For the pair of linear equations,
px + qy – 1 = 0
(p2 +93) (qx + py) – 2pq = 0, 2,9 70
The value of x + y is
Answers
Answer:
ANSWER
(i) px+qy=p−q ....(1)
qx−py=p+q ....(2)
Multiplying (1) by p and (2) by q we get (3) and (4) respectively as
p
2
x+pqy=p
2
−pq ....(3)
q
2
x−pqy=pq+q
2
...(4)
Adding (3) and (4) we get
(p
2
+q
2
)x=(p
2
+q
2
)
⇒x=1
Substituting value of x in (1), we get
p+qy=p−q
qy=−q
y=−1
Hence, x=1,y=−1
(ii) ax+by=c ...(1)
bx+ay=1+c ...(2)
Multiplying (1) by b and (2) by a we get (3) and (4) respectively as
abx+b
2
y=bc ...(3)
abx+a
2
y=a+ac ...(4)
Subtracting (4) from (3), we get
(b
2
−a
2
)y=(b−a)c−a
y=
(b
2
−a
2
)
(b−a)c−a
Substituting value of y in (i), we get
ax+b×
(b
2
−a
2
)
(b−a)c−a
=c
ax=c−
(b
2
−a
2
)
(b−a)bc−ab
ax=
(b
2
−a
2
)
c(b
2
−a
2
)−b
2
c+abc+ab
x=
a(b
2
−a
2
)
cb
2
−ca
2
−b
2
c+abc+ab
x=
a(b
2
−a
2
)
ab+abc−ca
2
x=
(b
2
−a
2
)
b−bc−ca
=
(b
2
−a
2
)
b+c(b−a)
(iii)
a
x
−
b
y
=0 ...(1)
ax+by=a
2
+b
2
....(2)
Multiplying (1) by b
2
a
b
2
x
−by=0 ....(3)
Adding (2) and (3) we get
a
b
2
x
+ax=a
2
+b
2
a
b
2
x+a
2
x
=a
2
+b
2
x(a
2
+b
2
)=a(a
2
+b
2
)
⇒x=a
Substituting value of x in (1)
a
a
−
b
y
=0
b
y
=1
⇒y=b
(iv) (a−b)x+(a+b)y=a
2
−2ab−b
2
....(1)
(a+b)(x+y)=a
2
+b
2
⇒(a+b)x+(a+b)y=a
2
+b
2
...(2)
Subtracting (2) from (1), we get
(a−b−(a+b))x=−2ab−2b
2
−2bx=−2b(a+b)
⇒x=(a+b)
Substituting value of x in (1), we get
(a−b)×(a+b)+(a+b)y=a
2
−2ab−b
2
a
2
−b
2
+(a+b)y=a
2
−2ab−b
2
(a+b)y=−2ab
⇒y=−
a+b
2ab
(v) 152x−378y=−74 ...(1)
−378x+152y=−604 ...(2)
Multiplying (1) by 378 and (2) by 152, we get
57456x−142884y=−27972 ...(3)
−57456x+23104y=−91808 ....(4)
Adding (3) and (4) we get
−119780y=−119780
⇒y=1
Putting this value of y in (1)
152x−378=−74
152x=378−74
x=
152
304
=2