Question

Solve the following linear equation.
 $ x + 7 - \dfrac{{8x}}{3} = \dfrac{{17}}{6} - \dfrac{{5x}}{2} $

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 + 7 - \dfrac{{8x}}{3} = \dfrac{{17}}{6} - \dfrac{{5x}}{2} \cdots \cdots \left( 1 \right) $ . We have to solve the equation $ \left( 1 \right) $ for $ x $ .
First we will add the term $ \dfrac{{5x}}{2} $ on both sides of equation $ \left( 1 \right) $ . Therefore, we get $
  \left( {x + 7 - \dfrac{{8x}}{3}} \right) + \dfrac{{5x}}{2} = \left( {\dfrac{{17}}{6} - \dfrac{{5x}}{2}} \right) + \dfrac{{5x}}{2} \\
   \Rightarrow x - \dfrac{{8x}}{3} + \dfrac{{5x}}{2} + 7 = \dfrac{{17}}{6} - \dfrac{{5x}}{2} + \dfrac{{5x}}{2} \\
   \Rightarrow \left( {x - \dfrac{{8x}}{3} + \dfrac{{5x}}{2}} \right) + 7 = \dfrac{{17}}{6} \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{{6x - 16x + 15x}}{6}} \right) + 7 = \dfrac{{17}}{6} \\
   \Rightarrow \dfrac{{5x}}{6} + 7 = \dfrac{{17}}{6} \cdots \cdots \left( 3 \right) \\
  $
Now we will subtract the number $ 7 $ from both sides of equation $ \left( 3 \right) $ . Therefore, we get $
  \left( {\dfrac{{5x}}{6} + 7} \right) - 7 = \dfrac{{17}}{6} - 7 \\
   \Rightarrow \dfrac{{5x}}{6} = \dfrac{{17 - 42}}{6} \\
   \Rightarrow \dfrac{{5x}}{6} = - \dfrac{{25}}{6} \cdots \cdots \left( 4 \right) \\
  $
Now we will multiply by the term $ \dfrac{6}{5} $ on both sides of equation $ \left( 4 \right) $ . Therefore, we get $
  \dfrac{{5x}}{6} \times \dfrac{6}{5} = - \dfrac{{25}}{6} \times \dfrac{6}{5} \\
   \Rightarrow x = - 5 \\
  $
Therefore, we can say that $ x = - 5 $ 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 = - 5 $ on LHS of the given equation. Therefore, we get LHS $ = - 5 + 7 - \dfrac{{8\left( { - 5} \right)}}{3} = - 5 + 7 + \dfrac{{40}}{3} = \dfrac{{ - 15 + 21 + 40}}{3} = \dfrac{{46}}{3} $ . Now we will put $ x = - 5 $ on the RHS of the given equation. Therefore, we get RHS $ = \dfrac{{17}}{6} - \dfrac{{5\left( { - 5} \right)}}{2} = \dfrac{{17}}{6} + \dfrac{{25}}{2} = \dfrac{{17 + 75}}{6} = \dfrac{{92}}{6} = \dfrac{{46}}{3} $ . 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.