Examples with solutions for Equations with Absolute Values: Absolute value on both sides of the equation

Exercise #1

∣x+2∣=∣x−2∣ |x+2|=|x-2|

Video Solution

Step-by-Step Solution

To solve the equation ∣x+2∣=∣x−2∣ |x+2|=|x-2| , we begin by considering the properties of absolute values.

The statement ∣A∣=∣B∣ |A| = |B| implies two cases:

  • Case 1: A=B A = B
  • Case 2: A=−B A = -B

For our problem, consider:

  • Case 1: x+2=x−2 x + 2 = x - 2
  • Case 2: x+2=−(x−2) x + 2 = -(x - 2)

Let's solve each case:

  • Case 1: x+2=x−2 x + 2 = x - 2
    Subtract x x from both sides: x+2−x=x−2−x x + 2 - x = x - 2 - x Reduce to: 2=−2 2 = -2 Since 2≠−2 2 \neq -2 , this case has no solution.
  • Case 2: x+2=−(x−2) x + 2 = -(x - 2)
    Expand the right-hand side: x+2=−x+2 x + 2 = -x + 2 Add x x to both sides: x+x+2=2 x + x + 2 = 2 This simplifies to: 2x+2=2 2x + 2 = 2 Subtract 2 from both sides: 2x=0 2x = 0 Solve for x x : x=0 x = 0

Thus, the solution to ∣x+2∣=∣x−2∣ |x+2|=|x-2| is x=0 x = 0 . The correct answer is the choice: x=0 x = 0 .

Answer

x=0 x=0

Exercise #2

∣x−1∣=∣2x+3∣ |x-1|=|2x+3|

Video Solution

Step-by-Step Solution

To solve the equation ∣x−1∣=∣2x+3∣ |x-1| = |2x+3| , follow these steps:

  • First, recall the property: for any real numbers a a and b b , ∣a∣=∣b∣ |a| = |b| implies a=b a = b or a=−b a = -b .
  • We will consider two cases based on this property:

Case 1: Assume x−1=2x+3 x-1 = 2x+3 .
Simplify the equation:
x−1=2x+3 x-1 = 2x+3
Subtract x x from both sides:
−1=x+3 -1 = x+3
Subtract 3 from both sides:
x=−4 x = -4 .

Case 2: Assume x−1=−(2x+3) x-1 = -(2x+3) .
Simplify the equation:
x−1=−2x−3 x-1 = -2x-3
Add 2x 2x to both sides:
3x−1=−3 3x-1 = -3
Add 1 to both sides:
3x=−2 3x = -2
Divide everything by 3:
x=−23 x = -\frac{2}{3} .

Therefore, the solutions to the equation ∣x−1∣=∣2x+3∣ |x-1| = |2x+3| are x=−4 x = -4 and x=−23 x = -\frac{2}{3} .

These solutions correspond to answer choice 4: x=−4 x = -4 , x=−23 x = -\frac{2}{3} .

Thus, x=−4 x = -4 and x=−23 x = -\frac{2}{3} .

Answer

x=−4 x=-4 , x=−23 x=-\frac{2}{3}

Exercise #3

∣−x+6∣=∣3x−2∣ |-x+6|=|3x-2|

Video Solution

Step-by-Step Solution

To solve the problem, follow these steps:

  • Case 1: Set −x+6=3x−2-x + 6 = 3x - 2.
  • Simplify and solve for xx:

−x+6=3x−2-x + 6 = 3x - 2
Add xx to both sides: 6=4x−26 = 4x - 2
Add 2 to both sides: 8=4x8 = 4x
Divide both sides by 4: x=2x = 2

  • Case 2: Set −x+6=−(3x−2)-x + 6 = -(3x - 2), which is −x+6=−3x+2-x + 6 = -3x + 2.
  • Simplify and solve for xx:

−x+6=−3x+2-x + 6 = -3x + 2
Add 3x3x to both sides: 2x+6=22x + 6 = 2
Subtract 6 from both sides: 2x=−42x = -4
Divide both sides by 2: x=−2x = -2

Finally, verify that both solutions satisfy the original absolute value equation:

  • When x=2x = 2: ∣−2+6∣=∣6−2∣⇒∣4∣=∣4∣|-2 + 6| = |6 - 2| \Rightarrow |4| = |4|, which holds true.
  • When x=−2x = -2: ∣2+6∣=∣−6−2∣⇒∣8∣=∣8∣|2 + 6| = |-6 - 2| \Rightarrow |8| = |8|, which holds true.

