Question

equation infinit number of salution 2x+3y=7 ; (k-1)x+(k+2)y=3k​

Answer 1

Answer:

Consider the given equations.

2x+3y=7

(k−1)x+(k+2)y=3k

The general equations

a

1

x+b

1

y=c

1

a

2

x+b

2

y=c

2

So,

a

1

=2,b

1

=3,c

1

=7

a

2

=k−1,b

2

=k+2,c

2

=3k

We know that the condition of infinite solution

a

2

a

1

=

b

2

b

1

=

c

2

c

1

Therefore,

k−1

2

=

k+2

3

=

3k

7

⇒

k−1

2

=

k+2

3

⇒2k+4=3k−3

⇒k=7

Hence, the value of k is