Problem 20 - Algebra for College Students

Question

Solve each equation. $$ \left|14-\frac{1}{3} x\right|=8 $$

Step-by-Step Solution

Answer

The solutions are \(x = 18\) and \(x = 66\).

Step 1: Understand the Absolute Value Equation

The given equation is \(\big|14-\frac{1}{3} x\big|=8\). An absolute value equation \(\big|A\big|=B\) implies two cases: \(A=B\) or \(A=-B\).

Step 2: Set Up the Two Cases

Set the expression inside the absolute value equal to both \(8\) and \(-8\): \(14 - \frac{1}{3} x = 8\) and \(14 - \frac{1}{3} x = -8\).

Step 3: Solve the First Case

Start with \(14 - \frac{1}{3} x = 8\).Subtract \(14\) from both sides:\(14 - \frac{1}{3} x - 14 = 8 - 14\)\(-\frac{1}{3} x = -6\).Multiply both sides by \(-3\) to solve for \(x\):\(x = 18\).

Step 4: Solve the Second Case

Next, solve \(14 - \frac{1}{3} x = -8\).Subtract \(14\) from both sides:\(14 - \frac{1}{3} x - 14 = -8 - 14\)\(-\frac{1}{3} x = -22\).Multiply both sides by \(-3\) to solve for \(x\):\(x = 66\).

Step 5: Verify the Solutions

Check both solutions by substituting back into the original equation.For \(x = 18\):\(\big|14 - \frac{1}{3} \times 18\big|=\big|14 - 6 \big|=\big|8 \big|=8\).For \(x = 66\):\(\big|14 - \frac{1}{3} \times 66\big|=\big|14 - 22\big|=\big|-8\big|=8\).Both solutions are valid.

Key Concepts

Solving Equations

To solve an absolute value equation, we generally start by understanding the structure of the equation. For instance, in the exercise given, we have \(\big|14-\frac{1}{3} x\big|=8\). The absolute value equation can be split into two separate linear equations because the absolute value \(\big|A\big|\) represents the distance of \(\big|A\big|\) from zero on the number line, meaning \(\big|A\big|=B\) implies \(|14-\frac{1}{3} x| = 8\) can be rephrased as \(14 - \frac{1}{3} x = 8\) and \(14 - \frac{1}{3} x = -8\). This method allows you to create two simpler equations from one absolute value equation, making it easier to solve.

Absolute Value Properties

Absolute values measure the distance a number is from zero, regardless of direction. This property is crucial because it leads us to establish two cases for our equation. Given \(\big|14-\frac{1}{3} x\big|=8\), we interpret it as:

• Case 1: \(14 - \frac{1}{3} x = 8\)
• Case 2: \(14 - \frac{1}{3} x = -8\)

This way, we account for both positive and negative distances that could give an absolute value of 8. This dual-case scenario is fundamental to solving absolute value equations correctly.

Linear Equations

Now each case from the absolute value equation becomes a linear equation, which we solve step-by-step. For \(14 - \frac{1}{3} x = 8\):
1. Subtract 14 from both sides:
\(14 - \frac{1}{3} x - 14 = 8 - 14\)
simplifies to \(-\frac{1}{3} x = -6\).
2. Multiply both sides by -3 to isolate x:
\(x = 18\).

For \(14 - \frac{1}{3} x = -8\):
1. Subtract 14 from both sides:
\(14 - \frac{1}{3} x - 14 = -8 - 14\)
simplifies to \(-\frac{1}{3} x = -22\).
2. Multiply both sides by -3 to solve for x:
\(x = 66\).
That's how we find both potential solutions, x=18 and x=66, by treating each case as a linear equation.

Verification of Solutions

Ensuring that your solutions are correct is an essential step. To verify, substitute each solution back into the original absolute value equation:

• For \(x = 18\):
  Substitute 18: \( \big|14 - \frac{1}{3} \times 18\big| \big| = \big|14 - 6 \big| = \big|8 \big| = 8 \)
• For \(x = 66\):
  Substitute 66: \(\big|14 - \frac{1}{3} \times 66\big| = \big|14 - 22\big|= \big|-8\big|= 8 \)

In both cases, the equation holds true, showing that both x=18 and x=66 are valid solutions. It’s always good practice to finish by verifying, as it confirms our steps were correct and the solutions satisfy the original equation.