# Find the Value of K for Which the Following Pair of Linear Equations Has Infinitely Many Solutions. 2x + 3y = 7, (K +1) X+ (2k -1) Y = 4k + 1 - Mathematics

Sum

Find the value of k for which the following pair of linear equations has infinitely many solutions.
2x + 3y = 7, (k +1) x+ (2k -1) y = 4k + 1

#### Solution

We have,

2x + 3y = 7 ⇒ 2x + 3y - 7 = 0
(k + 1) x + (2k - 1)y = 4k + 1 ⇒ (k + 1)x (2k -1)y - (4k + 1) = 0

For infinitely many solutions

a_1/a_2 = b_1/b_2 = c_1/c_2

⇒ (2)/(k+1) = (3)/((2k -1)) = (-7)/-(4k +1)

⇒ (2)/(k+1) = (3)/(2k -1)

⇒ 2(2k + 1) = 3 (k+1)

⇒4k - 2 = 3k + 3

⇒4k - 3k = 3 +2

k = 5

or

⇒ (2)/(k+1) = (-7)/-(4k +1)

⇒ 2(4k + 1) = 7 (k+1)

⇒ 8k + 2 = 7k + 2

⇒8k - 7k = 7- 2

k = 5

Hence, the value of k is 5 for which given equations have infinitely many solutions.

Concept: Pair of Linear Equations in Two Variables
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