Question

If \(\frac{a}{b} = \frac{4}{3}\), then find the value of \(\frac{9a+4b}{9a-4b}\).

• A. 2
• B. \(\frac{8}{5}\)
• C. \(\frac{16}{9}\)
• D. 3

Hint:

Write \(a=4k\) and \(b=3k\), substitute, then cancel \(k\) from the final ratio.

Answer 1

The Correct Option is A

Step 1: Represent \(a\) and \(b\) using a common multiplier.
Since \(\frac{a}{b} = \frac{4}{3}\), we can write \(a = 4k\) and \(b = 3k\) for some non-zero constant \(k\). This lets us replace both variables with a single unknown.

Step 2: Substitute into the numerator and denominator separately.
Numerator:
\[ 9a+4b = 9(4k)+4(3k) = 36k+12k = 48k \]
Denominator:
\[ 9a-4b = 9(4k)-4(3k) = 36k-12k = 24k \]

Step 3: Divide the two results.
\[ \frac{9a+4b}{9a-4b} = \frac{48k}{24k} \]

Step 4: Cancel the common factor \(k\).
Since \(k \ne 0\), it cancels top and bottom:
\[ \frac{48k}{24k} = \frac{48}{24} = 2 \]

Step 5: Rule out the other options.
Option (b) \(\frac{8}{5}\), option (c) \(\frac{16}{9}\) and option (d) 3 do not match this value, since substituting \(a=4k, b=3k\) leaves no room for any other result once \(k\) cancels.

Final Answer:
The value of the expression is 2.
\[ \boxed{2} \]