Problem 21 - Algebra and Trigonometry

Question

\(5-22=\) Solve the equation. $$ |x-1|=|3 x+2| $$

Step-by-Step Solution

Answer

The solutions are \(x = -\frac{3}{2}\) and \(x = -\frac{1}{4}\).

Step 1: Understand Absolute Value Equation

The equation \(|x-1| = |3x+2|\) contains absolute value expressions. Absolute value equations equalize the distance from zero on the number line, meaning we're dealing with two scenarios: when the expressions inside the absolute signs are both positive, or when one is negative.

Step 2: Case 1 - Both Expressions are Positive or Negative

For the first case, assuming both expressions are either positive or both negative, the equation simplifies to:\[x - 1 = 3x + 2\] Set the contents of the absolute values equal to each other.

Step 3: Solve Case 1

Rearrange the equation from Case 1:\[x - 1 = 3x + 2\]Subtract \(x\) from both sides: \[-1 = 2x + 2\]Subtract 2 from both sides:\[-3 = 2x\]Divide both sides by 2:\[x = -\frac{3}{2}\]

Step 4: Case 2 - One Expression is the Negative of the Other

In the second scenario, one expression is negative of the other:\[x - 1 = -(3x + 2)\] This represents one being the inverse of the other.

Step 5: Solve Case 2

Rewriting the equation for clarity:\[x - 1 = -3x - 2\]Add \(3x\) to both sides:\[4x - 1 = -2\]Add 1 to both sides:\[4x = -1\]Divide both sides by 4:\[x = -\frac{1}{4}\]

Step 6: Verify Solutions

Substitute \(x = -\frac{3}{2}\) into the original equation to verify it works: - The absolute values yield \(|-5/2| = |-5/2|\), which is correct.Similarly, substitute \(x = -\frac{1}{4}\) into the original equation to check: - The absolute values yield \(|-5/4| = |-5/4|\), which is accurate. Both solutions satisfy the original equation.

Key Concepts

Solving Equations

When we're dealing with equations, especially those involving absolute values, it's crucial to understand the underlying structure. An equation tells us that two expressions are equivalent under certain conditions. To solve an equation means to find all possible values of the variable that satisfy this condition. In the case of absolute value equations like \(|x-1| = |3x+2|\), we explore two main scenarios because absolute values measure distance from zero. One scenario occurs when both expressions inside the absolute values are equal, either both positive or both negative. The other scenario occurs when one expression is the negative of the other. To solve these, we break down the problem into these scenarios, solve each one separately, and then check which solutions are valid.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and arithmetic operations. In our exercise, the expressions \(x-1\) and \(3x+2\) inside the absolute value signs are algebraic expressions. These expressions can represent many different numbers depending on the value of the variable \(x\). To simplify and solve the equations arising from the absolute value, we treat each algebraic expression separately, considering its possible positive or negative nature depending on the value of \(x\).

• When \(x-1 = 3x+2\), we're dealing with an algebraic equation that demands rearranging like terms onto one side.
• Similarly, for \(x-1 = -(3x+2)\), the challenge lies in handling the negative signs effectively while rearranging.

Exploring these cases helps us understand how algebraic expressions interact within an equation.

Verification of Solutions

Verification of solutions is a critical step in solving equations, ensuring that the obtained values satisfy the original equation. After obtaining solutions for \(x\), such as \(x = -\frac{3}{2}\) and \(x = -\frac{1}{4}\), it's important to substitute these values back into the original equation \(|x-1| = |3x+2|\) to ensure they are correct.

• For \(x = -\frac{3}{2}\), substituting into the original equation gives us \(|-\frac{5}{2}| = |-\frac{5}{2}|\), verifying that it works.
• For \(x = -\frac{1}{4}\), checking yields \(|-\frac{5}{4}| = |-\frac{5}{4}|\), confirming its accuracy.

Verification is crucial because it guarantees that the solutions not only stem from the algebraic manipulations but also satisfy the terms dictated by absolute values in the original setup.