Question

What is the value of $ x $ that satisfies the equation $ \dfrac{7}{3}\left( x+\dfrac{9}{28} \right)=20 $ ?

Answer 1

Hint: We multiply both sides with $ \dfrac{3}{7} $ and then we separate the variable and the constants of the equation $ x+\dfrac{9}{28}=\dfrac{60}{7} $ . We apply the binary operation of subtraction for constants. We take the LCM form to complete the subtraction. The solutions of the variables and the constants will be added at the end to get the final answer to equate with 0. Then we solve the linear equation to find the value of $ x $ .

Complete step-by-step answer:
The given equation $ \dfrac{7}{3}\left( x+\dfrac{9}{28} \right)=20 $ is a linear equation of $ x $ .
We first multiply both sides with $ \dfrac{3}{7} $ and get $ x+\dfrac{9}{28}=20\times \dfrac{3}{7}=\dfrac{60}{7} $ .
Now we need to simplify the equation by solving the variables and the constants separately.
All the terms in the equation of $ x+\dfrac{9}{28}=\dfrac{60}{7} $ are either variable of $ x $ or a constant. We first separate the constants.
We take the constants all together to solve it.
 $ \begin{align}
  & x+\dfrac{9}{28}=\dfrac{60}{7} \\
 & \Rightarrow x=\dfrac{60}{7}-\dfrac{9}{28} \\
\end{align} $
The binary operation between the fractions is subtraction for which we need the LCM of the denominators.
The LCM of the numbers 7 and 28 is 28.
\[\begin{align}
  & 7\left| \!{\underline {\,
  7,28 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1,4 \,}} \right. \\
\end{align}\]
So, we get $ 7\times 4=28 $ .
Now we complete the subtraction and get
 $ \Rightarrow \dfrac{60}{7}-\dfrac{9}{28}=\dfrac{4\times 60-9}{28}=\dfrac{231}{28} $
Therefore, the final form is $ x=\dfrac{60}{7}-\dfrac{9}{28}=\dfrac{231}{28} $
The solution is $ x=\dfrac{231}{28} $ .
So, the correct answer is “$ x=\dfrac{231}{28} $”.

Note: We can verify the result of the equation $ \dfrac{7}{3}\left( x+\dfrac{9}{28} \right)=20 $ by taking the value of as $ x=\dfrac{231}{28} $ .
Therefore, the left-hand side of the equation $ \dfrac{7}{3}\left( x+\dfrac{9}{28} \right)=20 $ becomes
 $ \dfrac{7}{3}\left( \dfrac{231}{28}+\dfrac{9}{28} \right)=\dfrac{7}{3}\left( \dfrac{240}{28} \right)=\dfrac{240}{3\times 4}=20 $
Thus, verified for the equation $ \dfrac{7}{3}\left( x+\dfrac{9}{28} \right)=20 $ the solution is $ x=\dfrac{231}{28} $ .