Answer:
[tex]\sf a= {\dfrac{4y-3}{-2y+3}}[/tex] where p = 4, q = -3, r = -2, s = 3
Step-by-step explanation:
Given following:
[tex]\sf a= \dfrac{14}{3x-7}[/tex] and [tex]\sf x= \dfrac{7}{4y-3}[/tex]
To find a in terms of y then substitute x into the equation of a
[tex]\rightarrow \sf a= \dfrac{14}{3x-7}[/tex]
substitute
[tex]\rightarrow \sf a= \dfrac{14}{3(\dfrac{7}{4y-3})-7}[/tex]
simplify
[tex]\rightarrow \sf a= \dfrac{14}{\dfrac{21}{4y-3}-7}[/tex]
make the denominator same
[tex]\rightarrow \sf a= \dfrac{14}{\dfrac{21}{4y-3}-\dfrac{28y-21}{4y-3}}[/tex]
join fractions
[tex]\rightarrow \sf a= \dfrac{14}{\dfrac{21-(28y-21)}{4y-3}}[/tex]
simplify
[tex]\rightarrow \sf a= \dfrac{14}{\dfrac{21-28y+21}{4y-3}}[/tex]
use basic rules:
[tex]\rightarrow \sf a= \dfrac{14}{\dfrac{42-28y}{4y-3}}[/tex]
[tex]\rightarrow \sf a= 14 \div {\dfrac{42-28y}{4y-3}}[/tex]
[tex]\rightarrow \sf a= 14 \cdot {\dfrac{4y-3}{42-28y}}[/tex]
[tex]\rightarrow \sf a= {\dfrac{56y-42}{42-28y}}[/tex]
simplify (all the variables are factor of 14)
[tex]\rightarrow \sf a= {\dfrac{4y-3}{3-2y}}[/tex]
