We will demonstrate several methods. To emphasize how order of operations are reversed when undoing operations, we note that adding 8 would be last in a computation. Thus our first step is to subtract 8 (or add
\(-8\)) to both sides of our equation, resulting in
\(5(x-3)=50\text{.}\) Next we, divide both sides by
\(5\text{,}\) leaving
\(x-3=10\text{.}\) Finally, we add
\(3\) to get
\(x=13\text{.}\)
Note that if we start by dividing by 5,
PrincipleΒ A.2.21 tells us that we must also divide 8 by 5. The resulting equation,
\(x-3+\frac{8}{5}=\frac{58}{5}\) or
\(x-3+1.6=11.6\text{,}\) is correct but more difficult to solve.
Finally, we could solve this equation by first simplifying the left side. Our expression
\(5(x-3)+8=5x-15+8=5x-7.\) Replacing the original left side by this simplified version gives
\(5x-7=58\text{.}\) Adding
\(7\) to both sides gives
\(5x=65\) and dividing by 5 gives
\(x=13\text{.}\)