Answer: (a) \(x=3\) or \(x=-\frac12\); (b) \(-7\leq x\leq -1\).

Use the definition of an absolute value: split the equation or inequality into the relevant linear cases, then combine the intervals that satisfy the original statement.

(a) Start with

\(\frac{|4x-5|}{7}=1.\)

Multiplying by 7 gives

\(|4x-5|=7.\)

Therefore

\(4x-5=7 \quad\text{or}\quad 4x-5=-7.\)

So

\(x=3 \quad\text{or}\quad x=-\frac12.\)

(b) The inequality is

\(|3x+9|\leq |x-5|.\)

The two graphs meet where

\(|3x+9|=|x-5|.\)

Squaring both sides gives

\((3x+9)^2=(x-5)^2.\)

Using the difference of two squares,

\((3x+9-(x-5))(3x+9+x-5)=0.\)

Hence

\((2x+14)(4x+4)=0.\)

So the intersection values are

\(x=-7 \quad\text{and}\quad x=-1.\)

Between these two values, the graph of \(y=|3x+9|\) lies below or on the graph of \(y=|x-5|\). Therefore

\(-7\leq x\leq -1.\)