Question

Find the value of k, infinitely many solutions  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

a1 x+b1 y=c1

a2 x+b2 y=c2

So,

a1 =2,b1 =3,c1 =7

a2 =k−1,b2 =k+2,c2=3k

We know that the condition of infinite solution

a1/a2 = b1 / b2 = c1 / c2

Therefore,

2/k−1 = 3/k+2 = 7/3k

⇒ 2/k−1 = 3/k+2

⇒2k+4=3k−3

⇒k=7

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Answer 2

Answer:

Consider the given equations.

2x+3y=7

(k−1)x+(k+2)y=3k

 

The general equations

a1x+b1y=c1

a2x+b2y=c2

 

So,

a1=2,b1=3,c1=7

a2=k−1,b2=k+2,c2=3k

 

We know that the condition of infinite solution

a2a1=b2b1=c2c1

 

Therefore,

k−12=k+23=3k7

⇒k−12=k+23

⇒2k+4=3k−3

⇒k=7