# Find the Value Of K For Which Each of the Following System of Equations Has Infinitely Many Solutions : 2x + (K - 2)Y = K 6x + (2k - 1)Y - (2k + 5) - Mathematics

Find the value of k for which each of the following system of equations has infinitely many solutions :

2x + (k - 2)y = k

6x + (2k - 1)y - (2k + 5)

#### Solution

The given system of the equation may be written as

2x + (k - 2)y - k = 0

6x + (2k - 1)y - (2k + 5) = 0

The system of equation is of the form

a_1x + b_1y + c_1 = 0

a_2x + b_2y + c_2 = 0

Where a_1 = 2, b_1 = k - 2, c_1 = -k

And a_2 = 6, b_2 = 2k - 1, c_2 = -(2k + 5)

For a unique solution, we must have

a_1/a_2 = b_1/b_2 = c_1/c_2

=> 2/6 = (k -2)/(2k - 1) = (-k)/(-2(2k + 5))

=> 2/6 = (k -2)/(2k - 1) and (k - 2)/(2k -1) = k/(2k + 5)

=> 1/3 = (k -2)/(2k -1) and (k -2)(2k + 5) = k(2k - 1)

=> 2k - 1 = 3(k - 2) and 2k^2 + 5k - 4k - 10 = 2k^2 - k

=> 2k - 3k - 6 and k - 10 = -k

=> 2k - 3k = -6 + 1 and k + k = 10

=> -k  = -5 and 2k = 10

=> k = (-5)/(-1) and k = 10/2

=> k = 5 and  k = 5

K = 5  satisfies both the conditions

Hence, the given system of equations will have infinitely many solutions, if k = 5

Concept: Pair of Linear Equations in Two Variables
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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 3 Pair of Linear Equations in Two Variables
Exercise 3.5 | Q 17 | Page 73