What is the inverse of f(x)=(9z-3)/(9z+7) ?

The inverse of f(x)=(9z-3)/(9z+7) is (-3-7z)/(9(-1+z))

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inverse of f(x)=(9z-3)/(9z+7)

inverse

f (x) = \frac{9 z - 3}{9 z + 7}

\frac{- 3 - 7 z}{9 ( - 1 + z )}

Solution steps

x = \frac{9 z - 3}{9 z + 7}

Replace

z

with

x

z = \frac{9 x - 3}{9 x + 7}

Solve for

x, z = \frac{9 x - 3}{9 x + 7}

\frac{- 3 - 7 z}{9 ( - 1 + z )}

h (x) = 7 - x^{3}

h (x) = \frac{1}{4} x + \frac{5}{3}

f (x) = \frac{7 x + 5}{x}

f (x) = \frac{x - 4}{4 x + 14}

f (x) = x^{2} 3 x + 2

What is the inverse of f(x)=(9z-3)/(9z+7) ?
The inverse of f(x)=(9z-3)/(9z+7) is (-3-7z)/(9(-1+z))