Question 308935
Try this!
{{{f(x) = 4+5x}}}
{{{g(x) = 2x-1}}}
{{{highlight(f(g(x))) = f(2x-1)}}}={{{4+5*(2x-1) = 4+10x-5}}}={{{highlight(10x-1)}}}
{{{highlight_green(g(f(x))) = g(4+5x)}}}={{{2(4+5x)-1 = 8+10x-1}}}={{{highlight_green(10x+7)}}}