What is the inverse of h(x)=(2x+17)/(3x+1) ?

The inverse of h(x)=(2x+17)/(3x+1) is (17-x)/(3x-2)

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inverse of h(x)=(2x+17)/(3x+1)

inverse

h (x) = \frac{2 x + 17}{3 x + 1}

\frac{17 - x}{3 x - 2}

Solution steps

y = \frac{2 x + 17}{3 x + 1}

Replace

x

with

y

x = \frac{2 y + 17}{3 y + 1}

Solve for

y, x = \frac{2 y + 17}{3 y + 1}

\frac{17 - x}{3 x - 2}

f (x) = 2 (x + 1)^{2} - 3

6 x + 3

y = 3 x^{2}

3 x^{2} - 1

y = x^{\frac{1}{2}} + 4

What is the inverse of h(x)=(2x+17)/(3x+1) ?
The inverse of h(x)=(2x+17)/(3x+1) is (17-x)/(3x-2)