What is the inverse of (5s+3)/(s^2-1) ?

The inverse of (5s+3)/(s^2-1) is (5+sqrt(25-4s(-s-3)))/(2s),(5-sqrt(25-4s(-s-3)))/(2s)

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inverse of (5s+3)/(s^2-1)

inverse

\frac{5 s + 3}{s ^{2} - 1}

\frac{5 + 25 - 4 s ( - s - 3 )}{2 s}, \frac{5 - 25 - 4 s ( - s - 3 )}{2 s}

Solution steps

y = \frac{5 s + 3}{s ^{2} - 1}

Replace

s

with

y

s = \frac{5 y + 3}{y ^{2} - 1}

Solve for

y, s = \frac{5 y + 3}{y ^{2} - 1}

\frac{5 + 25 - 4 s ( - s - 3 )}{2 s}, \frac{5 - 25 - 4 s ( - s - 3 )}{2 s}

f (x) = \frac{1}{2} x - \frac{3}{7}

1 + \frac{4 x}{x + 3}

f (x) = - 3 + 3 2 x + 3

f (x) = \frac{4 x}{6 x - 5}

6.75 x + 12

What is the inverse of (5s+3)/(s^2-1) ?
The inverse of (5s+3)/(s^2-1) is (5+sqrt(25-4s(-s-3)))/(2s),(5-sqrt(25-4s(-s-3)))/(2s)