Question

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

Answer 1

Hint: Assume that A is the additive inverse of the expression $x-{\displaystyle \frac{1}{x}}$. 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 $A+x-{\displaystyle \frac{1}{x}}=0$
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 ${\displaystyle \frac{1}{x}}$ 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 $x-{\displaystyle \frac{1}{x}}$
Since we know that the sum of the number and its additive inverse is equal to the additive identity, i.e. 0, we have
$A+x-{\displaystyle \frac{1}{x}}=0$
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 ${\displaystyle \frac{1}{x}}$ on both sides of the equation, we get
$A+x={\displaystyle \frac{1}{x}}$
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
$A={\displaystyle \frac{1}{x}}-x$
Rewriting, we get
$A=-x+{\displaystyle \frac{1}{x}}$
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
$x-{\displaystyle \frac{1}{x}}-x+{\displaystyle \frac{1}{x}}=(x-x)+({\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{x}})=0+0=0$
Hence by definition, we have
$-x+{\displaystyle \frac{1}{x}}$ is the additive inverse of $x-{\displaystyle \frac{1}{x}}$
Hence our answer is verified to be correct.