[tex]f(x) = 5x + 1\\\\subtitute\ x = 1\ to\ the\ equation\\\\f(1) = 5\cdot1 + 1 = 5 + 1 = 6\\----------------------\\f(x)=\dfrac{2x+3}{5}\to y=\dfrac{2x+3}{5}\ \ \ \ |multiply\ both\ sides\ by\ 5\\\\5y=2x+3\ \ \ \ |subtract\ 3\ from\ both\ sides\\\\2x=5y-3\ \ \ \ \ |divide\ both\ sides\ by\ 2\\\\x=\dfrac{5y-3}{2}\\\\f^{-1}(x)=\dfrac{5\cdot(-1)-3}{2}=\dfrac{-5-3}{2}=\dfrac{-8}{2}=-4\\---------------------\\3y-7=y+5\ \ \ |add\ 7\ to\ both\ sides\\3y=y+12\ \ \ \ |subtract\ y\ from\ both\ sides\\2y=12\ \ \ \ |divide\ both\ sides\ by\ 2\\y=6[/tex]
Answer: f(1) when f(x) = 5x + 1 and 3y - 7 = y + 5
