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: Divide the target expression by $b$ instead of introducing a multiplier.
Start from
\[ \frac{9a+4b}{9a-4b} \]
Divide every term, top and bottom, by $b$:
\[ \frac{9\frac{a}{b}+4}{9\frac{a}{b}-4} \]

Step 2: Substitute the known ratio $\frac{a}{b} = \frac{4}{3}$.
First work out $9 \times \frac{a}{b}$:
\[ 9 \times \frac{4}{3} = \frac{36}{3} = 12 \]

Step 3: Plug this value into the expression.
\[ \frac{9\frac{a}{b}+4}{9\frac{a}{b}-4} = \frac{12+4}{12-4} \]

Step 4: Simplify the fraction.
\[ \frac{12+4}{12-4} = \frac{16}{8} = 2 \]

Step 5: Sanity check.
This matches the value obtained by putting real numbers, say $a=4$ and $b=3$: $9(4)+4(3)=48$ and $9(4)-4(3)=24$, giving $\frac{48}{24}=2$ again, so the answer is confirmed independent of method.
\[ \boxed{2} \]