Question

Additive inverse of the rational expression $x-\dfrac{1}{x}$ will be
[a] $x+\dfrac{1}{x}$
[b] $-x+\dfrac{1}{x}$
[c] $\dfrac{{{x}^{2}}-1}{x}$
[d] $-\dfrac{1}{x}+x$

Answer 1

Hint: Assume that A is the additive inverse of the expression $x-\dfrac{1}{x}$. Use the fact that the sum of the number and its additive inverse is equal to the additive identity,i.e. 0
Hence, prove that $A+x-\dfrac{1}{x}=0$
Use the fact that the addition and subtraction of equal things on both sides of an equation does not change the solution set of the equation. Hence add $\dfrac{1}{x}$ on both sides of the equation and subtract x from both sides of the equation. Hence find the value of A in terms of x. Verify your answer.

Complete step-by-step answer:
Let A be the additive inverse of the term $x-\dfrac{1}{x}$
Since we know that the sum of the number and its additive inverse is equal to the additive identity, i.e. 0, we have
$A+x-\dfrac{1}{x}=0$
We know that the addition of equal terms on both sides of the equation does not change the solution set of the equation.
Hence, adding $\dfrac{1}{x}$ on both sides of the equation, we get
$A+x=\dfrac{1}{x}$
We know that the subtraction of equal terms from both sides of the equation does not change the solution set of the equation
Hence, subtracting x from both sides of the equation, we get
$A=\dfrac{1}{x}-x$
Rewriting, we get
$A=-x+\dfrac{1}{x}$
Hence option[b] is correct

Note: Verification:
We know that the sum of a number and its additive inverse is equal to 0
Now, we have
$x-\dfrac{1}{x}-x+\dfrac{1}{x}=\left( x-x \right)+\left( \dfrac{1}{x}-\dfrac{1}{x} \right)=0+0=0$
Hence by definition, we have
$-x+\dfrac{1}{x}$ is the additive inverse of $x-\dfrac{1}{x}$
Hence our answer is verified to be correct.