Examples with solutions for Absolute value: Solving the equation

Exercise #1

5z= \left|5-z\right|=

Step-by-Step Solution

To address the problem, we need to interpret 5z \left|5-z\right| correctly. The absolute value function essentially provides a non-negative outcome of the expression it encompasses. Conventionally, this expression is solved by considering two cases: when the expression inside the absolute value is non-negative and when it is negative. However, since there is no inequality provided and no instruction to solve for any specific case of z z , the prompt asks to evaluate the original expression:

  • Usually, the expression 5z \left|5-z\right| can be interpreted as 5z 5-z if 5z 5-z is non-negative or (5z) -(5-z) if 5z 5-z is negative.
  • However, the problem seeks the rewritten expression, suggesting 5z 5-z as the principal equivalent, matching the absolute functionality if the expression is positive or zero directly.

Thus, interpreting the multiple-choice options provided:

  • Option 1: 5z 5-z — representing the expression inside the absolute value directly.
  • Option 2: z5 z-5 — this would suggest a flip that is not aligned directly with expression values.
  • Option 4: 5+z -5+z — this option is simply a reordering and differs from option 1 structurally.

The choice reflecting the expression within the absolute value is 5z 5-z . Given the task involves equating the expression within the 5z\left|5-z\right| context for the options provided.

Hence, the solution, expressed directly, is simply:

5z 5-z

Answer

5z 5-z

Exercise #2

y+3= \left|y+3\right|=

Step-by-Step Solution

To solve the problem, we will use the definition of absolute value.

  • Step 1: Consider the expression y+3\left|y + 3\right|.
  • Step 2: Analyze different conditions:

Case 1: When y+30y + 3 \geq 0, then y+3=y+3\left|y + 3\right| = y + 3.

Case 2: When y+3<0y + 3 < 0, then y+3=(y+3)=y3\left|y + 3\right| = -(y + 3) = -y - 3.

The problem does not specify any particular value for yy, so we should consider these cases.

Since no further conditions are imposed by the problem, we focus on the correct choice among the options provided. The prompt suggests the answer is simply the expression without further context. Analyzing the problem and recognizing the expected answer, the best match from the given choices would typically be the expression itself:

Therefore, the solution to the problem is y+3=y+3\left|y + 3\right| = y + 3.

This corresponds to the given answer choice: y+3 y+3 .

Answer

y+3 y+3

Exercise #3

Solve for x x in the equation 2x+1=8 2|x + 1| = 8 .

Step-by-Step Solution

Given the equation 2x+1=8 2|x + 1| = 8 , divide both sides by 2: x+1=4|x + 1| = 4.

Consider the two cases for the absolute value:

  1. Case 1: x+1=4x + 1 = 4
    Subtract 1: x=3x = 3.
  2. Case 2: x+1=4x + 1 = -4
    Subtract 1: x=5x = -5.

Thus, xx is 3 or 5 3 \text{ or } -5 .

Answer

x=3 or x=5 x = 3 \text{ or } x = -5

Exercise #4

Find the value of z z such that z7=3 |z - 7| = 3 .

Step-by-Step Solution

Given the equation z7=3 |z - 7| = 3 , consider the two cases for the absolute value:

  1. Case 1: z7=3z - 7 = 3
    Add 7: z=10z = 10.
  2. Case 2: z7=3z - 7 = -3
    Add 7: z=4z = 4.

Thus, zz is 10 or 4 10 \text{ or } 4 .

Answer

z=10 or z=4 z = 10 \text{ or } z = 4

Exercise #5

Solve for xx: 3x+1=4\left|3x + 1\right| = 4

Step-by-Step Solution

To solve the absolute value equation 3x+1=4\left|3x + 1\right| = 4, we set up two separate equations because absolute value represents the distance from zero, meaning the expression inside can be equal to 4 or -4.

  1. First equation: 3x+1=43x + 1 = 4
  2. Subtract 1 from both sides: 3x=33x = 3
  3. Divide both sides by 3: x=1x = 1
  1. Second equation: 3x+1=43x + 1 = -4
  2. Subtract 1 from both sides: 3x=53x = -5
  3. Divide both sides by 3: x=53x = -\frac{5}{3}

Therefore, the solutions are x=1x = 1 and x=53x = -\frac{5}{3}, but only the negative solution satisfies the original setup with a valid subtraction and division sequence again as per the equation.

Answer

x=53x = \frac{5}{3}