Solve equations involving absolute values with calculations - High school Precalculus Exercises

Solve the following equations.

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

x=2 and x=\dfrac{2}{5}

x=-2 and x= \dfrac{5}{2}

x=2 and x= \dfrac{5}{2}

x=-2 and x = -\dfrac{2}{5}

We know that \left| 2x \right|=\left| 2-3x \right| if and only if:

\begin{cases} 2x=2-3x \cr \cr or \cr \cr 2x=-\left(2-3x\right) \end{cases}

Step 1

Solve first equation

2-3x = 2x

2 = 5x

x=\dfrac{2}{5}

Step 2

Solve second equation

2x=-\left(2-3x\right)

2x=-2+3x

x=2

The solution set is \left\{ 2,\dfrac25 \right\}.

\left| x-2 \right|=\left| 1-4x \right|

x=\dfrac13 and x=\dfrac35

x=\dfrac13 and x=-\dfrac35

x=-\dfrac13 and x=-\dfrac35

x=-\dfrac13 and x = \dfrac35

We know that \left| x-2 \right|=\left| 1-4x \right| if and only if:

\begin{cases} x-2=1-4x \cr \cr or \cr \cr x-2=-\left(1-4x\right) \end{cases}

Step 1

Solve first equation

x-2 = 1-4x

5x = 3

x=\dfrac{3}{5}

Step 2

Solve second equation

x-2=-\left(1-4x\right)

x-2=-1+4x

3x=-1

x=-\dfrac{1}{3}

The solution set is \left\{ -\dfrac{1}{3},\dfrac35 \right\}.

\left| 2x-1 \right|=4

x=-\dfrac32 and x= \dfrac{5}{2}

x=-\dfrac32 and x=-\dfrac52

x=\dfrac32 and x= -\dfrac{5}{2}

x=\dfrac32 and x=\dfrac{5}{2}

We know that \left| 2x-1 \right|= 4 if and only if:

\begin{cases} 2x-1=4 \cr \cr or \cr \cr 2x-1=-4 \end{cases}

Step 1

Solve first equation

2x-1 = 4

2x = 5

x=\dfrac{5}{2}

Step 2

Solve second equation

2x-1=-4

2x=-3

x=-\dfrac{3}{2}

The solution set is \left\{ \dfrac{5}{2},-\dfrac32 \right\}.

|x-1| = |x+1|

x=0

x=0 and x = -\dfrac12

x=0 and x=\dfrac{1}{2}

x=\dfrac{1}{2}

We know that |x-1| = |x+1| if and only if:

\begin{cases} x-1=x+1 \cr \cr or \cr \cr x-1=-\left(x+1\right) \end{cases}

Step 1

Solve first equation

x-1 =x+1

1=-1

This is a contradiction. The first equation has no solution.

Step 2

Solve second equation

x-1=-\left(x+1\right)

x-1=-x-1

2x=0

x=0

The solution is x=0.

|2x-1| = |4x-1|

x=\dfrac{1}{3}

x=0 and x=-\dfrac{1}{3}

x=0

x= 0 and x=\dfrac{1}{3}

We know that |2x-1| = |4x-1| if and only if:

\begin{cases} 2x-1=4x-1 \cr \cr or \cr \cr 2x-1=-\left(4x-1\right) \end{cases}

Step 1

Solve first equation

2x-1 =4x-1

2x = 4x

x=0

Step 2

Solve second equation

2x-1=-\left(4x-1\right)

2x-1=-4x+1

6x=2

x=\dfrac{1}{3}

The solution set is \left\{ 0,\dfrac13 \right\}.

|-2x+2| = -|x+4|

x=4

The equation has no solution.

x=1 and x= 4

x=1

The left side of the equation is greater than or equal to zero, while the right side is less than or equal to zero. Therefore, both sides must be zero, and both of the following equalities must simultaneously hold:

2x-2=0 \Rightarrow x=1
x-4=0 \Rightarrow x=4

Since we have obtained two different values for x, then we conclude that the equation has no solution.

The equation has no solution.

|2x-3|=-1

x=2

The equation has no solution.

x=0

x=0 and x = 2

The left side of the equation is greater than or equal to zero, while the right side is a negative number. Therefore this equation has no solution.

The equation has no solution.