Question

if a and b represent positive real numbers, what is the inequality when solved for v?
\\(\frac{au}{2} - \frac{bv}{3} > 10\\)
\\(v < \frac{3au - 60}{2b}\\) \\(v < \frac{au - 60}{b}\\) \\(v > \frac{-3au - 60}{2b}\\) \\(v > \frac{3au - 60}{2b}\\)

Explanation:

Step1: Isolate the term with \( v \)

Start with the inequality \(\frac{au}{2} - \frac{bv}{3} > 10\). Subtract \(\frac{au}{2}\) from both sides:
\(-\frac{bv}{3} > 10 - \frac{au}{2}\)

Step2: Multiply both sides by -3 (reverse inequality)

Multiply each side by \(-3\). Since we multiply by a negative number, reverse the inequality sign:
\(bv < -30 + \frac{3au}{2}\) (or \(bv < \frac{3au}{2} - 30\))

Step3: Divide by \( b \) ( \( b > 0 \), so inequality sign remains)

Divide both sides by \( b \) ( \( b \) is positive, so the inequality direction stays the same):
\(v < \frac{\frac{3au}{2} - 30}{b}\)

Step4: Simplify the right-hand side

To combine the terms over a common denominator, rewrite \(-30\) as \(\frac{-60}{2}\):
\(v < \frac{\frac{3au - 60}{2}}{b}\)
Which simplifies to:
\(v < \frac{3au - 60}{2b}\)

Answer:

\( v < \frac{3au - 60}{2b} \) (corresponding to the first option: \( v < \frac{3au - 60}{2b} \))