if the point (2k-3,k+2) lies on the graph of equations 2x+3y+15=0. and find K​

2 thoughts on “if the point (2k-3,k+2) lies on the graph of equations 2x+3y+15=0. and find K​”

1. Answer:

   The answer will be -(15/7)

   Step-by-step explanation:

   Let put (2k-3,k+2) on the equation —

   2 (2k-3) + 3 (k+2) + 15 = 0

   => 4k – 6 + 3k + 6 + 15 = 0

   => 7k + 15 = 0

   => k = -(15/7)

   So, The value of k is -(15/7).

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2. Answer:

   K = [tex]\frac{-15}{7}[/tex]

   Step-by-step explanation:

   If (2k-3,k+2), lies on line 2x + 3y + 15 = 0

   x = 2k – 3

   y = k + 2

   Putting value of x and y

   2(2k – 3) + 3(k + 2) + 15 = 0

   4k – 6 + 3k + 6 = -15

   7k = -15

   k = [tex]\frac{-15}{7}[/tex]

   Hope this helps

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