Prove: For every $\epsilon > 0$, there exists a $\delta >0$ such that $1 - \delta < x < 1 + \delta$ implies that $2 - \epsilon < 7 -5x< 2 + \epsilon$

Question

So far this is what I have

Let $$ \delta = \frac{\epsilon}{5} $$

So, if we start with $1 - \delta < x < 1 + \delta$

\begin{align} &\Rightarrow -5 + 5\delta < -5x < -5 -5\delta \quad \text{(multiply by} -5) \\ &\Rightarrow 2 + 5\delta < 7 -5x < 2 - 5\delta \quad \text{(add} \,\, 7) \\ &\Rightarrow 2 + 5 * \frac{\epsilon}{5} < 7 - 5x < 2 - 5*5\epsilon \quad \text{(substitute)}\\ &\Rightarrow 2 + \epsilon < 7 - 5x < 2 - \epsilon \quad \text{(simplify)} \end{align}

Is there something about inequalities that I am forgetting that my signs are opposite or am I doing something completely wrong?

Thanks in advance.

Answer 1

When you multiply an inequality by a negative number, the direction of the inequality is reversed, so that

$$ 1-\delta < x < 1+\delta $$

implies

$$ -5+5\delta > -5x > -5-5\delta $$

Answer 2

When you multiply two sides of an inequality by a negative number, the direction of inequality is reversed.

Example, 1 < 2, Multiplying by -5 yields -5 > -10 and not -5 < -10