Find all values of $y$ that satisfy the equation \[\frac{y}{3} + 1 = \frac{y + 3}{y}.\]
y/3 + 1 = (y + 3)/y
==> (y + 3)/3 = (y + 3)/y
==> y = 3
The only solution is y = 3.
+515 \(\frac{y}{3}+1=\frac{y+3}{y}\)
\(y^2+3y=3\left(y+3\right)\) (multiply both sides by 3y)
\(y^2+3y=3y+9\) (distribute right side)
\(y^2=9\) (subtract both sides by 3y)
\(y=3,\:y=-3\) (final answer)
