How to find $y$ in $-\frac12+\frac y2=1$ without combining $-\frac12+\frac y2$? I am trying to find the value of a variable. The problem given was$$
\frac{1}{2}(-1+y) = 1. 
$$
I understand that $\dfrac{1}{2}(-1+y) = 1$ is the same as $-\dfrac{1}{2} + \dfrac{y}{2} =1$. How do I get $y$ by itself starting from $-\dfrac{1}{2} + \dfrac{y}{2} =1$? I tried to multiply $-\dfrac{1}{2}$ by $-2$ and then multiply $1$ by $-2$:$$
-\frac{1}{2} × (-2) + \frac{y}{2} = 1 × (-2).
$$
That leaves me with $\dfrac{y}{2} = -2$. Then I multiplied $\dfrac{y}{2}$ by $2$ and also multiplied $-2$ by $2$:
$$
\frac{y}{2} × 2 = (-2) × 2.
$$
That leaves me with $y = -4$.
But this is not correct. The answer should be $3$. What step am I missing or getting wrong?

As Ethan Bolker pointed out in the comments, without having to divide each summand by $2$ in the expression $$\frac{y-1}{2}=1,$$ 
one can multiply both sides by two to get $$\require{cancel}\frac{y-1}{\cancel{2}}\cdot\cancel{2}=1\cdot 2\iff y-1=2\iff y=3.$$
Note: I use the symbol $\iff$ to mean “if and only if,” or “this is equivalent to.”
Regarding your work, you forgot to multiply the $y/2$ term in the step when you multiplied both sides by $-2$: the expression $$-\frac{1}{2}\cdot -2 +\color{red}{\frac{y}{2}}=1\cdot -2,$$
should be $$-2\left(-\frac{1}{2}+\frac{y}{2}\right)=1\cdot -2\iff -\frac{1}{2}\cdot -2 +\color{blue}{\frac{y}{2}\cdot -2}=-2\iff 1-y=-2$$
Can you take it from here? 
