How do you solve $ \dfrac{3}{8}x = 6 $ ?

Hint: Here in this we have to solve the given equation and it is in the form of algebraic expression. Solving this equation, we have to find the unknown value x. The LHS of the given equation is in the form of fraction. by using the simple multiplication and division we find the value of x.

Complete step-by-step answer:
The given equation is an algebraic equation. The algebraic equation is a combination of variable and constant. Here the equation is in the form of fraction in LHS. So we use multiplication and division or arithmetic operations and solve for further
Now consider the given equation
 $ \dfrac{3}{8}x = 6 $
Multiply by 8 on the both sides of the equation we get
 $ \Rightarrow \dfrac{3}{8} \times 8x = 6 \times 8 $
On multiplying we get
 $ \Rightarrow 3x = 48 $
Now divide the above equation by 3 we get
 $ \Rightarrow x = 16 $
Therefore, the value of x is 16.
The above equation can be solved in the other procedure also.
Now consider the given equation
 $ \dfrac{3}{8}x = 6 $
Take 6 to LHS and the equation is rewritten as
 $ \Rightarrow \dfrac{3}{8}x - 6 = 0 $
Now let we take 3 as a common and we have
 $ \Rightarrow 3\left( {\dfrac{1}{8}x - 2} \right) = 0 $
The number 3 can’ be equal to 0 therefore $ \left( {\dfrac{1}{8}x - 2} \right) = 0 $
Now let us get LCM for the number 1 and 8. The LCM for 1 and 8 is 8.
Therefore we have
 $ \Rightarrow \dfrac{{\dfrac{x}{8} \times 8 - 2 \times 8}}{8} = 0 $
On simplifying we get
 $ \Rightarrow \dfrac{{x - 16}}{8} = 0 $
The numerator value is equal to 0, so we have
 $ \Rightarrow x - 16 = 0 $
Therefore $ \Rightarrow x = 16. $

We can also verify the given question by substituting the value of x.
Consider $ \dfrac{3}{8}x = 6 $ . Substitute the value of x as 16 so we have
 $ \Rightarrow \dfrac{3}{8}(16) = 6 $
On simplification we get
 $
   \Rightarrow 3(2) = 6 \\
   \Rightarrow 6 = 6 \;
  $
Hence LHS is equal to RHS.
So, the correct answer is “16”.

Note: If the algebraic expression contains only one unknown, we determine the value by using simple multiplication and division. The function contains a fraction then there is no change in solving the algebraic expression. The tables of multiplication should be known to solve these kinds of problems.