Question

Solve the following linear equation.
 $ x - \dfrac{1}{5} = \dfrac{x}{3} + \dfrac{1}{4} $

Answer 1

Hint: To solve the given linear equation, we will remove the variable term from the right-hand side. Usually we put the variable term on the left-hand side and all other (constant) terms on the right-hand side. We will use basic knowledge of mathematical operations like addition, subtraction, multiplication and division.

Complete step-by-step answer:
In this problem, we have a single linear equation in one variable $ x $ . The given equation is $ x - \dfrac{1}{5} = \dfrac{x}{3} + \dfrac{1}{4} \cdots \cdots \left( 1 \right) $ . We have to solve the equation $ \left( 1 \right) $ for $ x $ .
First we will subtract the term $ \dfrac{x}{3} $ from both sides of equation $ \left( 1 \right) $ . Therefore, we get $
  \left( {x - \dfrac{1}{5}} \right) - \dfrac{x}{3} = \left( {\dfrac{x}{3} + \dfrac{1}{4}} \right) - \dfrac{x}{3} \\
   \Rightarrow \left( {x - \dfrac{x}{3}} \right) - \dfrac{1}{5} = \left( {\dfrac{x}{3} - \dfrac{x}{3}} \right) + \dfrac{1}{4} \\
   \Rightarrow \left( {x - \dfrac{x}{3}} \right) - \dfrac{1}{5} = \dfrac{1}{4} \cdots \cdots \left( 2 \right) \\
  $
Let us simplify the LHS of equation $ \left( 2 \right) $ by taking LCM (least common multiple). Therefore, we get $
  \left( {\dfrac{{3x - x}}{3}} \right) - \dfrac{1}{5} = \dfrac{1}{4} \\
   \Rightarrow \dfrac{{2x}}{3} - \dfrac{1}{5} = \dfrac{1}{4} \cdots \cdots \left( 3 \right) \\
  $
Now we will add the term $ \dfrac{1}{5} $ on both sides of equation $ \left( 3 \right) $ . Therefore, we get $
  \left( {\dfrac{{2x}}{3} - \dfrac{1}{5}} \right) + \dfrac{1}{5} = \dfrac{1}{4} + \dfrac{1}{5} \\
   \Rightarrow \dfrac{{2x}}{3} = \dfrac{1}{4} + \dfrac{1}{5} \cdots \cdots \left( 4 \right) \\
  $
Let us simplify the RHS of equation $ \left( 4 \right) $ by taking LCM (least common multiple). Therefore, we get $
  \dfrac{{2x}}{3} = \dfrac{{5 + 4}}{{20}} \\
   \Rightarrow \dfrac{{2x}}{3} = \dfrac{9}{{20}} \cdots \cdots \left( 5 \right) \\
  $
Now we will multiply by the term $ \dfrac{3}{2} $ on both sides of equation $ \left( 5 \right) $ . Therefore, we get $
  \dfrac{{2x}}{3} \times \dfrac{3}{2} = \dfrac{9}{{20}} \times \dfrac{3}{2} \\
   \Rightarrow x = \dfrac{{27}}{{40}} \\
  $
Therefore, we can say that $ x = \dfrac{{27}}{{40}} $ is the solution of the given equation.

Note: In this type of problem, we can verify the answer by putting the value of $ x $ in the given equation. Let us put $ x = \dfrac{{27}}{{40}} $ on the LHS of the given equation. Therefore, we get LHS $ = \dfrac{{27}}{{40}} - \dfrac{1}{5} = \dfrac{{19}}{{40}} $ . Now we will put $ x = \dfrac{{27}}{{40}} $ on the RHS of the given equation. Therefore, we get RHS $ = \dfrac{{\dfrac{{27}}{{40}}}}{3} + \dfrac{1}{4} = \dfrac{9}{{40}} + \dfrac{1}{4} = \dfrac{{9 + 10}}{{40}} = \dfrac{{19}}{{40}} $ . Hence, the answer is verified. The general form of the linear equation in one variable $ x $ is given by $ ax + b = 0 $ where $ a $ and $ b $ are real numbers. The power of variable $ x $ is $ 1 $ . So, it is called a linear equation. A linear equation in one variable has exactly one solution.