Given that $ \frac{4 p+9 q}{p}=\frac{5 q}{p-q} $ and $ p $ and $ q $ are both positive. The value of $\frac{p}{q}$ is
Given:
\( \frac{4 p+9 q}{p}=\frac{5 q}{p-q} \) and \( p \) and \( q \) are both positive.
To do:
We have to find the value of $\frac{p}{q}$.
Solution:
$\frac{4 p+9 q}{p}=\frac{5 q}{p-q}$
$(4p+9q)(p-q)=5q(p)$ (On cross multiplication)
$4p^2-4pq+9pq-9q^2=5pq$
$4p^2=9q^2$
$\frac{p^2}{q^2}=\frac{3^2}{2^2}$
$\frac{p}{q}=\frac{3}{2}$ (Taking square root on both sides)
The value of $\frac{p}{q}$ is $\frac{3}{2}$.
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