For what value of k, will the following system of equations have no solutions
(3k-1) x + 3y = 2
(k²+1) x + (k-2) y =5
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Answered by
0
Answer:
The given system of equations is
(3k+1)x+3y−2=0
(k
2
+1)x+(k−2)y−5=0
This is of the form a
1
x+b
1
y+c
1
=0
a
2
x+b
2
y+c
2
=0,
where, a
1
=3k+1,b
1
=3,c
1
=−2
and a
1
=k
2
+1,b
2
=k−2,c
2
=−5
For no solution, we must have
a
2
a
1
=
b
2
b
1
=
c
2
c
1
The given system of equations will have no solution, if
k
2
+1
3k+1
=
k−2
3
=
−5
−2
⇒
k
2
+1
3k+1
=
k−2
3
and
k−2
3
=
5
2
Now,
k
2
+1
3k+1
=
k−2
3
⇒(3k+1)(k−2)=3(k
2
+1)
⇒3k
2
−5k−2=3k
2
+3
⇒−5k−2=3
⇒−5k=5
⇒k=−1
Clearly,
k−2
3
=
5
2
for k=−1
Hence, the given system of equations will have no solution for k=−1.
Answered by
0
I think the second equation have no solution.....
Step-by-step explanation:
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