Question 1105411
{{{g(x)=px+q}}} , {{{f(x)= 4x-3}}}
{{{"g ["}}}{{{(x)}}}{{{"]"=p(px+q)+q=p^2x+pq+q=p^2x+(p+1)q}}}
If {{{"g ["}}}{{{(x)}}}{{{"]"=f(x)}}} for all values of x,
meaning that {{{p^2x+(p+1)q=4x-3}}} for all values of x,
then {{{system(p^2=4,"and",(p+1)q=-3)}}} .
There are two solutions for {{{p^2=4}}} : {{{system(p=2,"or",p=-2)}}} .
{{{system(p=2,(p+1)q=-3)}}} --> {{{system(p=2,(2+1)q=-3)}}} --> {{{system(p=2,3q=-3)}}} --> {{{system(p=2,q=-3/3)}}} --> {{{highlight(system(p=2,q=-1))}}}
 
{{{system(p=-2,(p+1)q=-3)}}} --> {{{system(p=-2,(-2+1)q=-3)}}} --> {{{system(p=-2,(-1)q=-3)}}} --> {{{system(p=-2,q=-3/(-1))}}} --> {{{highlight(system(p=-2,q=1))}}}