Answer
$\dfrac{7}{6p^{2}+q}\div\dfrac{14}{18p^{2}+3q}=\dfrac{3}{2}$
Work Step by Step
$\dfrac{7}{6p^{2}+q}\div\dfrac{14}{18p^{2}+3q}$
Take out common factor $3$ from the denominator of the second fraction:
$\dfrac{7}{6p^{2}+q}\div\dfrac{14}{18p^{2}+3q}=\dfrac{7}{6p^{2}+q}\div\dfrac{14}{3(6p^{2}+q)}=...$
Evaluate the division of the two rational expressions and simplify by removing repeated terms in the numerator and the denominator:
$...=\dfrac{21(6p^{2}+q)}{14(6p^{2}+q)}=\dfrac{21}{14}=\dfrac{3}{2}$
