Question

2x + 3y + 6 = 0 and 4x + ky + 12 =0 have a unique solution
​

Answer 1

Answer:

yes there is

we have to put them

Step-by-step explanation:

Consider the given equations.

4x+ky+8=0 ……. (1)

2x+3y+7=0 …… (2)

The general equations,

a

1

x+b

1

y+c

1

=0

a

2

x+b

2

y+c

2

=0

Therefore,

a

1

=4,b

1

=k,c

1

=8

a

2

=2,b

2

=3,c

2

=7

Since, the condition of unique solution is

a

2

a

1



=

b

2

b

1

Therefore,

2

4



=

3

k

3

k



=2

k



=6

Hence, k is any real number except 6.

Answer 2

Answer:

k is any number except 6

Step-by-step explanation:

here a1=2

a2=4

b1=4

b2=k

now, to have an unique solution we must have

a1/a2not equal to b1/b2

that is 2/4not equal to 4/k

=>2k not equal to 12

=>k not equal to 6

now

equation 1*2 and subtract equation 2 from 1..we got (6-k) y=o

here either 6=k ( it is not possible)

so k is any nmbr except 6