Either the positive or negative of the expression inside the absolute value must be equal to the right-hand side of the equation.

So, we set up the two corresponding equations and solve them independently:

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

• If $|z-2| = -(z-2),$ we obtain: \begin{align} |z-2| &= 3 \\[5pt] -(z-2) &= 3 \\[5pt] -z+2 &= 3 \\[5pt] -z+2 -2 &= 3 - 2 \\[5pt] -z &= 1 \\[5pt] z &= -1 \end{align}

Therefore, the solutions are $z = -1$ and $z = 5.$

Either the positive or negative of the expression inside the absolute value must be equal to the right-hand side of the equation.

So, we set up the two corresponding equations and solve them independently:

• If $|7z| = 7z,$ we obtain: \begin{align} |7z| & = 42 \\[5pt] 7z & = 42 \\[5pt] \dfrac{7z}{7} & = \dfrac{42}{7} \\[5pt] z &= 6 \end{align}

• If $|7z| = -7z,$ we obtain: \begin{align} |7z| & = 42 \\[5pt] -7z & = 42 \\[5pt] \dfrac{-7z}{-7} & = \dfrac{42}{-7} \\[5pt] z &=-6 \end{align}

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

Either the positive or negative of the expression inside the absolute value must be equal to the right-hand side of the equation.

So, we set up the two corresponding equations and solve them independently:

• If $\left\vert \dfrac t8 \right\vert = \dfrac t8,$ we obtain: \begin{align} \left\vert \dfrac t8 \right\vert & = 9 \\[5pt] \dfrac t8 & = 9 \\[5pt] 8\cdot \dfrac t8 & = 8\cdot 9 \\[5pt] t &=72 \end{align}

• If $\left\vert \dfrac t8 \right\vert = -\dfrac t8,$ we obtain: \begin{align} \left\vert \dfrac t8 \right\vert & = 9 \\[5pt] -\dfrac t8 & = 9 \\[5pt] (-8)\cdot \left(-\dfrac t8 \right) & = (-8)\cdot 9 \\[5pt] t &=-72 \end{align}

Therefore, in ascending order, the solutions are $t = \bbox[4pt,border: 1px solid lightgray]{-72}$ and $t = \bbox[4pt,border: 1px solid lightgray]{72}.$

Either the positive or negative of the expression inside the absolute value must be equal to the right-hand side of the equation.

So, we set up the two corresponding equations and solve them independently:

• If $|3z+7| = 3z+7,$ we obtain: \begin{align} |3z+7| &= 5 \\[5pt] 3z+7 &= 5 \\[5pt] 3z &= -2 \\[5pt] \dfrac{3z}{3} &= \dfrac{-2}{3} \\[5pt] z &= -\dfrac 23 \end{align}

• If $|3z+7| = -(3z+7),$ we obtain: \begin{align} |3z+7| &= 5 \\[5pt] -(3z+7) &= 5 \\[5pt] -3z-7 &= 5 \\[5pt] -3z &= 12 \\[5pt] \dfrac{-3z}{-3} &= \dfrac{12}{-3} \\[5pt] z &= -4 \end{align}

Therefore, in ascending order, the solutions are $z = \bbox[4pt,border: 1px solid lightgray]{-4}$ and $z = \bbox[4pt,border: 1px solid lightgray]{-\dfrac 23}.$

Either the positive or negative of the expression inside the absolute value must be equal to the right-hand side of the equation.

So, we set up the two corresponding equations and solve them independently:

• If $|3x-4| = 3x-4,$ we obtain: \begin{align} |3x-4| &= 1 \\[5pt] 3x-4 &= 1 \\[5pt] 3x &= 5 \\[5pt] \dfrac{3x}{3} &= \dfrac{5}{3} \\[5pt] x &= \dfrac 53 \end{align}

• If $|3x-4| = -(3x-4),$ we obtain: \begin{align} |3x-4| &= 1 \\[5pt] -(3x-4) &= 1 \\[5pt] -3x+4 &= 1 \\[5pt] -3x &= -3\\[5pt] \dfrac{-3x}{-3} &= \dfrac{-3}{-3} \\[5pt] x &= 1 \end{align}

Therefore, in ascending order, the solutions are $x = \bbox[4pt,border: 1px solid lightgray]{1}$ and $x = \bbox[4pt,border: 1px solid lightgray]{\dfrac 53}.$

Either the positive or negative of the expression inside the absolute value must be equal to the right-hand side of the equation.

So, we set up the two corresponding equations and solve them independently:

• If $|2-k| = 2-k,$ we obtain: \begin{align} |2-k| & = 9 \\[5pt] 2-k & = 9 \\[5pt] -k &= 7 \\[5pt] k &= -7 \end{align}

• If $|2-k| = -(2-k),$ we obtain: \begin{align} |2-k| &= 9 \\[5pt] -(2-k) &= 9 \\[5pt] -2+k &= 9 \\[5pt] k &= 11 \end{align}

Therefore, in ascending order, the solutions are $k = \bbox[4pt,border: 1px solid lightgray]{-7}$ and $k = \bbox[4pt,border: 1px solid lightgray]{11}.$

Either the positive or negative of the expression inside the absolute value must be equal to the right-hand side of the equation.

