Question

Find the value of k so that the system of equations has no solution. 2x+ky=1, 6x+4y=3

Answer 1

Answer:

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For what value of k the following system of linear equations has no solution?

3x+y=1

(2k−1)x+(k−1)y=2k+1

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ANSWER

We know that the system of equations

a

1

x+b

1

y=c

1

a

2

x+b

2

y=c

2

has no solution, if

a

2

a

1

=

b

2

b

1

=

c

2

c

1

Here, a

1

=3,b

1

=1,c

1

=1,a

2

=2k−1,b

2

=k−1,c

2

=2k+1

So, the given system of equations will have no solution, if

2k−1

3

=

k−1

1

=

2k+1

1

2k−1

3

=

k−1

1

and

k−1

1

=

2k+1

1

Now,

2k−1

3

=

k−1

1

⇒3k−3=2k−1⇒k=2

Step-by-step explanation:

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Answer 2

Answer:

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Step-by-step explanation:

value of k is 4/3