Question
Solve the inequality for \( v \). \[ v+\frac{7}{8}>\frac{1}{2} \] Simplify your answer as much as possible. \( \square \) \( \square<\square \quad \square>\square \quad \square \leq \square \)
Ask by Delgado Park. in the United States
Jan 27,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
\( v > -\frac{3}{8} \)
Solution
To solve the inequality for \( v \):
\[
v + \frac{7}{8} > \frac{1}{2}
\]
**Step 1: Subtract \(\frac{7}{8}\) from both sides to isolate \( v \).**
\[
v > \frac{1}{2} - \frac{7}{8}
\]
**Step 2: Find a common denominator to subtract the fractions.**
The common denominator for \( \frac{1}{2} \) and \( \frac{7}{8} \) is 8.
\[
\frac{1}{2} = \frac{4}{8}
\]
So, the inequality becomes:
\[
v > \frac{4}{8} - \frac{7}{8}
\]
**Step 3: Subtract the fractions.**
\[
v > \frac{4 - 7}{8} = \frac{-3}{8}
\]
**Final Solution:**
\[
v > -\frac{3}{8}
\]
**In Interval Notation:**
\[
v \in \left( -\frac{3}{8}, \infty \right)
\]
**On a Number Line:**
- **\( \square \)** \( \square < \square \quad \square > \square \quad \square \leq \square \)
This indicates that \( v \) is greater than \(-\frac{3}{8}\).
Reviewed and approved by the UpStudy tutoring team
Like
error msg
Explain 
Simplify this solution Beyond the Answer
To solve the inequality \( v+\frac{7}{8}>\frac{1}{2} \), we first subtract \( \frac{7}{8} \) from both sides. This gives us: \[ v > \frac{1}{2} - \frac{7}{8} \] Next, we need a common denominator to simplify the right side. The common denominator of \( 2 \) and \( 8 \) is \( 8 \). We can rewrite \( \frac{1}{2} \) as \( \frac{4}{8} \): \[ v > \frac{4}{8} - \frac{7}{8} \] \[ v > \frac{4 - 7}{8} \] \[ v > \frac{-3}{8} \] So the solution to the inequality is: \( v > -\frac{3}{8} \) In terms of the output format: \( \square \) \( -\frac{3}{8} < v \quad v > -\frac{3}{8} \)
