Respuesta :
[tex] \bf \cfrac{4p-5q}{6p+q}=\cfrac{1}{4}\implies 16p-20q=6p+q\implies 10p=21q
\\\\\\
p=\cfrac{21q}{10}\implies \cfrac{p}{q}=\cfrac{21}{10}\implies \stackrel{\textit{multiplying both sides by }\frac{2}{7}}{\cfrac{2p}{7q}=\cfrac{2(21)}{7(10)}}
\\\\\\
\cfrac{2p}{7q}=\cfrac{42}{70}\implies \cfrac{2p}{7q}=\stackrel{simplified}{\cfrac{3}{5}} [/tex]
Start from
[tex] \cfrac{4p-5q}{6p+q} = \cfrac{1}{4} [/tex]
and cross multiply the denominators (i.e. multiply both sides by [tex] 4(6p+q) [/tex]
The result is
[tex] 4(4p-5q) = 6p+q [/tex]
Expand the left hand side:
[tex] 16p-20q = 6p+q [/tex]
Bring all terms involving p to the left, and all terms involving q to the right:
[tex] 16p-6p = 20q+q \implies 10p=21q [/tex]
Divide both sides by 21q:
[tex] \cfrac{10p}{21q} = 1 [/tex]
Now we have a ratio between multiples of p and q. It's not exactly the one we want, though. Nevertheless, we can keep multiplying both sides by approriate constants in order to get the ratio we want:
Divide both sides by 5:
[tex] \cfrac{2p}{21q} = \cfrac{1}{5} [/tex]
Multiply both sides by 3:
[tex] \cfrac{2p}{7q} = \cfrac{3}{5} [/tex]
