∣5−z∣=
\( \left|5-z\right|= \)
\( \left|y+3\right|= \)
Solve for \( x \) in the equation \( 2|x + 1| = 8 \).
Find the value of \( z \) such that \( |z - 7| = 3 \).
Solve for \(x\): \(\left|3x + 1\right| = 4\)
To address the problem, we need to interpret 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 , the prompt asks to evaluate the original expression:
Thus, interpreting the multiple-choice options provided:
The choice reflecting the expression within the absolute value is . Given the task involves equating the expression within the context for the options provided.
Hence, the solution, expressed directly, is simply:
To solve the problem, we will use the definition of absolute value.
Case 1: When , then .
Case 2: When , then .
The problem does not specify any particular value for , 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 .
This corresponds to the given answer choice: .
Solve for in the equation .
Given the equation , divide both sides by 2: .
Consider the two cases for the absolute value:
Thus, is .
Find the value of such that .
Given the equation , consider the two cases for the absolute value:
Thus, is .
Solve for :
To solve the absolute value equation , 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.
Therefore, the solutions are and , but only the negative solution satisfies the original setup with a valid subtraction and division sequence again as per the equation.