Question

How do you solve $\dfrac{20}{x}=-\dfrac{5}{2}$?

Answer 1

Hint: We solve the given linear equation by simplifying the equation. We cross-multiply the equations. Then we apply the binary operation of division to get the value of $x$. We use the G.C.D of the denominator and the numerator to divide both of them. We get the simplified form when the G.C.D is 1.

Complete step by step answer:
The given equation $\dfrac{20}{x}=-\dfrac{5}{2}$ is a linear equation of $x$.
We apply cross-multiplication to multiply $x$ with $-5$ and 20 with 2.
\[\begin{align}
  & \dfrac{20}{x}=-\dfrac{5}{2} \\
 & \Rightarrow x\left( -5 \right)=2\times 20 \\
\end{align}\]
We complete the multiplication to get \[-5x=40\].
Now we divide both sides with $-5$ and get
\[\begin{align}
  & \dfrac{-5x}{-5}=\dfrac{40}{-5} \\
 & \Rightarrow x=-\dfrac{40}{5} \\
\end{align}\]
We need to find the simplified form of the proper fraction \[\dfrac{40}{5}\].
Simplified form is achieved when the G.C.D of the denominator and the numerator is 1.
This means we can’t eliminate any more common root from them other than 1.
For any fraction $\dfrac{p}{q}$, we first find the G.C.D of the denominator and the numerator. If it’s 1 then it’s already in its simplified form and if the G.C.D of the denominator and the numerator is any other number d then we need to divide the denominator and the numerator with d and get the simplified fraction form as $\dfrac{{}^{p}/{}_{d}}{{}^{q}/{}_{d}}$.
For our given fraction \[\dfrac{40}{5}\], the G.C.D of the denominator and the numerator is 5.
Now we divide both the denominator and the numerator with 5 and get $\dfrac{{}^{40}/{}_{5}}{{}^{5}/{}_{5}}=\dfrac{8}{1}=8$.
Therefore, the solution of $\dfrac{20}{x}=-\dfrac{5}{2}$ is $x=-8$.

Note:
We also could have formed a factorisation of the equation \[-5x=40\]. We take the constant 5 common out of the reformed equation \[5x+40=0\].
Therefore, $5x+40=5\left( x+8 \right)=0$.
The multiplication of two terms gives 0 where one of the terms is positive and non-zero. This gives that the other term has to be zero.
So, $\left( x+8 \right)=0$ which gives $x=-8$ as the solution.