Question

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

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 $ \dfrac{{x - 5}}{3} = \dfrac{{x - 3}}{5} \cdots \cdots \left( 1 \right) $ . We have to solve the equation $ \left( 1 \right) $ for $ x $ .
First we will rewrite the equation $ \left( 1 \right) $ . Therefore, we get $ \dfrac{x}{3} - \dfrac{5}{3} = \dfrac{x}{5} - \dfrac{3}{5} \cdots \cdots \left( 2 \right) $ .
Now we will subtract the term $ \dfrac{x}{5} $ on both sides of equation $ \left( 2 \right) $ . Therefore, we get $
  \left( {\dfrac{x}{3} - \dfrac{5}{3}} \right) - \dfrac{x}{5} = \left( {\dfrac{x}{5} - \dfrac{3}{5}} \right) - \dfrac{x}{5} \\
   \Rightarrow \left( {\dfrac{x}{3} - \dfrac{x}{5}} \right) - \dfrac{5}{3} = \left( {\dfrac{x}{5} - \dfrac{x}{5}} \right) - \dfrac{3}{5} \\
   \Rightarrow \left( {\dfrac{x}{3} - \dfrac{x}{5}} \right) - \dfrac{5}{3} = - \dfrac{3}{5} \cdots \cdots \left( 3 \right) \\
  $
Let us simplify the LHS of equation $ \left( 3 \right) $ by taking LCM (least common multiple). Therefore, we get $
  \left( {\dfrac{{5x - 3x}}{{15}}} \right) - \dfrac{5}{3} = - \dfrac{3}{5} \\
   \Rightarrow \dfrac{{2x}}{{15}} - \dfrac{5}{3} = - \dfrac{3}{5} \cdots \cdots \left( 4 \right) \\
  $
Now we will add the number $ \dfrac{5}{3} $ on both sides of equation $ \left( 4 \right) $ . Therefore, we get $
  \left( {\dfrac{{2x}}{{15}} - \dfrac{5}{3}} \right) + \dfrac{5}{3} = - \dfrac{3}{5} + \dfrac{5}{3} \\
   \Rightarrow \dfrac{{2x}}{{15}} = - \dfrac{3}{5} + \dfrac{5}{3} \\
   \Rightarrow \dfrac{{2x}}{{15}} = - \dfrac{9}{{15}} + \dfrac{{25}}{{15}} \\
   \Rightarrow \dfrac{{2x}}{{15}} = \dfrac{{16}}{{15}} \cdots \cdots \left( 5 \right) \\
  $
Now we will multiply by the term $ \dfrac{{15}}{2} $ on both sides of equation $ \left( 5 \right) $ . Therefore, we get $
  \dfrac{{2x}}{{15}} \times \dfrac{{15}}{2} = \dfrac{{16}}{{15}} \times \dfrac{{15}}{2} \\
   \Rightarrow x = 8 \\
  $
Therefore, we can say that $ x = 8 $ is a 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 = 8 $ on the LHS of the given equation. Therefore, we get LHS $ \dfrac{{8 - 5}}{3} = \dfrac{3}{3} = 1 $ . Now we will put $ x = 8 $ on the RHS of the given equation. Therefore, we get RHS $ \dfrac{{8 - 3}}{5} = \dfrac{5}{5} = 1 $ . 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.