for which value of k will the pair of equations 3x+y=1 (2k-1)x+(k-1)y=2k+1
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Answer:
The given pair of equations are:
3x+y=1...(1)
(2k−1)x+(k−1)y=2k+1..(2)
Now rearranging eq1 and eq2 will get
3x+y−1=0...(3)
(2k−1)x+(k−1)y−(2k+1)=0..(4)
Now compare with
a
1
=3,b
1
=1,c
1
=−1
a
2
=2k−1,b
2
=k−1,c
2
=−(2k+1)
Now we get
a
2
a
1
=
2k−1
3
,
b
2
b
1
=
k−1
1
,
c
2
c
1
=
−(2k+1)
−1
Now will take
a
2
a
1
=
b
2
b
1
⇒
2k−1
3
=
k−1
1
⇒3k−3=2k−1
⇒3k−2k=−1+3
⇒k=2
Hence k=2 is the value.
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