Inverse Functions

[lexieaj]

I'm having trouble finding the inverse function for 1+3x/5-2x...I struggle getting the variable out of the denominator. Please Help. I know somewhere I should be factoring, but just returned to school after 6 years, and am a little rusty.
Thanks,
Lexie

 

[soroban]

Hello, Lexie!

\(\displaystyle \text{Find the inverse of: }\:f(x) \;=\;y \;=\;\frac{1+3x}{5-2x}\)

\(\displaystyle \text{(1) Switch }x\text{'s and }y\text{'s.}\)

. . .\(\displaystyle x \;=\;\frac{1+3y}{5-2y}\)


\(\displaystyle \text{(2) Solve for }y.\)

. . \(\displaystyle \text{Multiply by }(5-2y)\!:\;\;x(5-2y) \;=\;1+3y \quad\Rightarrow\quad 5x - 2xy \;=\;1 + 3y \quad\Rightarrow\quad -2xy - 3y \;=\;1 - 5x\)

. . \(\displaystyle \text{Factor: }\;-y(2x+3) \;=\;1-5x \quad\Rightarrow\quad y \;=\;\frac{1-5x}{-(2x+3)} \quad\Rightarrow\quad y \;=\;\frac{5x-1}{2x+3}\)

\(\displaystyle \text{Therefore: }\:f^{-1}(x) \;=\;\frac{5x-1}{21x+3}\)

 

[lexieaj]

Thanks so much for your help! I just returned to college and it has been difficult. I appreciate it!