Thus, both x=2x = 2 and x=−2x = -2 are valid solutions to the equation.

The solutions to the problem are x=−2\boldsymbol{x=-2} and x=2\boldsymbol{x=2}.

Therefore, the correct answer choice is x=−2\boldsymbol{x=-2} , x=2\boldsymbol{x=2}.

Answer

x=−2 x=-2 , x=2 x=2

Exercise #4

∣x+3∣=∣2x+6∣ |x+3|=|2x+6|

Video Solution

Step-by-Step Solution

To solve the equation ∣x+3∣=∣2x+6∣ |x+3| = |2x+6| , we need to consider the properties of absolute values and analyze the cases for different ranges of x x .

The equation ∣x+3∣=∣2x+6∣ |x+3| = |2x+6| implies two possibilities based on absolute value properties:

  • x+3=2x+6 x+3 = 2x+6
  • x+3=−(2x+6) x+3 = -(2x+6)

Let's solve each case:

Case 1: x+3=2x+6 x+3 = 2x+6

Simplify the equation:
x+3=2x+6 x + 3 = 2x + 6
Move x x to the other side:
3=x+6 3 = x + 6
Subtract 6 from both sides:
x=−3 x = -3

Case 2: x+3=−(2x+6) x+3 = -(2x+6)

Simplify the equation:
x+3=−2x−6 x+3 = -2x-6
Add 2x 2x to both sides:
3x+3=−6 3x + 3 = -6
Subtract 3 from both sides:
3x=−9 3x = -9
Divide both sides by 3:
x=−3 x = -3

In both cases, we find that x=−3 x = -3 . However, we need to verify if x=−3 x = -3 satisfies the original equation:

Substitute x=−3 x = -3 into the original equation:
∣x+3∣=∣2x+6∣ |x+3| = |2x+6|
∣−3+3∣=∣2(−3)+6∣ |-3+3| = |2(-3)+6|
∣0∣=∣−6+6∣ |0| = |-6+6|
0=0 0 = 0

Therefore, x=−3 x = -3 satisfies the equation. The solution to the problem is x=−3 x = -3 .

In conclusion, the correct answer is x=−3 x = -3 .

Answer

x=−3 x=-3

Exercise #5

∣x+4∣=∣2x+20∣ |x+4|=|2x+20|

Video Solution

Step-by-Step Solution

To solve the given absolute value equation ∣x+4∣=∣2x+20∣ |x+4| = |2x+20| , we consider the properties of absolute values:

  • If ∣A∣=∣B∣ |A| = |B| , then A=B A = B or A=−B A = -B .

Applying this, we consider two cases:

  • Case 1: x+4=2x+20 x+4 = 2x+20

  • Case 2: x+4=−(2x+20) x+4 = - (2x+20)

Let's solve each case:

Case 1:

x+4=2x+20 x+4 = 2x+20

Rearrange the equation:
x+4=2x+20  ⟹  x−2x=20−4 x+4 = 2x+20 \implies x - 2x = 20 - 4
−x=16-x = 16
So, x=−16 x = -16 .

Case 2:

x+4=−(2x+20) x+4 = - (2x+20)

Distribute the negative sign:
x+4=−2x−20 x+4 = -2x-20

Rearrange the equation:
x+2x=−20−4 x + 2x = -20 - 4
3x=−24 3x = -24
x=−8 x = -8 .

Therefore, the solutions to the equation are x=−8 x = -8 and x=−16 x = -16 .

To verify, plug back x=−8 x = -8 and x=−16 x = -16 into the original equation:

  • For x=−16 x = -16 : ∣x+4∣=∣−16+4∣=12 |x+4| = |-16 + 4| = 12 and ∣2x+20∣=∣2(−16)+20∣=12 |2x+20| = |2(-16) + 20| = 12 . Both sides equal.

  • For x=−8 x = -8 : ∣x+4∣=∣−8+4∣=4 |x+4| = |-8 + 4| = 4 and ∣2x+20∣=∣2(−8)+20∣=4 |2x+20| = |2(-8) + 20| = 4 . Both sides equal.

Both solutions satisfy the original equation. Therefore, the correct answer is x=−16 x = -16 and x=−8 x = -8 .

Thus, the solution to the absolute value equation is:

The solutions are x=−8\mathbf{x = -8} and x=−16\mathbf{x = -16}.

Answer

x=−8 x=-8 , x=−16 x=-16

Exercise #6

∣2x−6∣=∣−x+4∣ |2x-6|=|-x+4|

Video Solution

Answer

x=313 x=3\frac{1}{3} , x=2 x=2