9(p+q+1)=9(2p−q)
=>p+q+1=2p−q
=>2p−p=q+q+1
=>p=2q+1
Substituting(i),wehave,
5q=2q+1
=>q=
3
1
Also,p=5q=5(
3
1
)=
3
5
∴p=
3
5
andq=
3
1
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