
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 . . . . . . . $\displaystyle \left(\text{-}\infty,\,\text{-}\tfrac{1}{7}\right)\:\cup\:\left[1,\,\infty\right)$
. . . . . $\displaystyle \begin{array}{ccccc} === & \circ & --- & \bullet & === \\ & \text{-}\frac{1}{7} && 1 \end{array}$