We can rearrange the equation to isolate the absolute value on the left side:

First, we add $3$ to both sides of the equation:

\begin{align*} 2|x-4| - 3 &= 7 \\[5pt] 2|x-4| - 3 + 3 &= 7 + 3 \\[5pt] 2|x-4| &= 10 \end{align*}

Then, we divide both sides of the equation by $2{:}$

\begin{align*} \require{cancel} 2|x-4| &= 10 \\[5pt] \dfrac{2|x-4|}{2} &= \dfrac{10}{2} \\[5pt] \dfrac{\cancel{2}|x-4|}{\cancel{2}} &= 5 \\[5pt] |x-4| &=5 \end{align*}

Therefore, $k=5.$

We can rearrange the equation to isolate the absolute value on the left side:

First, we subtract $2$ from both sides of the equation: \begin{align*} 2-|x+2| = -7 \\[5pt] 2-|x+2| - 2 = -7 -2 \\[5pt] -|x+2| = -9 \\[5pt] \end{align*}

Then, we multiply both sides of the equation by $-1{:}$ \begin{align*} \require{cancel} -|x+2| & = -9 \\[5pt] (-1)\cdot \left(-|x+2| \right) & = (-1)\cdot (-9) \\[5pt] (\cancel{-}1)\cdot \left(\cancel{-}|x+2| \right) & = 9 \\[5pt] |x+2| &=\bbox[4pt,border: 1px solid lightgray]{9} \end{align*}

We can rearrange the equation to isolate the absolute value on the left side:

First, we subtract $13$ from both sides of the equation: \begin{align*} -4|1-5x|+13 &=9 \\[5pt] -4|1-5x|+13 - 13&=9 - 13 \\[5pt] -4|1-5x|&= -4 \end{align*}

Then, we divide both sides of the equation by $-4{:}$ \begin{align*} \require{cancel} -4|1-5x|&= -4 \\[5pt] \dfrac{-4|1-5x|}{-4}&= \dfrac{-4}{-4} \\[5pt] \dfrac{\cancel{-4}|1-5x|}{\cancel{-4}}&= 1 \\[5pt] |1-5x|&= \bbox[4pt,border: 1px solid lightgray]{1} \end{align*}

An absolute value equation has no solution if, when the absolute value is isolated on the left side, the right-hand side is negative.

With that in mind, let's look at each equation in turn:

• Equation I has a solution. The absolute value is already isolated on the left-hand side, and the right-hand side is non-negative.

• Equation II has no solution. To see this, we first isolate the absolute value on the left-hand side as follows: \begin{align} |x+2| + 5 &= 3 \\[5pt] |x+2| &= -2 \end{align} Since the right-hand side is negative, this equation has no solution.

• Equation III has no solution. The absolute value is already isolated on the left-hand side, and the right-hand side is negative.

Therefore, the answer is "II and III only."

An absolute value equation has no solution if, when the absolute value is isolated on the left side, the right-hand side is negative.

With that in mind, let's look at each equation in turn:

• Equation I has no solution. The absolute value is already isolated on the left-hand side, and the right-hand side is negative.

• Equation II has no solution. To see this, we first isolate the absolute value on the left-hand side as follows: \begin{align} 2|1-3x| +7&= -1 \\[5pt] 2|1-3x| &= -8 \\[5pt] |1-3x| & = -4 \end{align} Since the right-hand side is negative, this equation has no solution.

• Equation III has a solution. The absolute value is already isolated on the left-hand side, and the right-hand side is non-negative.

Therefore, the answer is "I and II only."

First, we isolate the absolute value on the left side of the equation: \begin{align*} |2x-5|-7&= 4 \\[5pt] |2x-5| &= 11 \end{align*}

Now, we have the following two cases:

• If $|2x-5| = 2x-5,$ we obtain: \begin{align} |2x-5| &= 11 \\[5pt] 2x-5 &= 11 \\[5pt] 2x &= 16 \\[5pt] x &= 8 \end{align}

• If $|2x-5| = -(2x-5),$ we obtain: \begin{align} |2x-5| &= 11 \\[5pt] -(2x-5) &= 11 \\[5pt] -2x + 5 &= 11 \\[5pt] -2x &= 6 \\[5pt] x &= -3 \end{align}

Therefore, in ascending order, the solutions are $x=\bbox[4pt,border: 1px solid lightgray]{-3}$ and $x=\bbox[4pt,border: 1px solid lightgray]{8}.$

First, we isolate the absolute value on the left side of the equation: \begin{align*} 5|x+1| - 17 &= 18 \\[5pt] 5|x+1| &= 35 \\[5pt] |x+1| &= 7 \end{align*}

Now, we have the following two cases:

• If $|x+1| = x+1,$ we obtain: \begin{align} x+1 &= 7 \\[5pt] x+1 &= 7 \\[5pt] x &= 6 \end{align}

• If $|x+1| = -(x+1),$ we obtain: \begin{align} |x+1| &= 7 \\[5pt] -(x+1) &= 7 \\[5pt] -x - 1 &= 7 \\[5pt] -x &= 8 \\[5pt] x &= -8 \end{align}

Therefore, in ascending order, the solutions are $x=\bbox[4pt,border: 1px solid lightgray]{-8}$ and $x=\bbox[4pt,border: 1px solid lightgray]{6}.$

First, we isolate the absolute value on the left side of the equation:

\begin{align*} 4|5z+4| + 7 &= -9 \\[5pt] 4|5z+4| &= -16 \\[5pt] |5z+4| &= -4 \end{align*}

Now, notice that the right-hand side is negative. Therefore, the equation has no solutions.