What is the inverse of (x^2+2x)/(2x^2-7x) ?

The inverse of (x^2+2x)/(2x^2-7x) is (2+7x)/(2x-1)

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inverse of (x^2+2x)/(2x^2-7x)

inverse

\frac{x ^{2} + 2 x}{2 x ^{2} - 7 x}

\frac{2 + 7 x}{2 x - 1}

Solution steps

y = \frac{x ^{2} + 2 x}{2 x ^{2} - 7 x}

Replace

x

with

y

x = \frac{y ^{2} + 2 y}{2 y ^{2} - 7 y}

Solve for

y, x = \frac{y ^{2} + 2 y}{2 y ^{2} - 7 y}

\frac{2 + 7 x}{2 x - 1}

f (r) = \frac{( - 16 + r )}{4}

f (x) = 4 + x^{2}, x \geq 0

\frac{6 s + 3}{s ^{2}}

f (x) = \frac{1}{x ^{2} + 1}

f (x) = 12 x^{2} - 14 x

What is the inverse of (x^2+2x)/(2x^2-7x) ?
The inverse of (x^2+2x)/(2x^2-7x) is (2+7x)/(2x-1)