Letβs say we want to solve the equation
\begin{equation*}
x+3=7
\end{equation*}
Well, we need to get \(x\) by itself. The \(3\) is in the way of that, so we need to move it. Since the \(3\) is attached to the \(x\) by addition, we will use subtraction to move it, since subtraction cancels addition.
\begin{align*}
x+3\amp= 7\\
x+3 {\color{red}{-3}}\amp=7{\color{red}{-3}}\\
x \amp=4
\end{align*}
Letβs do another. Suppose we want to solve the equation
\begin{equation*}
3x=7
\end{equation*}
Well, we need to get \(x\) by itself. The \(3\) is in the way of that, so we need to move it. Since the \(3\) is attached to the \(x\) by multiplication, we will use division to move it, since division cancels multiplication.
\begin{align*}
3x\amp= 7\\
\frac{3x}{\color{red}{3}}\amp=\frac{7}{\color{red}{3}}\\
x \amp=\frac{7}{3}
\end{align*}
One more. Suppose we want to solve the equation
\begin{equation*}
x^3=7
\end{equation*}
Well, we need to get \(x\) by itself. The \(3\) is in the way of that, so we need to move it. Since the \(3\) is attached to the \(x\) as the power, we will use a root to move it, since the cube root cancels a cube.
\begin{align*}
x^3\amp= 7\\
\sqrt[{\color{red}{3}}]{x^3}\amp=\sqrt[{\color{red}{3}}]{7}\\
x \amp=\sqrt[3]{7}
\end{align*}