Question 73977
*[Tex \Large y=\frac{3+2x}{2-5x}]Start with given equation
*[Tex \Large x=\frac{3+2y}{2-5y}]Switch the x and y variables
Now solve for y
*[Tex \Large (2-5y)x=(\frac{3+2y}{2-5y})(\frac{2-5y}{1})]Multiply both sides by (2-5y)
*[Tex \Large 2x-5xy=3+2y]Distribute the x on the left side
*[Tex \Large 2x-3=2y+5xy]Get all y terms to one side
*[Tex \Large 2x-3=y(2+5x)]Factor out a y
*[Tex \Large \frac{2x-3}{2+5x}=y\frac{(2+5x)}{(2+5x)}]Divide both sides by (2+5x)
*[Tex \Large y=\frac{-3+2x}{2+5x}]Here's our inverse function. In other words
*[Tex \Large f^{-1}(x)=\frac{-3+2x}{2+5x}]
<p>
Check:
If we let x=1 and plugged it into *[Tex \Large y=\frac{3+2x}{2-5x}] we would get
*[Tex \Large y=\frac{3+2}{2-5}]
*[Tex \Large y=\frac{-5}{3}]
If we plug in  *[Tex \Large \frac{-5}{3}] in for x into our inverse equation *[Tex \Large y=\frac{-3+2x}{2+5x}] we should get y=1 out.
*[Tex \Large y=\frac{-3+2(-5/3)}{2+5(-5/3)}]
*[Tex \Large y=\frac{-3-\frac{10}{3}}{2-\frac{25}{3}}]
*[Tex \Large y=\frac{\frac{-9-10}{3}}{\frac{6-25}{3}}]
*[Tex \Large y=\frac{\frac{-19}{3}}{\frac{-19}{3}}]
*[Tex \Large y=\frac{-19}{3}{\frac{3}{-19}]
*[Tex \Large y=1]So this verifies that *[Tex \Large y=\frac{-3+2x}{2+5x}] is our answer. Hope all of this makes sense.