Question

find the value of k if x=3 y=9 is a solution of the equation kx +y=3​

Answer 1

Answer:

Solution

Given

kx−y=2

6x−2y=3

kx−y−2=0

6x−2y−3=0

We know that for a

1

x+b

1

y+c

1

=0,a

2

x+b

2

y+c

2

=90

a

2

a

1

=

b

2

b

1



=

c

2

c

1

⇒ for no solution

a

1

=k,b=−1,c

1

=−2

a

2

=b,b

2

=−1,c

2

=−3

6

k

=

−2

−1

=

−3

−2

⇒

6

k

=

2

1



=

3

2

for no solution

6

k

=

2

1

nd

2

1



=

3

2

⇒k=

2

6

⇒

k=3

for infinite solution

b

1

a

1

=

b

2

b

1

=

c

2

c

1

b

k

=

−2

−1

=

−3

−2

+2

+1



=

−3

+2

=

2

1



=

3

2

Here is no value of k the system has infinite solution.

Answer 2

$\huge\fbox\pink{Given:}$

• x = 3

• y = 9

$\huge\fbox\pink{To Find:}$

• Solution of kx+y=3

$\huge\fbox\pink{Solution:}$

Now,Just substituting the values ,

k(3)+9 = 3

3k+9 = 3

3k = 3-9

3k = -6

k= -6/3

k = -2

Therefore,The value of k is -2.