Question

find K for which the following pair have unique number of solution?
5x+Ky=7
3x+2y=8​

Answer 1

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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:

5x+ky=7

x=3 and y=-4

Putting the values of x and y

5 (3)+k (-4)=7

15-4k=7

15-7=4k

8=4k

k=8/4

k=2

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