Question

Given that (4p-5q)/(6p+q) = 1/4 , find the value of 2p : 7q in its lowest term.

Answer 1

$\bf \cfrac{4p-5q}{6p+q}=\cfrac{1}{4}\implies 16p-20q=6p+q\implies 10p=21q \\\\\\ p=\cfrac{21q}{10}\implies \cfrac{p}{q}=\cfrac{21}{10}\implies \stackrel{\textit{multiplying both sides by }\frac{2}{7}}{\cfrac{2p}{7q}=\cfrac{2(21)}{7(10)}} \\\\\\ \cfrac{2p}{7q}=\cfrac{42}{70}\implies \cfrac{2p}{7q}=\stackrel{simplified}{\cfrac{3}{5}}$

Answer 2

Start from

$\cfrac{4p-5q}{6p+q} = \cfrac{1}{4}$

and cross multiply the denominators (i.e. multiply both sides by $4(6p+q)$

The result is

$4(4p-5q) = 6p+q$

Expand the left hand side:

$16p-20q = 6p+q$

Bring all terms involving p to the left, and all terms involving q to the right:

$16p-6p = 20q+q \implies 10p=21q$

Divide both sides by 21q:

$\cfrac{10p}{21q} = 1$

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:

$\cfrac{2p}{21q} = \cfrac{1}{5}$

Multiply both sides by 3:

$\cfrac{2p}{7q} = \cfrac{3}{5}$