Question

How do you solve $\left( \dfrac{x}{2} \right)+1+\left( \dfrac{3x}{4} \right)=-9$?

Answer 1

Hint: In this problem we need to solve the given equation i.e., we need to calculate the value of $x$ for which the given equation is satisfied. To solve the given equation, we will first rearrange terms in the given equation so that all the variables are at one place and constants are at one place. Now on LHS we will simplify the fractions by taking LCM and adding them. After that we will do cross multiplication and simplify the equation to get the result.

Complete step by step solution:
Given equation, $\left( \dfrac{x}{2} \right)+1+\left( \dfrac{3x}{4} \right)=-9$.
Rearranging the terms in the above equation, so all the variables are at one place and all the constants are at one place. Then the above equation is modified as
$\begin{align}
  & \Rightarrow \dfrac{x}{2}+\dfrac{3x}{4}=-9-1 \\
 & \Rightarrow \dfrac{x}{2}+\dfrac{3x}{4}=-10 \\
\end{align}$
In the above equation we have the terms $2$, $4$ in the denominators. Adding the fraction in LHS by considering the LCM of the denominators, then we will get
$\begin{align}
  & \Rightarrow \dfrac{2\times x+1\times 3x}{4}=-10 \\
 & \Rightarrow \dfrac{2x+3x}{4}=-10 \\
\end{align}$
Doing the cross multiplication and simplifying the equation, then we will have
$\begin{align}
  & \Rightarrow 5x=-10\times 4 \\
 & \Rightarrow 5x=-40 \\
\end{align}$
Dividing the above equation with $5$ on both sides, then we will get
$\begin{align}
  & \Rightarrow \dfrac{5x}{5}=-\dfrac{40}{5} \\
 & \Rightarrow x=-8 \\
\end{align}$
Hence the solution of the given equation $\left( \dfrac{x}{2} \right)+1+\left( \dfrac{3x}{4} \right)=-9$ is $x=-8$.

Note: We can also check whether the obtained solution is correct or wrong. When we substitute the calculated solution in the given equation, if the equation is satisfied then our solution is correct otherwise the solution is wrong.
Substituting $x=-8$ in $\left( \dfrac{x}{2} \right)+1+\left( \dfrac{3x}{4} \right)=-9$, then we will get
$\begin{align}
  & \Rightarrow \left( \dfrac{-8}{2} \right)+1+\left( \dfrac{3\times -8}{4} \right)=-9 \\
 & \Rightarrow -4+1+\left( 3\times -2 \right)=-9 \\
 & \Rightarrow -3-6=-9 \\
 & \Rightarrow -9=-9 \\
 & \Rightarrow LHS=RHS \\
\end{align}$
Hence the calculated solution is correct.