Ed filled $2/3$ of his radiator with antifreeze and then added $4$ more quarts (a gallon) of antifreeze. After draining half the antifreeze, he needed $11$ quarts of antifreeze to fill the radiator to capacity. How many gallons of antifreeze can the radiator hold?
To me the equation was : $$x - \left(\frac{\frac{2}{3}x+4}{2}\right)+11=x,$$ but the contest solutions on one online link said it was only $$\left(\frac{2}{3}x+4\right)+11=x$$ where $x$ is total capacity in gallons. I don't see why. Can someone explain ?
Both of these solutions are wrong.
Let $x$ denote the capacity of the radiator in gallons.
We start with $\frac23x$ and add $1$ gallon, so we have $\frac23x+1$. Then we drain half, so we're left with $\frac12\left(\frac23x+1\right)$. Then we add $\frac{11}4$ more gallons, and the result is the full capacity, i.e. $x$. Thus
$$ \frac12\left(\frac23x+1\right)+\frac{11}4=x\;. $$