Simplifying $s=w_1-c$, where $c=\frac{1}{2}(w_1-t_1+\frac{w_2-t_2}{R})$ I have to solve this easy equation, but can't find the same answer as the correction of the exercise.
We have two equations that I want to simplify (plugging $c$ into $s$)
$$c=\dfrac{1}{2}\left(w_1-t_1+\dfrac{w_2-t_2}{R}\right)$$
$$s=w_1-c$$
I find:
$$s=w_1-\dfrac{1}{2}\left(w_1-t_1+\dfrac{w_2-t_2}{R}\right)=\dfrac{1}{2}\left(w_1+t_1-\dfrac{w_2+t_2}{R}\right)$$
The correction actually says:
$$s=w_1-\dfrac{1}{2}\left(w_1-t_1+\dfrac{w_2-t_2}{R}\right)=\dfrac{1}{2}\left(w_1-t_1-\dfrac{w_2-t_2}{R}\right)$$
I must be wrong but I don't know why. Could somebody explain this to me?

You first expand a bit to simplify for the coefficient of $w_1$:
$$w_1-\frac{1}{2}\left(w_1-t_1+\frac{w_2-t_2}{R}\right) = w_1-\frac{1}{2}w_1-\frac{1}{2}\left(-t_1+\frac{w_2-t_2}{R}\right)$$
$$= \frac{1}{2}w_1-\frac{1}{2}\left(-t_1+\frac{w_2-t_2}{R}\right)$$
Factoring $\frac{1}{2}w_1$ by $-\frac{1}{2}$, the expression simplifies to:
$$-\frac{1}{2}\left(-w_1-t_1+\frac{w_2-t_2}{R}\right)$$
And factoring by $-1$ yields:
$$\frac{1}{2}\left(\color{blue}{+}w_1\color{blue}{+}t_1\color{blue}{-}\frac{w_2-t_2}{R}\right)$$
You apparently made a mistake in the numerator of the final fraction. It should be $w_2-t_2$ rather than $w_2+t_2$, but I think it may have been a typo or an accident on your part. The rest is correct.
