The value of "y" is 3
Given that,
[tex]y+\frac{4}{12}-y-\frac{1}{12}=\frac{y}{2}[/tex]
We have to solve for "y"
Let us simplify the given expression and solve for"y"
[tex]y+\frac{4}{12}-y-\frac{1}{12}=\frac{y}{2}[/tex]
On cross multiplication in L.H.S we get,
[tex]\frac{12 y+4-12 y-1}{12}=\frac{y}{2}[/tex]
Cancelling denominator 12 from both sides, we get
12y + 4 -12y - 1 = y
Again cancelling +12y and -12y we get
4 - 1 = y
y = 3
Thus the value of "y" is 3
