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**Hint:**We recall the linear equations and basic properties of linear equations. We solve the given linear equation by eliminating the constant term $6$ at the left hand side so that only the variable term remains at the left hand side. We finally divide both sides by a coefficient of $x$ that is 3 to get only $x$ at the left hand side. \[\]

**Complete step by step answer:**

We know from algebra that the general linear equation in one variable $x$ and constants $a\ne 0,b$ is given by

\[ax=b\]

The term with which the variable is multiplied is called variable term and with whom is not multiplied is called a constant term. We also know that if we add, subtract, multiply or divide the same number on both sides of the equation, equality holds. It is called balancing the equation. It means for some term $c$ we have,

\[\begin{align}

& ax+c=b+c \\

& ax-c=b-c \\

& ax\times c=b\times c \\

& \dfrac{ax}{c}=\dfrac{b}{c} \\

\end{align}\]

When we are asked to solve for $x$ in an equation it means we have to find the value or values of $x$ for which the equation satisfies. We are given the following linear equation

\[3x+6=21\]

We see that we are given an equation with one variable $x$ and a variable term $3x$ which is only on the left hand side. So in order to eliminate the constant term that is 6 at the left hand side we subtract both sides by 6 to have

\[\begin{align}

& \Rightarrow 3x+6-6=21-6 \\

& \Rightarrow 3x=15 \\

\end{align}\]

We see now only the variable term remains at the left hand side of the equation. We divide both side of the above step coefficient of $x$ that is 3 to have;

\[\begin{align}

& \Rightarrow \dfrac{3x}{3}=\dfrac{15}{3} \\

& \Rightarrow x=5 \\

\end{align}\]

So the solution of the given linear equation is $x=5$. \[\]

**Note:**

We alternatively solve by first dividing both sides of the given equation $3x+6=21$ by 3 to have

\[\begin{align}

& \Rightarrow 3x+6=21 \\

& \Rightarrow \dfrac{3x+6}{3}=\dfrac{21}{3} \\

\end{align}\]

We separate the numerator using the property for same denominators $\dfrac{a+b}{d}=\dfrac{a}{d}+\dfrac{b}{d}$ in the above step to have

\[\begin{align}

& \Rightarrow \dfrac{3x}{3}+\dfrac{6}{3}=\dfrac{21}{3} \\

& \Rightarrow x+2=7 \\

\end{align}\]

We subtract both sides of the above equation by 2 to have the solution as;

\[\begin{align}

& \Rightarrow x+2-2=7-2 \\

& \Rightarrow x=5 \\

\end{align}\]

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