If x = 5t and y = 1/3( t +1), then the value of t for x = 3y is :

X = 5t

$y = \frac{1}{3}( t +1)$

X = 3Y 

We have, X = 3Y      ... (A) 

So, put the value of X from equation (A) in X  = 5t 

= 3Y = 5t      ... (B)

Now, put the value of $y = \frac{1}{3}( t +1)$ in equation (B) we have,

= 3 × \(\frac{1}{3}\) × ( t+1) = 5t

= t +1 = 5t 

= 4t = 1 

= t = \(\frac{1}{4}\)