If y varies directly s the square root of (x + 1) and y = 6 when x = 3, find x when y = 9

A.

8

B.

7

C.

6

D.

5

Correct answer is A

y \(\alpha\) \sqrt{x + 1}\), y = k\sqrt{x + 1}\)

6 = k\(\sqrt{3 + 1}\)

6 = k\(\sqrt{4}\)

6 = 2k

k = \(\frac{6}{2}\) = 3

y = \(\sqrt{(x + 1)}\)

9 = 3\(\sqrt{(x + 1)}\)(divide both side by 3)

\(\frac{9}{3}\) = \(\frac{3\sqrt{x + 1}}{3}\)

3 = \(\sqrt{x + 1}\)(square both sides)

9 = x + 1

x = 9 - 1

x = 8