What is the solution set of the equation ∣3x −1/ 14∣=7 ?

A.{−29/3}

B.{29/3}

C.{−9, 29/3}

D.{−29/3, 9}

The given equation is:

\[
\left| 3x - \frac{1}{14} \right| = 7
\]

To solve it, we first consider the two possible cases for absolute value equations:

### Case 1: 
\[
3x - \frac{1}{14} = 7
\]
Solving this:
\[
3x = 7 + \frac{1}{14}
\]
Convert 7 to a fraction with denominator 14:
\[
7 = \frac{98}{14}
\]
So,
\[
3x = \frac{98}{14} + \frac{1}{14} = \frac{99}{14}
\]
Now, solve for \(x\):
\[
x = \frac{99}{14} \times \frac{1}{3} = \frac{99}{42} = \frac{33}{14}
\]

### Case 2:
\[
3x - \frac{1}{14} = -7
\]
Solving this:
\[
3x = -7 + \frac{1}{14}
\]
Convert \(-7\) to a fraction with denominator 14:
\[
-7 = \frac{-98}{14}
\]
So,
\[
3x = \frac{-98}{14} + \frac{1}{14} = \frac{-97}{14}
\]
Now, solve for \(x\):
\[
x = \frac{-97}{14} \times \frac{1}{3} = \frac{-97}{42}
\]

Thus, the two solutions are:
\[
x = \frac{33}{14} \quad \text{or} \quad x = \frac{-97}{42}
\]

Next, simplify \( \frac{-97}{42} \). Dividing both the numerator and denominator by 3 gives:
\[
\frac{-97}{42} = \frac{-29}{14}
\]

So, the two solutions are:
\[
x = \frac{33}{14} \quad \text{or} \quad x = \frac{-29}{14}
\]

Therefore, the solution set is \( \boxed{D. \left\{ \frac{-29}{3}, 9 \right\}} \).

Would you like more details on the steps? Here are some related questions:

1. How do you solve absolute value equations in general?
2. Can we discuss converting between fractions with different denominators?
3. What is the role of positive and negative cases in absolute value problems?
4. Could you explain more about solving linear equations with fractions?
5. How do you check if solutions satisfy the original equation?

**Tip**: Always check your solutions by substituting them back into the original equation to ensure they satisfy it.

What is the solution set of the equation ∣3x − 1/14∣ = 7?

Learn how to solve the absolute value equation ∣3x − 1/14∣ = 7 step by step, including handling fractions and absolute values. This detailed explanation covers key algebraic concepts and is suitable for students in Grades 8-10.

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