Problem 51 - Elementary and Intermediate Algebra

Question

Solve each equation. \(|2-x|=|3 x+2|\)

Step-by-Step Solution

Answer

The solutions are \(x = 0\) and \(x = -2\).

Step 1: Understanding the Absolute Values

Given the equation \(|2-x|=|3x+2|\), we need to understand that the absolute value \(|a|\) represents the distance of \(a\) from zero on the number line, so \(|a| = a\) if \(a \geq 0\), and \(|a| = -a\) if \(a < 0\). Thus, our original equation is stating two expressions with uncertainties of direction (positive or negative).

Step 2: Setting Up Two Cases

When solving an absolute value equation, we must consider all potential sign combinations. This gives us two primary cases to solve. Case 1: \(2-x = 3x+2\) and Case 2: \(2-x = -(3x+2)\).

Step 3: Solving Case 1

For Case 1, solve the equation \(2-x = 3x+2\):\[\begin{align*}2 - x &= 3x + 2 \2 - 2 &= 3x + x \0 &= 4x \x &= 0.\end{align*}\]So, the solution for this case is \(x = 0\).

Step 4: Solving Case 2

For Case 2, solve the equation \(2-x = -(3x+2)\):\[\begin{align*}2 - x &= -3x - 2 \2 + 2 &= -3x + x \4 &= -2x \x &= -2.\end{align*}\]So, the solution for this case is \(x = -2\).

Step 5: Verifying Solutions

We need to check both solutions in the original equation. For \(x = 0\), the equation becomes \(|2-0| = |3(0)+2|\) or \(|2| = |2|\), which is true. For \(x = -2\), the equation becomes \(|2-(-2)| = |3(-2)+2|\) or \(|4| = |-4|\), which is also true. Hence, both solutions are valid.

Key Concepts

Understanding the Absolute Value Property

Absolute value is a critical concept in algebra, especially when solving equations involving it. The absolute value of a number is its distance from zero on the number line, without any regard to direction. This means that for any given number, either positive or negative, its absolute value is always non-negative.
The mathematical representation of the absolute value is \(|a|\), which equals to \(a\) if \(a \geq 0\), and becomes \(-a\) if \(a < 0\).
This property implies that the equation \(|2-x| = |3x+2|\) expresses two conditions that |2-x| could either equal \(3x+2\) or its negative counterpart. Understanding this property helps to set the grounds for solving absolute value equations effectively.

• The absolute value reflects the length, not the direction.
• It ensures that \(|a| \geq 0\) for any real number.

Employing the Case Problem-Solving Method

When dealing with absolute value equations, we use the case problem-solving method. This method involves considering different scenarios based on the nature of absolute values. Because the expression inside the absolute values could either be positive or negative, we have two possible cases to evaluate.
In the given equation \(|2-x| = |3x+2|\), we split the equation into two scenarios based on the absolute value properties:

• Case 1: Treat \(2-x = 3x+2\) as equation without absolute value.
• Case 2: Consider \(2-x = -(3x+2)\), which flips the sign of the second absolute value expression.

Solving both cases is crucial since solutions can exist in either case. Thus, the method ensures that no potential solution is overlooked. This consideration helps address the uncertainty of signs while solving the equation.
By exploring both sign possibilities, we systematically address all potential solutions.

Ensuring the Validity of Solutions in Algebra

Once you have found solutions by solving each case separately, the next step is to verify their correctness. Verifying solutions in algebra involves checking if they satisfy the original equation. This is a critical step to ensure their validity.
For the solutions \(x = 0\) and \(x = -2\), substitute them back into the original equation \(|2-x| = |3x+2|\):

• For \(x = 0\), plug it into the equation to get \(|2-0| = |3(0)+2|\), simplifying to \(|2| = |2|\), which holds true.
• For \(x = -2\), substitute to obtain \(|2-(-2)| = |3(-2)+2|\), or \(|4| = |-4|\), both are equal in magnitude.

Both values satisfy the equation, making them valid solutions. This verification task is essential in algebra to ensure that we have accurately solved the equation and that no errors were made during calculations.
Remember, solutions are only valid if they meet the original statements of the equation.