Question

find the value of k so that the following pair of linear equations have 1) no solution 2) a unique solution (3k+1)x + 3y -2=0,(k^2 + 1)x + (k-2)y -5=0​

Answer 1

Step-by-step explanation:

Consider the given equations.

2x+3y=7

(k−1)x+(k+2)y=3k

The general equations

a1

x+b1

y=c1

a2

x+b2y=c2

So,

a1 =2,b1=3,c1=7a2=k−1,b2=k+2,c2=3k

We know that the condition of infinite solution

a 2a 1 = b 2b 1 = c 2c 1

Therefore,

k−12 = k+23= 3k+7⇒k−12= k+23⇒2k+4=3k−3⇒k=7

Hence, the value of k is 7.

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