Interchange the x and y-variables.

\(\displaystyle x= 3(3-9y)\)

Solve for y.  Distribute the three through both terms of the binomial.

\(\displaystyle x=9-27y\)

Subtract 9 on both sides.

\(\displaystyle x-9=9-27y-9\)

The equation becomes:  \(\displaystyle x-9=-27y\)

Divide by negative 27 on both sides.

\(\displaystyle \frac{x-9}{-27}=\frac{-27y}{-27}\)

Simplify both sides.

The answer is:  \(\displaystyle y=-\frac{1}{27}x+\frac{1}{3}\)

Interchange the x and y-variables.

\(\displaystyle x=-4y-32\)

Solve for y.  Add 32 on both sides.

\(\displaystyle x+32=-4y-32+32\)

The equation becomes:

\(\displaystyle x+32=-4y\)

Divide by negative four on both sides.

\(\displaystyle \frac{x+32}{-4}=\frac{-4y}{-4}\)

Simplify both sides.

The answer is:  \(\displaystyle y=-\frac{1}{4}x-8\)

Interchange the x and y variables.

\(\displaystyle x=2-(2y-3)\)

Solve for y. 

Subtract two from both sides.

\(\displaystyle x-2=2-(2y-3)-2\)

\(\displaystyle x-2=-(2y-3)\)

Divide by negative one on both sides.

\(\displaystyle \frac{x-2}{-1}=\frac{-(2y-3)}{-1}\)

\(\displaystyle -x+2 =2y-3\)

Add three on both sides.

\(\displaystyle -x+2+3 =2y-3+3\)

\(\displaystyle -x+5 =2y\)

Divide by two on both sides.

\(\displaystyle \frac{-x+5 }{2}=\frac{2y}{2}\)

The answer is:  \(\displaystyle y=-\frac{1}{2}x+\frac{5}{2}\)

Interchange the x and y variables.

\(\displaystyle x=-3(6y+3)\)

Solve for y.  Distribute the negative three across both terms in the binomial.

\(\displaystyle x=-3(6y)+(-3)(3)\)

Simplify the equation.

\(\displaystyle x=-18y-9\)

Add 9 on both sides.

\(\displaystyle x+9=-18y-9+9\)

\(\displaystyle x+9=-18y\)

Divide by negative 18 on both sides.

\(\displaystyle \frac{x+9}{-18}=\frac{-18y}{-18}\)

The answer is:  \(\displaystyle y=-\frac{1}{18}x -\frac{1}{2}\)

Interchange the x and y-variables.

\(\displaystyle x=-6y+3\)

Solve for y.

Subtract three from both sides.

\(\displaystyle x-3=-6y+3-3\)

\(\displaystyle x-3=-6y\)

Divide by negative 6 on both sides.

\(\displaystyle \frac{x-3}{-6}=\frac{-6y}{-6}\)

The answer is:  \(\displaystyle y=-\frac{1}{6}x+\frac{1}{2}\)

Swap the x and y variables.

\(\displaystyle x=-6y-10\)

Solve for y.  Add 10 on both sides.

\(\displaystyle x+10=-6y-10+10\)

\(\displaystyle x+10=-6y\)

Divide by negative six on both sides.

\(\displaystyle \frac{x+10}{-6}=\frac{-6y}{-6}\)

The answer is:  \(\displaystyle y=-\frac{1}{6}x-\frac{5}{3}\)

In order to solve for the inverse, interchange the x and the y variables, and solve for y.

\(\displaystyle y+2x = -4\)

Subtract \(\displaystyle 2x\) from both sides of the equation.

\(\displaystyle y+2x-2x =-2x -4\)

The answer is:  \(\displaystyle y=-2x -4\)

Given the graph of \(\displaystyle f(x)\), the graph of its inverse, \(\displaystyle f^{-1}(x)\) is the reflection of the former about the line \(\displaystyle y = x\). This line is in dark green below; critical points are reflected as shown:

The blue figure is \(\displaystyle f^{-1}(x)\), recreated below:

The inverse function \(\displaystyle f^{-1}(x)\) of a function \(\displaystyle f(x)\) can be found as follows:

Replace \(\displaystyle f(x)\) with \(\displaystyle y\):

\(\displaystyle f(x) = \frac{1}{5}x +7\)

\(\displaystyle y= \frac{1}{5}x +7\)

Switch the positions of \(\displaystyle y\) and \(\displaystyle x\):

\(\displaystyle x= \frac{1}{5}y +7\)

or,

\(\displaystyle \frac{1}{5}y +7 = x\)

Solve for \(\displaystyle y\) - that is, isolate it on one side.

Subtract 7:

\(\displaystyle \frac{1}{5}y +7 - 7 = x - 7\)

\(\displaystyle \frac{1}{5}y = x - 7\)

Multiply by 5, distributing on the right:

\(\displaystyle 5 \cdot \frac{1}{5}y =5 \cdot (x - 7)\)

\(\displaystyle y =5 \cdot x - 5 \cdot 7\)

\(\displaystyle y =5x-35\)

Replace \(\displaystyle y\) with \(\displaystyle f^{-1}(x)\):

\(\displaystyle f^{-1}(x) =5x-35\)

The inverse function \(\displaystyle f^{-1}(x)\) of a function \(\displaystyle f(x)\) can be found as follows:

Replace \(\displaystyle f(x)\) with \(\displaystyle y\):

\(\displaystyle f(x) = 5^{x} + 5\)

\(\displaystyle y = 5^{x} + 5\)

Switch the positions of \(\displaystyle y\) and \(\displaystyle x\):

\(\displaystyle x = 5^{y} + 5\), 

or

\(\displaystyle 5^{y} + 5 = x\)

Solve for \(\displaystyle y\) - that is, isolate it on one side - as follows:

First, subtract 5 from both sides:

\(\displaystyle 5^{y} + 5- 5 = x - 5\)

\(\displaystyle 5^{y} = x - 5\)

Take the base-5 logarithm of both sides:

\(\displaystyle \log_{5} 5^{y} = \log_{5} (x - 5)\)

A property of logarithms states that \(\displaystyle \log_{a} a^{N}= N\), so

\(\displaystyle y = \log_{5} (x - 5)\)

Replace \(\displaystyle y\) with \(\displaystyle f^{-1}(x)\):

\(\displaystyle f^{-1}(x) = \log_{5} (x - 5)\)