find the value of k in which the following each system of linear equation has no solution 2x+5y-3=0,8x+ky-10=0​

1 thought on “find the value of k in which the following each system of linear equation has no solution 2x+5y-3=0,8x+ky-10=0​”

1. Step-by-step explanation:

   Given:–

   2x+5y-3=0

   8x+ky-10=0

   The system of linear equation has no solution.

   To find:–

   Find the value of k ?

   Solution:–

   Given pair of linear equations in two variables are

   2x+5y-3=0———-(1)

   On Comparing this with a1x+b1y+c1=0

   a1=2 ,b1=5 and c1=-3

   8x+ky-10=0———-(2)

   On Comparing this with a2x+b2y+c2=0

   a2 = 8, b2=k and c2=-10

   Given that

   The system of linear equation has no solution

   We know that

   The system of linear equation has no solution if a1/a2=b1/b2≠c1/c2

   => 2/8=5/k≠-3/-10

   => 2/8=5/k

   On applying cross multiplication then

   => k×2 = 8×5

   => 2k = 40

   =>k = 40/2

   => k=20

   Answer:–

   The value of k for the given problem is 20

   Used formula:–

   • The system of linear equations a1x+b1y+c1=0 and a2x+b2y+c2=0 has no solution if a1/a2=b1/b2≠c1/c2
   • The lines are inconsistent lines or parallel lines

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