Question

How do you solve $-\dfrac{3}{4x}+\dfrac{3}{8}=\dfrac{27}{32}$?

Answer 1

Hint: We separate the variables and the constants of the equation $-\dfrac{3}{4x}+\dfrac{3}{8}=\dfrac{27}{32}$. We apply the algebraic operation of subtraction for constants. The solutions of the variables and the constants will be added at the end to get the final answer to equate with 0 using cross-multiplication. Then we solve the linear equation to find the value of $x$.

Complete step-by-step solution:
We have to find the solution for the equation $-\dfrac{3}{4x}+\dfrac{3}{8}=\dfrac{27}{32}$.
The given equation $-\dfrac{3}{4x}+\dfrac{3}{8}=\dfrac{27}{32}$ is a linear equation of $x$. we need to simplify the equation by solving the constants separately.
All the terms in the equation of $-\dfrac{3}{4x}+\dfrac{3}{8}=\dfrac{27}{32}$ are either variable of $x$ or a constant.
We take the constants all together to solve it.
$\begin{align}
  & -\dfrac{3}{4x}+\dfrac{3}{8}=\dfrac{27}{32} \\
 & \Rightarrow -\dfrac{3}{4x}=\dfrac{27}{32}-\dfrac{3}{8} \\
\end{align}$
We take the constants fractions together.
The LCM is 32. So, we get $\dfrac{27}{32}-\dfrac{3}{8}=\dfrac{27-12}{32}=\dfrac{15}{32}$.
Now we apply cross multiplication in the equation $-\dfrac{3}{4x}=\dfrac{15}{32}$ to get
\[\begin{align}
  & -\dfrac{3}{4x}=\dfrac{15}{32} \\
 & \Rightarrow -\dfrac{3\times 32}{4\times 15}=\dfrac{x}{1} \\
 & \Rightarrow x=-\dfrac{24}{15} \\
\end{align}\]
Therefore, the final solution becomes \[x=-\dfrac{24}{15}\].

Note: We can verify the result of the equation $-\dfrac{3}{4x}+\dfrac{3}{8}=\dfrac{27}{32}$ by substitute the value of x as \[x=-\dfrac{24}{15}\].
Therefore, the left-hand side of the equation $-\dfrac{3}{4x}+\dfrac{3}{8}$ becomes
$-\dfrac{3}{4x}+\dfrac{3}{8}=-\dfrac{3}{4\times \left( -\dfrac{24}{15} \right)}+\dfrac{3}{8}=\dfrac{45}{96}+\dfrac{3}{8}=\dfrac{45+36}{96}=\dfrac{27}{32}$
Thus, verified for the equation $-\dfrac{3}{4x}+\dfrac{3}{8}=\dfrac{27}{32}$ the solution is \[x=-\dfrac{24}{15}\].