Question

Example 16 : For what values of k will the following pair of linear equations have
infinitely many solutions?
kx + 3y - (-3) = 0
12x + ky-k=0​

Answer 1

Answer:

Given 

kx+3y−(k−3)=0

Comparing with a1x+b1y+c1=0

∴a1=k, b1=3, c=−(k−3)

12x+ky−k=0

Comparing with a1x+b1y+c1=0

∴a1=12, b1=k, c=−k

Since equation has infinite number of solutions

So, a2a1=b2b1=c2c1

12k=k3=kk−3

12k=k3

Answer 2

1. Answer:
2. Step-by-step explanation:
3. Sin 65=0.90
4. Sin 25= 0.42
5. Cos 65= 0.42
6. Cos 25 = 0.90
7. 1.32/1.32 = 1
8. Step-by-step explanation:
9. please mark your calendar.
10. ok..