Originally Posted by

**Soroban** Hello, BobRoss!

This one is tricky and dangerous.

There are several approaches.

I'll show you mine . . .

Like you, I split it into two inequalities: .$\displaystyle -7 \:<\:\frac{1}{x}\:\text{ and }\:\frac{1}{x} \:\le\:1$

Note the "and"; we want *both* to be true.

We'd like to multiply through by , but we must be *very* careful!

What we get depends on whether $\displaystyle x$ is positive or negative.

Suppose $\displaystyle x$ is positive.

Multiply both inequalities by $\displaystyle x.$

We have: .$\displaystyle \text{-}7x \:<\:1 \quad\Rightarrow\quad x \:>\:\text{-}\tfrac {1}{7}$

. . . and: .$\displaystyle 1 \:\le\:x \quad\Rightarrow\quad x \:\ge\:1$

To make both statements true, we use: .$\displaystyle \boxed{x\:\ge\:1}$

Suppose $\displaystyle x$ is negative.

Multiply both inequalities by negative $\displaystyle x.$

We have: .$\displaystyle \text{-}7x \:>\:1 \quad\Rightarrow\quad x \:<\:\text{-}\tfrac{1}{7}$

. . . and: .$\displaystyle 1 \:\ge\:x \quad\Rightarrow\quad x \:\le\:1 $

To make both statements true, we use: .$\displaystyle \boxed{x \:<\:\text{-}\tfrac{1}{7}}$

Solution: .$\displaystyle \left(x\:<\:\text{-}\tfrac{1}{7}\right)\:\cup\:\left(x\:\ge\:1\right)$

n . . . . . . . $\displaystyle \left(\text{-}\infty,\,\text{-}\tfrac{1}{7}\right)\:\cup\:\left[1,\,\infty\right)$

. . . . . $\displaystyle \begin{array}{ccccc} === & \circ & --- & \bullet & === \\ & \text{-}\frac{1}{7} && 1 \end{array}$