Math, asked by ashraykundapura, 9 months ago

Help me with this please​

Attachments:

Answers

Answered by saheedhussain21
0

Answer:

Sorry I cannot

Step-by-step explanation:

Answered by atharvashinde8b
0

Answer:

got only 2 ans

Step-by-step explanation:

Ans. ) Let the equation (i) be 2x + 3y – 7 = 0  

Let the equation (ii) be (k – 1)x + (k + 2)y – 3k = 0  

We have,

For infinite number of solutions

2k–1=3k+2=−7−3k

2k–1=3k+2=73k

Now, the following cases arises

Case 1:

2k–1=3k+2

2(k+2)=3(k–1)⇒2k+4=3k−3

k=7

Case 2:

3k+2=73k

(k+2)=9k⇒7k+14=9k

k=7

Case 3:

2k–1=73k

7k–7=6k

k=7

Similar questions