Question

If the point (2K-3, K+2) lies on the graph of the equation 2x+3y+15=0, find the value of k.?

Answer 1

Since the point lies on the graph, it must satisfy the equation.

Hence put the value of x and y i.e., (2k – 3, k + 2) in the equation 2x + 3y + 15 = 0

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

4k - 6 + 3k + 6 + 15 = 0

7k = -15

k = -15/7

Answer 2

Answer:

$\frac{-15}{7}$

Step-by-step explanation:

We are given that (2K-3,K+2) lies on the line 2x+3y+15=0

Since the given point lies on the line

So, it must satisfy the equation of line .

$2(2K-3)+3(K+2)+15=0$

$4K-6+3K+6+15=0$

$7K+15=0$

$K = \frac{-15}{7}$

Hence The value of K is $\frac{-15}{7}$