Question

How do you solve easy questions involving both fractions and variables?

Answer 1

Hint: In the above question we are asked about the concept for solving a particular type of equations, which includes both the fractions and the variables. For explaining the concept, we need to take the example of the equation of such type. So we can choose the equation $-\dfrac{4}{5}x+\dfrac{2}{3}=1$. For solving it, we have to subtract $\dfrac{2}{3}$ from both the sides to get the equation as \[-\dfrac{4}{5}x=1-\dfrac{2}{3}\]. Then we have to simplify the RHS by taking the LCM so as to get the equation \[-\dfrac{4}{5}x=\dfrac{1}{3}\]. Finally, on multiplying the equation by \[-\dfrac{5}{4}\], we will get the solution of the equation.

Complete step by step answer:
According to the question, we need to explain the method of solving the equations which involve both of the fractions and the variables. Therefore, for taking the example of such an equation, we consider the linear equation given as
$\Rightarrow -\dfrac{4}{5}x+\dfrac{2}{3}=1$
For solving the above equation, we subtract $\dfrac{2}{3}$ from both the sides to get
\[\begin{align}
  & \Rightarrow -\dfrac{4}{5}x+\dfrac{2}{3}-\dfrac{2}{3}=1-\dfrac{2}{3} \\
 & \Rightarrow -\dfrac{4}{5}x=1-\dfrac{2}{3} \\
\end{align}\]
Now, we simplify the RHS of the above equation by taking the LCM to get
\[\begin{align}
  & \Rightarrow -\dfrac{4}{5}x=\dfrac{3-2}{3} \\
 & \Rightarrow -\dfrac{4}{5}x=\dfrac{1}{3} \\
\end{align}\]
Now, we multiply both sides of the above equation by \[5\] to get
\[\begin{align}
  & \Rightarrow -\dfrac{4}{5}x\left( 5 \right)=\dfrac{5}{3} \\
 & \Rightarrow -4x=\dfrac{5}{3} \\
\end{align}\]
Finally, we divide both the sides of the above equation by \[-4\] to get
\[\begin{align}
  & \Rightarrow \dfrac{-4x}{-4}=\dfrac{5}{3\left( -4 \right)} \\
 & \Rightarrow x=-\dfrac{5}{12} \\
\end{align}\]

Hence, we have explained the solution of the equations containing both the fractions and variables and have obtained the solution as $x=-\dfrac{5}{12}$.

Note: The solution of these types of equations involves many calculations. Therefore there are many chances of committing mistakes. To check these mistakes, we must substitute back the final solution into the given equation and confirm whether LHS is equal to RHS.