Question

lines on the grapii.
solutions i.e. represent conicident lines.
(i) Kr + 3y = K - 3; 12x + Ky = K
(ii) 10x + 5y - (K-5) = 0; 20x + 10y - K = 0
Example 9. For what value of K will the following pair of linear equations have
(Basic) (C.B.S.E. 2008, 2011​

Answer 1

Answer:

10x+5y−(k−5)=0and

20x+10y−k=0

These given equations are in the form:

a

1

x+b

1

y+c

1

=0 and

a

2

x+b

2

y+c

2

=0

Where

a

1

=10,b

1

=5 and c

1

=−(k−5)

a

2

=20,b

2

=10 and c

2

=−k

System has infinitely many solutions.

20

10

=

10

5

=

−k

−(k−5)

=

k

k−5

5k=10k−50

or k=10

The value of k is 10