Question

Point (3, 4) lies on the graph of the equation 3y = kx + 7. The value of k is:

Answer 1

Since Point (3,4) lies on the graph of equation

3y=kx+7--(1)

So Put x=3 and y=4 in equation (1).

Therefore, (1)=>3x4=3xk+7

12-7=3k

5=3k

k=5/3

Therefore the. required Value of k is 5/3 .

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Answer 2

Hi,

Answer: B

Explanation: 3y = kx + 7

Here, x = 3 and y = 4

Hence,

(3×4) = (kx3) + 7

12 = 3k+7

3k = 12–7

3k = 5

k = 5/3