Problem 26
Question
Solve each equation. \(\left|\frac{x}{7}\right|=2\)
Step-by-Step Solution
Verified Answer
The solutions are \( x = 14 \) and \( x = -14 \).
1Step 1: Understand the Absolute Value Equation
The absolute value equation given is \( \left| \frac{x}{7} \right| = 2 \). This means that the expression inside the absolute value, \( \frac{x}{7} \), can be either 2 or -2 for the equation to hold true.
2Step 2: Set Up the Two Possible Equations
We need to solve for \( x \) in both scenarios: \( \frac{x}{7} = 2 \) and \( \frac{x}{7} = -2 \).
3Step 3: Solve the First Equation
For \( \frac{x}{7} = 2 \), multiply both sides by 7 to isolate \( x \): \[ x = 2 \times 7 = 14 \].
4Step 4: Solve the Second Equation
For \( \frac{x}{7} = -2 \), multiply both sides by 7 to isolate \( x \): \[ x = -2 \times 7 = -14 \].
5Step 5: Conclusion
The solutions to the equation \( \left| \frac{x}{7} \right| = 2 \) are \( x = 14 \) and \( x = -14 \).
Key Concepts
Solving EquationsAlgebraic ExpressionsStep-by-Step Solutions
Solving Equations
Solving equations is a fundamental skill in algebra. It involves finding the value of a variable that makes an equation true. When dealing with absolute value equations like \( |\frac{x}{7}| = 2 \), we need to consider what the absolute value function represents. Absolute value measures the distance a number is from zero on the number line, regardless of direction. Therefore, both positive and negative values can emerge as solutions.
To solve \( |\frac{x}{7}| = 2 \), we express the equation as two distinct cases. The first case is straightforward, where the expression inside the absolute value \( \frac{x}{7} \) equals 2. The second case occurs when \( \frac{x}{7} \) equals -2. This bifurcation is necessary because of the nature of absolute values. Each case is then individually solved, leading to two possible solutions.
To solve \( |\frac{x}{7}| = 2 \), we express the equation as two distinct cases. The first case is straightforward, where the expression inside the absolute value \( \frac{x}{7} \) equals 2. The second case occurs when \( \frac{x}{7} \) equals -2. This bifurcation is necessary because of the nature of absolute values. Each case is then individually solved, leading to two possible solutions.
Algebraic Expressions
An algebraic expression involves numbers, variables, and arithmetic operations. In our exercise, the algebraic expression is \( \frac{x}{7} \). To work with an expression wrapped in absolute value, we must distill it into simpler forms.
Algebraic expressions allow us to codify mathematical statements and solve for unknowns. Simplifying expressions involves rearranging and manipulating terms to isolate the variable, like \( x \). For an absolute value equation, breaking it into possible distinct algebraic expressions (\( \frac{x}{7} = 2 \) and \( \frac{x}{7} = -2 \)) helps in managing the inherent duality of absolute values.
Algebraic expressions allow us to codify mathematical statements and solve for unknowns. Simplifying expressions involves rearranging and manipulating terms to isolate the variable, like \( x \). For an absolute value equation, breaking it into possible distinct algebraic expressions (\( \frac{x}{7} = 2 \) and \( \frac{x}{7} = -2 \)) helps in managing the inherent duality of absolute values.
Step-by-Step Solutions
Approaching mathematics with a step-by-step methodology paves the way for clarity and understanding. By dissecting complex problems into manageable parts, students can navigate through the solution process more effortlessly.
Here's how the problem is tackled:
Here's how the problem is tackled:
- Identify: Recognize the equation \( |\frac{x}{7}| = 2 \) as one involving absolute values.
- Translate: Convert the single absolute value equation into two linear equations: \( \frac{x}{7} = 2 \) and \( \frac{x}{7} = -2 \).
- Calculate: Solve each linear equation by performing inverse operations like multiplication to isolate \( x \).
- Conclude: Combine solutions from both scenarios: \( x = 14 \) and \( x = -14 \).
Other exercises in this chapter
Problem 26
Factor each polynomial. $$ 18 y^{2} z^{2}+12 y^{2} z^{3}-24 y^{4} z^{3} $$
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Factor difference of two squares. \(400 x^{2} z^{4}-a^{4}\)
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Solve each compound inequality, if possible. Graph the solution set (if one exists) and write it using interval notation. $$ x \leq-4 \text { and } x \geq-7 $$
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Solve each equation. $$ \frac{x}{2}+\frac{x}{3}=10 $$
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