Chapter A, Problem 26E

To determine

To solve the below inequality in terms of intervals and illustrate the solution set on the real number line -

  $-3<\frac{1}{x}\le 1$

Answer to Problem 26E

The solution of the inequality is $x<-\frac{1}{3}$ or $x\ge 1$ and the solution set on a real number line -

  

Explanation of Solution

Given: Inequality: $-3<\frac{1}{x}\le 1$

Formula Used:

An inequality compares two values, showing if one is less than, greater than, or simply not equal to another value.

Real number line is the line whose points are the real numbers. 

Calculation:

Given: Inequality equation is $-3<\frac{1}{x}\le 1$

The above inequality can be broken down into:

  $-3<\frac{1}{x}$ and $\frac{1}{x}\le 1$

Solving the first inequality, we have:

  $-3<\frac{1}{x}$

Adding $3$ to both the sides, we have:

  $\begin{array}{l}-3+3<\frac{1}{x}+3\\ 0<\frac{1}{x}+3\\ 0<\frac{1+3x}{x}\\ \frac{x}{1+3x}>0\end{array}$

To solve the above inequality, we needto find the different intervals for which the inequality gives a value greater than $0$ .

When $x<-\frac{1}{3}$ :

  $x$ is negative and $\left(1+3x\right)$ is negative

Thus, $\frac{x}{1+3x}>0$

So, $x<-\frac{1}{3}$ isone of the solutions.

When $-\frac{1}{3}<x<0$ :

  $x$ is negative and $\left(1+3x\right)$ is positive

Thus, $\frac{x}{1+3x}<0$

So, $-\frac{1}{3}<x<0$ is not one of the solutions.

When $x>0$ :

  $x$ is positiveand $\left(1+3x\right)$ is positive

Thus, $\frac{x}{1+3x}>0$

So, $x>0$ is one of the solutions.

Combining all the solutions, we have the solution set as:

  $x<-\frac{1}{3}$ or $x>0$

Solving the second inequality, we have:

  $\frac{1}{x}\le 1$

Subtract $1$ from both the sides, we have:

  $\begin{array}{l}\frac{1}{x}-1\le 1-1\\ \frac{1}{x}-1\le 0\\ \frac{1-x}{x}\le 0\\ \frac{x-1}{x}\ge 0\end{array}$

To solve the above inequality, we need to find the different intervals for which the inequality gives a value less than or equal to $0$ .

When $x<0$ :

  $x$ is negative and $\left(x-1\right)$ is negative

Thus, $\frac{x-1}{x}>0$

So, $x<0$ is one of the solutions.

When $x=0$ :

  $x$ is zero and $\left(x-1\right)$ is negative

Thus, $\frac{x-1}{x}$ is undefined

So, $x=0$ is not one of the solutions.

When $0<x<1$ :

  $x$ is positive and $\left(x-1\right)$ is negative

Thus, $\frac{x-1}{x}<0$

So, $0<x<1$ is not one of the solutions.

When $x=1$ :

  $x$ is positive and $\left(x-1\right)$ is zero

Thus, $\frac{x-1}{x}>0$

So, $x=1$ is one of the solutions.

When $x>1$ :

  $x$ is positive and $\left(x-1\right)$ is positive

Thus, $\frac{x-1}{x}>0$

So, $x>1$ is one of the solutions.

Combining all the solutions, we have the solution set as:

  $x<0$ or $x\ge 1$

Now, combining solutions from both the inequalities, we have:

  $x<-\frac{1}{3}$ or $x>0$ and $x<0$ or $x\ge 1$

Thus, solution set is $x<-\frac{1}{3}$ or $x\ge 1$

Drawing the above inequality on a real number line, we have:

  

Conclusion:

Hence, the solution of the inequality is $x<-\frac{1}{3}$ or $x\ge 1$ and the solution set on a real number line -