## Algebra 1: Common Core (15th Edition)

$c=3$
We start with the given equation: $\frac{1}{2}(6c-4)=4+c$ We multiply the entire equation by $2$ to clear the fraction: $$2\left[\frac{1}{2}(6c-4)=4+c\right]$$$$6c-4=8+2c$$ We add $4-2c$ to both sides of the equation: $6c-4+4-2c=8+2c+4-2c$ We combine like terms: $4c=12$ We divide by $4$ on each side of the equation: $$\frac{4c}{4}=\frac{12}{4}$$$$c=3$$