How do you solve $\dfrac{1}{2}\left( x-3 \right)=\dfrac{1}{3}\left( 2x+1 \right)$ ?

Hint: To solve $\dfrac{1}{2}\left( x-3 \right)=\dfrac{1}{3}\left( 2x+1 \right)$ , we have to apply distributive property by multiplying the term outside the bracket to the terms inside the bracket on both sides.

Complete step-by-step answer:
We have to solve $\dfrac{1}{2}\left( x-3 \right)=\dfrac{1}{3}\left( 2x+1 \right)$ . Let us apply distributive property on both sides.
$\Rightarrow \dfrac{1}{2}x-\left( \dfrac{1}{2}\times 3 \right)=\left( \dfrac{1}{3}\times 2x \right)+\left( \dfrac{1}{3}\times 1 \right)$
Let us simplify the terms inside the parenthesis.
$\Rightarrow \dfrac{x}{2}-\dfrac{3}{2}=\dfrac{2x}{3}+\dfrac{1}{3}$
We can see that the denominator of LHS is the same. Hence, we can write the above equation as
$\Rightarrow \dfrac{x-3}{2}=\dfrac{2x}{3}+\dfrac{1}{3}$
Let us take 2 from LHS to RHS.
$\Rightarrow x-3=2\left( \dfrac{2x}{3}+\dfrac{1}{3} \right)$
We can now apply distributive property again on LHS.
$\Rightarrow x-3=\dfrac{2\times 2x}{3}+\dfrac{2\times 1}{3}$
Let us solve the RHS.
$\Rightarrow x-3=\dfrac{4x}{3}+\dfrac{2}{3}$
Now, we have to collect terms in x in LHS and constants in RHS.
$\Rightarrow x-\dfrac{4x}{3}=\dfrac{2}{3}+3$
Let us take the LCM on both sides.
$\Rightarrow \dfrac{3x-4x}{3}=\dfrac{2+9}{3}$
Let us solve both the sides.
$\Rightarrow \dfrac{-x}{3}=\dfrac{11}{3}$
We can see that the denominators in LHS and RHS are similar. Let us cancel them out.
$\Rightarrow -x=11$
Let us take the negative sign from the LHS to the RHS.
$\Rightarrow x=-11$
Hence, the solution of $\dfrac{1}{2}\left( x-3 \right)=\dfrac{1}{3}\left( 2x+1 \right)$ is -11.

Note: Students must be very thorough with the algebraic rules and properties. We commonly use distributive property in most of the questions of this type. Distributive property can be explained by considering the following example. Let us consider $a\left( b+c \right)$ . We have to multiply a with both the terms inside the parenthesis. Therefore, we will get \[a\left( b+c \right)=ab+ac\] . In problems like the given questions, we can reach a solution only by finding the value of x.