So, we set up the two corresponding equations and solve them independently:

• If $|4-3x| = 4-3x,$ we obtain: \begin{align} |4 - 3x| &= 14 \\[5pt] 4 - 3x &= 14 \\[5pt] -3x &= 10 \\[5pt] \dfrac{-3x}{-3} &= \dfrac{10}{-3}\\[5pt] x &= -\dfrac{10}3 \end{align}

• If $|4-3x| = -(4-3x),$ we obtain: \begin{align} |4 - 3x| &= 14 \\[5pt] -(4 - 3x) &= 14 \\[5pt] -4 + 3x &= 14 \\[5pt] 3x &= 18 \\[5pt] \dfrac{3x}{3} &= \dfrac{18}{3}\\[5pt] x &= 6 \end{align}

Therefore, the solutions are $x=-\dfrac{10}3$ and $x=6.$

Either the positive or negative of the expression inside the absolute value must be equal to the right-hand side of the equation.

So, we set up the two corresponding equations and solve them independently:

• If $\left| \dfrac{1}{2}x + 3 \right| = \dfrac{1}{2}x + 3,$ we obtain: \begin{align*} \left| \dfrac{1}{2}x + 3 \right| & = 5\\[5pt] \dfrac{1}{2}x + 3 & = 5\\[5pt] \dfrac{1}{2}x & = 2 \end{align*} Multiplying both sides of the equation by $2$ gives \begin{align*} \require{cancel} \dfrac{1}{2}x & = 2\\[5pt] (2) \cdot \left(\dfrac{1}{2}x\right) & = (2) \cdot (2) \\[5pt] \left(\cancel{2}\right) \cdot \left(\cancel{\dfrac{1}{2}}x\right) & = 4\\[5pt] x & = 4. \end{align*}

• If $\left| \dfrac{1}{2}x + 3 \right| = -\left( \dfrac{1}{2}x + 3 \right),$ we obtain: \begin{align*} \left| \dfrac{1}{2}x + 3 \right| & = 5 \\[5pt] -\left( \dfrac{1}{2}x + 3 \right) & = 5 \\[5pt] -\dfrac{1}{2}x - 3 & = 5\\[5pt] -\dfrac{1}{2}x & = 8 \end{align*} Multiplying both sides of the equation by $-2$ gives \begin{align*} \require{cancel} -\dfrac{1}{2}x & = 8\\[5pt] (-2) \cdot \left(-\dfrac{1}{2}x\right) & = (-2) \cdot \left(8\right) \\[5pt] \left(\cancel{-2}\right) \cdot \left(\cancel{-\dfrac{1}{2}}x \right) & = -16\\[5pt] x & = -16. \end{align*}

Therefore, in ascending order, the solutions are $x = \bbox[4pt,border: 1px solid lightgray]{-16}$ and $x = \bbox[4pt,border: 1px solid lightgray]{4}.$

Either the positive or negative of the expression inside the absolute value must be equal to the right-hand side of the equation.

So, we set up the two corresponding equations and solve them independently:

• If $\left| 4-\dfrac32x \right| = 4-\dfrac32x,$ we obtain: \begin{align*} \require{cancel} \left| 4 - \dfrac32x \right| &= 2 \\[5pt] 4 - \dfrac32x &= 2 \\[5pt] -\dfrac32x &= -2 \end{align*} Multiplying both sides of the equation by $-\dfrac23$ gives \begin{align*} -\dfrac32x &= -2 \\[5pt] \left(-\dfrac23\right)\cdot\left(-\dfrac32x\right) &= \left(-\dfrac23\right)\cdot (-2) \\[5pt] \left(\cancel{-\dfrac23}\right)\cdot\left(\cancel{-\dfrac32}x\right) &=\dfrac43\\[5pt] x &=\dfrac43. \end{align*}

• If $\left| 4-\dfrac32x \right| = -\left(4 - \dfrac32x\right),$ we obtain: \begin{align*} \left| 4 - \dfrac32x \right| &= 2 \\[5pt] -\left( 4 - \dfrac32x \right) &= 2 \\[5pt] -4 + \dfrac32x &= 2 \\[5pt] \dfrac32x &= 6 \end{align*} Multiplying both sides of the equation by $\dfrac23$ gives \begin{align*} \dfrac32x &= 6 \\[5pt] \left(\dfrac23\right)\cdot \left(\dfrac32x\right) &= \left(\dfrac23\right) \cdot (6) \\[5pt] \left(\cancel{\dfrac23}\right)\cdot \left(\cancel{\dfrac32}x\right) &=\dfrac{12}{3} \\[5pt] x &= 4. \end{align*}

Therefore, the solutions are $x=\dfrac43$ and $x=4.$