Question

How do you solve $\dfrac{1}{x} + \dfrac{2}{{x - 5}} = 0$?

Answer 1

Hint:In order to determine the value of variable$x$ in the above equation. Combine the terms on the left hand side of the equation by taking LCM of the terms and use the rules of transposing terms to transpose terms having $(x)$on the Left-hand side and constant value terms on the Right-Hand side of the equation. Solving like terms will lead to your required result.

Complete step by step solution:
We are given a Mathematical expression in one variable ($x$)
$\dfrac{1}{x} + \dfrac{2}{{x - 5}} = 0$.and we have to solve this equation for variable ($x$).

$ \Rightarrow \dfrac{1}{x} + \dfrac{2}{{x - 5}} = 0$

Now combining like terms on both of the sides.

Taking LCM of both the terms which comes to be $(x)(x - 5)$our equation becomes,
$
\Rightarrow \dfrac{{1(x - 5) + 2(x)}}{{(x)(x - 5)}} = 0 \\
\Rightarrow \dfrac{{x - 5 + 2x}}{{(x)(x - 5)}} = 0 \\
$

Simplifying more, we get
$ \Rightarrow \dfrac{{ - 5 + 3x}}{{(x)(x - 5)}} = 0$

Now multiplying both sides with $(x)(x - 5)$
$
\Rightarrow \dfrac{{ - 5 + 3x}}{{(x)(x - 5)}}.(x)(x - 5) = 0 \times (x)(x - 5) \\
\Rightarrow - 5 + 3x = 0 \\
$

Terms having $x$will on the Left-Hand side of the equation and constant terms on the right-hand side.

Let’s recall one basic property of transposing terms that on transposing any term from one side to another the sign of that term get reversed .In our case,$ - 5$on the left hand side will become $ + 5$on the right hand side .

After transposing terms our equation becomes
$ \Rightarrow 3x = 5$

Dividing both sides of the equation with the coefficient of x
$
\Rightarrow 3x = 5 \\

\Rightarrow \dfrac{{3x}}{3} = \dfrac{5}{3} \\

\Rightarrow x = \dfrac{5}{3} \\
$

Therefore, the solution to the equation $\dfrac{1}{x} + \dfrac{2}{{x - 5}} = 0$is equal to $x = \dfrac{5}{3}$.

Note:Linear Equation: A linear equation is a equation which can be represented in the form of $ax + c$where $x$is the unknown variable and a,c are the numbers known where $a \ne 0$.If $a = 0$then the equation will become a constant value and will no more be a linear equation .

The degree of the variable in the linear equation is of the order 1.
Every Linear equation has 1 root.