Question

If x varies directly as 3y+1 and x=9 when y=1, then what is the value of x when y=5

Answer 1

x = -5

And then ,

3y + 1

3(1) + 1 = 4

Answer 2

Answer:  The required value of x is 36.

Step-by-step explanation:  Given that x varies directly as (3y + 1) and x = 9 when y = 1.

We are to find the value of x when y = 5.

According to the given information, we have

$x\propto (3y+1)\\\\\Rightarrow x=k(3y+1)~~~~~~~~~~~~~~~~~~~~~~~[\textup{where k is the constant of proportionality}]~~~~~~~~~~~~~~~~~~(i)$

Since x = 9 when y = 1, so substituting these values in equation (i), we get

$9=k(3\times1+1)\\\\\Rightarrow 9=4k\\\\\Rightarrow k=\dfrac{9}{4}.$

So, equation (i) implies that

$x=\dfrac{9}{4}(3y+1).$

Therefore, when y = 5, then

$x=\dfrac{9}{4}(3\times5+1)=\dfrac{9}{4}\times16=9\times4=36.$

Thus, the required value of x is 36.