First one
y = -4x + 1 subtract 1 tfrom both sides
y - 1 = - 4x divide both sides by -4
y - 1 / -4 = x
(1 - y) / 4 = x swap x and y
(1 - x) / 4 = y
-(1/4)x + 1/4 = y = f(x) so....they are inverses
Second one
y = [ -20 + 8x ] / 5 multiply both sides by 5
5y = -20 + 8x add 20 to both sides
5y + 20 = 8x divide both sides by 8
[ 5y + 20 ] / 8 = x swap x and y
[5x + 20 ] / 8 = y = f(x) .....also inverses
Inverse
1.
\(\begin{array}{|rcll|} \hline g(f(x)) & \overset{?}{=} & x \quad | \quad f(x) = -\dfrac{1}{4}x + \dfrac{1}{4} \\ g(-\dfrac{1}{4}x + \dfrac{1}{4} ) & \overset{?}{=} & x \\ -4\left( -\dfrac{1}{4}x + \dfrac{1}{4} \right) +1 & \overset{?}{=} & x \\ x-1+1 & \overset{?}{=} & x \\ x & \overset{!}{=} & x \quad \text{the functions are inverses }\\ \hline \end{array}\)
2.
\(\begin{array}{|rcll|} \hline g(f(x)) & \overset{?}{=} & x \quad | \quad f(x) = \dfrac{5x+20}{8} \\ g(\dfrac{5x+20}{8}) & \overset{?}{=} & x \\ \dfrac{-20+8\left( \dfrac{5x+20}{8} \right) }{5}& \overset{?}{=} & x \\ -4 + \dfrac{5x+20}{5} & \overset{?}{=} & x \\ -4 + x + 4 & \overset{?}{=} & x \\ x & \overset{!}{=} & x \quad \text{the functions are inverses }\\ \hline \end{array}\)