for what value of P the pair of linear equations(p+2)x-(2p+1)y=3(2p-1)and2x-3y=7 has a unique solution
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4
The given system of equations is: (p+2)x− (2p+1)y=3(2p−1)2x−3y=7 These equations can be written as:(p + 2) x − (2p + 1) y − (2p − 1) = 0and 2x − 3y − 7 = 0 These equations are of the form a1x + b1y + c1 = 0 a2x + b2y + c2= 0
Where
a1 = p + 2,
b1 = −(2p + 1), c1 = −3 (2p − 1) and
a2 = 2,
b2 = −3,
c2 = −7
For a unique solution, we must have
a1a2 ≠ b1b2
⇒ p + 22 ≠ −(2p + 1)−3
⇒ p + 22 ≠ 2p + 13
⇒ 4p + 2 ≠ 3p + 6
⇒ 4p − 3p ≠ 6 − 2
⇒ p ≠ 4
Where
a1 = p + 2,
b1 = −(2p + 1), c1 = −3 (2p − 1) and
a2 = 2,
b2 = −3,
c2 = −7
For a unique solution, we must have
a1a2 ≠ b1b2
⇒ p + 22 ≠ −(2p + 1)−3
⇒ p + 22 ≠ 2p + 13
⇒ 4p + 2 ≠ 3p + 6
⇒ 4p − 3p ≠ 6 − 2
⇒ p ≠ 4
Answered by
1
Answer:
Step-by-step explanation:
The given pair of linear equation :
⇒ (p+2)x−(2p+1)y=3(2p−1) ------ ( 1 )
⇒ 2x−3y=7 ------- ( 2 )
On comparing with equations,
a
1
x+b
1
y+c
1
=0
and a
2
x+b
2
y+c
2
=0
We get, a
1
=(p+2),b
1
=−(2p+1),c=−[3(2p−1)]
a
2
=2,b
2
=−3,c=−7
⇒
a
2
a
1
=
2
p+2
and
b
2
b
1
=
−3
−(2p+1)
A pair of linear equations has a unique solution, if
⇒
a
2
a
1
=
b
2
b
1
⇒
2
(p+2)
=
−3
−(2p+1
⇒ −3(p+2)
=2(−2p−1)
⇒ −3p−6
=−4p−2
⇒ −3p+4p
=−2+6
⇒ p
=4
Therefore, the given system will have unique solution for all real values of p other than 4.
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