Find all values of y that satisfy the equation:
\([\frac{y}{3} + 1 = \frac{y + 3}{y}] \)
There is only one value of y that satisfies the equation:
y ==3
[3/3+1 ==(3+3)/3]
2 == 2
Hmm! What about y = -3?
Hi Guest,
So...
\(\frac{y}{3}+1=\frac{\left(y+3\right)}{y}\)
\(y^2+3y=3\left(y+3\right)\)
\(y=3,\:y=-3\)
The number of values of y: \(2\)
Thank you so much! I was confused and needed help. So thanks!
