Solving Inequalities with Absolute Values

[Dethrone]

1. if $|x-2|<A$, then $|2x-4|<3$

My steps:

$$|2x-4|<3$$
$$2|x-2|<3$$
$$|x-2|<3/2$$

Hence, $A>3/2$, why does the answer say that $A\ge 3/2$?

2. if $|x-a|<5$, then $|2x-3|<A$

No idea how to do this one...I tried to manipulate the right inequality into a form such as $|2x-4|$, but I was unsuccessful, any hints?

 

[chisigma]

1. if $|x-2|<A$, then $|2x-4|<3$

My steps:

$$|2x-4|<3$$
$$2|x-2|<3$$
$$|x-2|<3/2$$

Hence, $A>3/2$, why does the answer say that $A\ge 3/2$?

2. if $|x-a|<5$, then $|2x-3|<A$

No idea how to do this one...I tried to manipulate the right inequality into a form such as $|2x-4|$, but I was unsuccessful, any hints?

About the number 1 is pretty obvious that if $A = \frac{3}{2}$ and $|x-2| < \frac{3}{2}$ is also $|x-2| < A$...

About the number 2 is...

$\displaystyle |x - a| < 5 \implies a - 5 < x < a+5 \implies 2 a - 10 < 2 x < 2 a + 10 \implies 2 a - 13 < 2 x - 3 < 2 a + 7 \implies |2 x - 3| < A$

... where $\displaystyle A = \text {max} \{|2 a - 13|, |2 a + 7| \}$...

Kind regards

$\chi$ $\sigma$

 

[Dethrone]

I follow that, but the answer for a) in my textbook is $a\ge 3/2$. Is that correct? I got $a>3/2$ as my answer.

I understand how you did the second one, but can you explain how I can obtain the answer $a \ge 11$ from that?