Turning Inequality with $x$ into Inequality with $|x|$

Question

My textbook jumps from $$-3<x-7<-1$$ to $$|x-7| \ge 1$$ I don't understand how the second statement was derived from the first.

Answer 1

Let $y=x-7$.

If $$-3 < y < -1$$

then we know that $y$ is negative for sure.

Also, by multiplying $-1$.

$$1 < -y < 3$$

Since $y$ is negative $-y = |y|$.

$$1 < |y| <3$$

which implies that $|y|\geq 1$.

Answer 2

Because if $$-3<x-7<-1$$ then $$x-7<-1,$$ from which we get $$x-7<-1$$ or $$x-7>1,$$ which is $$|x-7|>1.$$

Let $X$ be some inequality on $\mathbb R$ with one variable and $T(X)$ be the truth set of $X$.

We'll say that an inequality $A$ implies an inequality $B$

(we'll write this so: $A\Rightarrow B$)

iff $T(A)\subset T(B).$

In our case let $A$ be $-3<x-7<-1$ and $B$ be $|x-7|>1$.

Since $T(A)\subset T(B)$, we see that $A\Rightarrow B$.