Solving a linear equation

Question

I need to solve: $$\frac{1}{2}\left[10\beta+(1-\beta)(-10)\right]-\frac{c}{i}= 5-c$$ for $\beta$ to get to: $$\beta = 1 – \left( 1 - \frac{1}{i}\right)\frac{c}{10}$$ But i get stuck somewhere in the middle with no idea what to do. does anyone knows how this is done?

Answer 1

$$ \frac{1}{2}(10\beta -10(1-\beta))-\frac{c}{i}=5-c $$ $$ 5(\beta -(1-\beta))-\frac{c}{i}=5-c $$ $$ 5(2\beta -1)-\frac{c}{i}=5-c $$ $$ 10\beta -5-\frac{c}{i}=5-c $$ $$ 10\beta =10-c+ \frac{c}{i} $$ $$ \beta =1-\frac{c}{10}+ \frac{c}{10i}=1-\frac{c}{10}\left(1-\frac{1}{i}\right)$$

Answer 2

Let us start from $$\frac{1}{2}\left[10\beta+(1-\beta)(-10)\right]-\frac{c}{i}= 5-c$$ Add $\frac{c}{i}$ to both sides, so $$\frac{1}{2}\left[10\beta+(1-\beta)(-10)\right]= 5-c+\frac{c}{i}$$ Multiply both sides by $2$ so $$\left[10\beta+(1-\beta)(-10)\right]= 2(5-c+\frac{c}{i})$$ Develop what is inside the brackets, so $$20\beta-10= 2(5-c+\frac{c}{i})$$ Add $10$ to each side so $$20\beta= 2(5-c+\frac{c}{i})+10$$ Expand the rhs ... and you will be almost done.

I am sure that you can take from here.