Problem 102 - Intermediate Algebra

Question

Solve. $$ |6-4 x|=|x+2| $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{4}{5}\) and \(x = \frac{8}{3}\).

Step 1: Understand the Absolute Value Expression

The problem features two absolute value expressions: \(|6 - 4x|\) and \(|x + 2|\). The goal is to find \(x\) values that make these expressions equal. This means \(6 - 4x\) could equal \(x + 2\) or the negative of \(x + 2\), which results in two separate equations.

Step 2: Set up the Positive Equation

First, set the expression inside the absolute value signs equal to one another without considering any negative signs: \(6 - 4x = x + 2\). Solve this equation for \(x\).

Step 3: Solve the Positive Equation

Rearrange the equation \(6 - 4x = x + 2\) to combine like terms. Subtract \(x\) from both sides to get \(6 - 4x - x = 2\). Simplify to get \(6 - 5x = 2\). Then, subtract 6 from both sides, yielding \(-5x = -4\). Solve for \(x\) by dividing both sides by -5 to get \(x = \frac{4}{5}\).

Step 4: Set up the Negative Equation

Next, set the expression inside \(|6 - 4x|\) equal to the negative of the expression in \(|x + 2|\): \(6 - 4x = -(x + 2)\). Solve this equation for \(x\).

Step 5: Solve the Negative Equation

First, distribute the negative sign in \(6 - 4x = -x - 2\) to get \(6 - 4x = -x - 2\). Add \(x\) to both sides to obtain \(6 - 3x = -2\). Now subtract 6 from both sides of the equation, resulting in \(-3x = -8\). Divide both sides by -3 to solve for \(x\), getting \(x = \frac{8}{3}\).

Step 6: Verify Solutions

To ensure both solutions are correct, substitute \(x = \frac{4}{5}\) and \(x = \frac{8}{3}\) back into the original equation \(|6 - 4x| = |x + 2|\) and verify both sides are equal. Both solutions should satisfy the original absolute equation.

Key Concepts

Intermediate Algebra

Intermediate algebra serves as the bridge between basic arithmetic and advanced algebraic concepts. When working with absolute value equations, like the one given in the problem

• |6 - 4x| = |x + 2|

we use skills from intermediate algebra to analyze and solve the problem. Absolute value equations often involve understanding expressions that can take either positive or negative values.
For instance, the absolute value of a number refers to its distance from zero on the number line, regardless of direction. This means you are dealing with two possible scenarios for each term inside the absolute value. By comprehending these underlying concepts, students build a solid foundation to approach more complex algebraic challenges.

Solving Equations

Solving absolute value equations requires a systematic approach to handle the possible scenarios implied by the absolute value. With the equation

• |6 - 4x|=|x + 2|

we need to address the equations in two distinct forms because absolute values create two possibilities for each expression.
First, we consider the positive equation:

• 6 - 4x = x + 2

Solving this involves aligning like terms by isolating variables on one side and constants on the other. By solving, we find

• x = \(\frac{4}{5}\)

Next, we move to the negative scenario where

• 6 - 4x = -(x + 2)

This requires distributing the negative sign and simplifying the equation.
Eventually, we find the other solution is

• x = \(\frac{8}{3}\)

By treating absolute values as indicators of these dual pathways, we masterfully navigate through problem-solving techniques fundamental to algebra.

Verification of Solutions

Verification of solutions in absolute value equations is a critical step to ensure accuracy. Once potential solutions are found through solving, we must confirm they actually satisfy the original equation.
In our problem, the solutions obtained were

• x = \(\frac{4}{5}\)
• x = \(\frac{8}{3}\)

To verify, substitute these values back into the initial equation

• |6 - 4x| = |x + 2|

Start with

• x = \(\frac{4}{5}\)

Calculate both sides to ensure the left side equals the right side. If they do, it's an indicator that the solution holds true. Repeat the process with

• x = \(\frac{8}{3}\)

This step provides confidence that the equations and calculations were performed correctly and that all solutions are valid.
Verification reinforces understanding and highlights the importance of accuracy in mathematical procedures.