Solve an absolute value equation with operations Algebra I

Solve the following equations using operations.

\left| 2-x \right|=4

S=\left \{-4;-2\right \}

S=\left \{-6;-2\right \}

S=\left \{-2;6\right \}

S=\left \{-2\right \}

\left | 3-2x \right | = 7

S = \{-1, 4 \}

S = \{2, -5 \}

S = \{-2, 5 \}

S = \{1, -4 \}

\left | 3x-2 \right | = 13

S = \left\{\dfrac {-11} {3}, -5 \right\}

S = \left\{\dfrac {-10} {3}, 4 \right\}

S=\left\{\dfrac{10}{3}, 5\right\}

S=\left\{\dfrac{-11}{3}, 5\right\}

\left | 4x-2 \right | = 10

S = \{- 3{,}5 \}

S = \{3{,}5\}

S=\{-2,-3\}

S=\{-2{,}3\}

\left | 1-2x \right | = 5

S=\{-2{,}3\}

S = \{- 2{,}4 \}

S = \{1.3 \}

S = \{- 1{,}4 \}

\left | 2-3x \right | = -4

S = \{2, 5 \}

S=\{3, 4\}

No solution S=\varnothing

S = \{1, 5 \}

\left | 2-3x \right | =4

S = \left\{\dfrac {-2} {3}, -2\right \}

S=\left\{\dfrac{-2}{3}, 2\right\}

S = \left\{\dfrac {2} {3}, -2\right \}

S = \left\{\dfrac {2} {3}, -3\right \}