Q7 of 39 Page 1

find the value of k for which the following pair of linear equations have infinitely many solutions.

2x + 3y = 7


(k+1)x + (2k-1)y = 4k + 1


We have,


2x + 3y = 7


⇒ 2x + 3y – 7 = 0 ….. (1)


(k+1)x + (2k-1)y = 4k + 1


⇒ (k +1)x + (2k – 1)y – (4k+1) = 0 ….. (2)


For the equations of the form:


a1x+ b1 y + c1 = 0


a2 x+ b2y + c2 = 0


The condition for having infinitely many solutions is:



For the equations (1) and (2),





Now,


2(2k-1) = 3(k+1)


⇒ 4k – 2 = 3k + 3


⇒ k = 5


or


2(4k+1) = 7(k+1)


⇒ 8k + 2 = 7k + 7


⇒ k = 5


Hence, the value k =5 will give infinitely many solutions for the given set of equations.


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