Problem 64 - Algebra 2

Question

Solve each equation. Check your solutions. \(|x+1|=x\)

Step-by-Step Solution

Answer

No solutions.

Step 1: Set Up the Equations from the Absolute Value

The given equation is \(|x+1| = x\). Since it is an absolute value equation, we consider two cases.**Case 1:** If \(x+1 \geq 0\), then \(|x+1| = x+1\).**Case 2:** If \(x+1 < 0\), then \(|x+1| = -(x+1) = -x - 1\).

Step 2: Solve Case 1 where \(x+1 \geq 0\)

For Case 1, we have:\[x+1 = x\]Subtract \(x\) from both sides:\[1 = 0\]This is a contradiction, implying there are no solutions in this case.

Step 3: Solve Case 2 where \(x+1 < 0\)

For Case 2, we have:\[-x-1 = x\]Add \(x\) to both sides:\[-1 = 2x\]Divide both sides by 2:\[x = -\frac{1}{2}\]Check that this satisfies the condition of Case 2:\(-\frac{1}{2}+1 < 0\) is false, so no solutions apply for Case 2.

Step 4: Verify Conditions for Solutions

In both cases, no valid solution satisfies both the algebraic equation and the condition derived from the absolute value. Therefore, the equation has no solutions.

Key Concepts

Understanding Algebraic Equations with Absolute Values

Algebraic equations are mathematical statements that show the equality of two expressions. In absolute value equations, like \(|x+1| = x\), we need to consider that the absolute value translates a number into its non-negative form.
The absolute value equation splits into two scenarios or cases, due to this nature:

• Case 1: The expression inside the absolute value is zero or positive.
• Case 2: The expression is negative.

To tackle the problem \(|x+1| = x\), we set up the scenarios as:

• Case 1: If \(x+1 \geq 0\), then \(|x+1| = x+1\)
• Case 2: If \(x+1 < 0\), then \(|x+1| = -(x+1) = -x-1\)

Each must be solved like a regular algebraic equation. This way, the solution can be verified against the original expressions' conditions.

Significance of Solution Verification

Verification of a solution is an integral part of solving algebraic equations, especially those involving absolute values. It ensures that the calculated results are consistent with the original equation.
Once we obtain values for \(x\) in each case, we check them to reinforce three essential checks:

• Does the solution actually solve the original equation?
• Does it meet the initial assumptions or conditions of the case?
• Do these solutions fit within the problem's logical structure?

For \(|x+1| = x\), verification showed:

• In Case 1, \(1 = 0\) is a contradiction, so no solution is valid.
• In Case 2, \(x = -\frac{1}{2}\) fails as it contradicts the case condition \(x+1 < 0\).

Verification is crucial as it helps identify mistakes or contradictions within solutions.

Recognizing Contradictions in Equations

Contradictions occur in algebraic equations when the mathematical results contradict the established conditions or mathematical truths. When solving \(|x+1| = x\), Case 1 produced \(1 = 0\), an evident contradiction.
This signifies:

• The equation has no solutions satisfying \(x+1 \geq 0\) turning into \(|x+1| = x+1\).

Case 2 resulted in \(x = -\frac{1}{2}\), which seemingly solves \(-x-1 = x\), yet it does not satisfy \(x+1 < 0\). Here lies another contradiction as the solution cannot be applied.
Recognizing contradictions:

• Helps determine if an equation has no solutions.
• Assures the consistency of logic in mathematical argumentation.

Spotting contradictions early prevents erroneous conclusions and sharpens problem-solving skills.