Solve an absolute value equation with operations - High school Algebra I Exercises

Solve the following equations using operations.

\left| 2-x \right|=4

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

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

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

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

The following equation:

|2-x|=4

is equivalent to:

\begin{cases} 2-x=4 \cr \cr or \cr \cr 2-x=-4 \end{cases}

Step 1

Solve the first equation

2-x = 4

First, subtract 2 from both sides:

2-x-2 = 4-2

-x = 2

Then, multiply both sides by -1:

x=-2

Step 2

Solve the second equation

2-x = -4

First, subtract 2 from both sides:

2-x-2 = -4-2

-x=-6

Then multiply both sides by -1:

x = 6

The solutions are -2 and 6.

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

S = \{-1, 4 \}

S = \{1, -4 \}

S = \{-2, 5 \}

S = \{2, -5 \}

The following equation:

| 3-2x | = 7

is equivalent to:

\begin{cases} 3-2x=7 \cr \cr or \cr \cr 3-2x=-7 \end{cases}

Step 1

Solve the first equation

3-2x=7

First, subtract 3 from both sides:

3-2x-3 = 7-3

-2x = 4

Then, divide both sides by -2:

x=-2

Step 2

Solve the second equation

3-2x = -7

First, subtract 3 from both sides:

3-2x-3 = -7-3

-2x =-10

Then, divide both sides by -2:

x =5

The solutions are -2 and 5.

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

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

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

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

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

The following equation:

|3x-2|=13

is equivalent to:

\begin {cases} 3x-2=13 \cr \cr or \cr \cr 3x-2=-13 \end {cases}

Step 1

Solve the first equation

3x-2=13

First, add 2 to both sides:

3x-2+2 = 13+2

3x=15

Then, divide both sides by 3:

x=5

Step 2

Solve the second equation

3x-2 = -13

First, add 2 to both sides:

3x-2+2 = -13+2

3x =-11

Then, divide both sides by 3:

x=\dfrac{-11}{3}

The solutions are 5 and \dfrac{-11}{3}.

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

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

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

S=\{-2,-3\}

S = \{3{,}5\}

The following equation:

|4x-2 | = 10

is equivalent to:

\begin {cases}4x-2=10 \cr \cr or \cr \cr 4x-2=-10 \end {cases}

Step 1

Solve the first equation

4x-2=10

First, add 2 to both sides:

4x-2+2=10+2

4x=12

Then, divide both sides by 4:

x=3

Step 2

Solve the second equation

4x-2 = -10

First, add 2 to both sides:

4x-2 + 2 = -10 + 2

4x=-8

Then, divide both sides by 4:

x =-2

The solutions are -2, and 3.

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

S = \{1.3 \}

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

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

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

The following equation:

| 1-2x | = 5

is equivalent to:

\begin {cases} 1-2x = 5 \cr \cr or \cr \cr 1-2x= -5 \end {cases}

Step 1

Solve the first equation

1-2x=5

First, subtract 1 from both sides:

1-2x-1 = 5-1

-2x = 4

Then, divide both sides by -2:

x=-2

Step 2

Solve the second equation

1-2x = -5

First, subtract 1 from both sides:

1-2x-1 = -5-1

-2x = -6

Then, divide both sides by -2:

x = 3

The solutions are -2, and 3.

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

S=\{3, 4\}

S = \{1, 5 \}

No solution S=\varnothing

S = \{2, 5 \}

The absolute value is always positive. Thus \left | 2-3x \right | cannot be negative. Therefore, \left | 2-3x \right | = -4 cannot have a solution.

There is no solution.

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

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

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

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

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

The following equation:

| 2-3x | = 4

is equivalent to:

\begin {cases} 2-3x = 4 \cr \cr or \cr \cr 2-3x = -4 \end {cases}

Step 1

Solve the first equality

2-3x = 4

First, subtract 2 from both sides:

2-3x-2 = 4-2

-3x = 2

Then, divide both sides by -3:

x = \dfrac{-2}{3}

Step 2

Solve the second equality

2-3x = -4

First, subtract 2 from both sides:

2-3x-2 = -4-2

-3x = -6

Then, divide both sides by -3:

x = 2

The solutions are \dfrac{-2}{3} and 2.