Question

How do solve \[2{{(x-5)}^{-1}}+\dfrac{1}{x}=0\]?

Answer 1

Hint: We are given an expression and we have to solve for \[x\]. We will simplify the given expression further, such as \[{{(x-5)}^{-1}}\] can be written as \[\dfrac{1}{(x-5)}\] and so on. The simplified expression is then written in terms of \[x\] and we continue to solve until we find the value of \[x\]. Hence, we will have solved the expression for \[x\].

Complete step-by-step answer:
According to the given question, we are given an expression which we have to solve and find the value of \[x\].
The expression we have is,
\[2{{(x-5)}^{-1}}+\dfrac{1}{x}=0\]----(1)
We will begin to solve the expression. We will first have to simplify the given expression and then write the simplified form of the expression in terms of \[x\].
We know that, any number raised to the power -1 means that the number is in the denominator, so we have the expression as,
\[\Rightarrow \dfrac{2}{(x-5)}+\dfrac{1}{x}=0\]---(2)
As we can see that the equation (2) has two fractions. But, we cannot add them simply as their denominators are different.
So, our next task is to make the denominators the same and for this we will be using Least Common Multiple (LCM).
LCM of ‘x-5’ and ‘x’ will be the product of each other, that is, \[(x-5)x\].
Multiplying the numerator with the appropriate factor, we will get the expression as,
\[\Rightarrow \dfrac{2x+(x-5)}{(x-5)x}=0\]
Solving the above expression further, we will get,
\[\Rightarrow \dfrac{3x-5}{(x-5)x}=0\]-----(3)
The denominator of the above expression in LHS is cross multiplied to 0 and gets equal to 0. So, the expression we get is,
\[\Rightarrow 3x-5=0\]----(4)
Now, adding 5 on both sides of the equation (5), we will get,
\[\Rightarrow 3x=5\]
Writing the expression in terms of ‘x’ alone, we will have the value of ‘x’ as,
\[\Rightarrow x=\dfrac{5}{3}\]
Therefore, the value of \[x=\dfrac{5}{3}\].

Note: The expression should be carefully reduced without missing any terms in the process. Also, it should be done in proper steps. The value of ‘x’ should be carefully reduced as well, if possible. Since the R.H.S was 0, we were able to simplify easily. If we had non-zero numbers, then we would have obtained a quadratic equation in x and x would have had 2 values.