Functions: Fractional functions
Inverse of linear fractional function
We have seen that determining the inverse function is the same as isolating the variable #x# in a formula of the form #y=\ldots#. Now we will investigate how to do that for linear fractional functions.
Procedure We determine the inverse function of the linear fractional function #\green{y}=\frac{a\blue{x}+b}{c\blue{x}+d}# with #a#, #b#, #c# and #d# as numbers. 
Example #\green{y}=\frac{2\blue{x}5}{3\blue{x}+2}# 

Step 1  Multiply by the denominator of the fraction: #c\blue{x}+d#.  #\green{y} \left(3\blue{x}+2\right)=2\blue{x}5# 
Step 2  Expand the brackets.  #3\blue{x}\green{y}+2 \green{y}=2\blue{x}5# 
Step 3  By means of reduction move the terms without #x# to the right and the terms with a #x# to the left hand side.  #3\blue{x}\green{y}2\blue{x}=2 \green{y}5# 
Step 4  Move #x# outside brackets.  #\blue x \left(3 \green{y}2\right)=2 \green{y}5# 
Step 5  Divide by what's in between the brackets, so that we only have #x# at the left hand side.  #\blue x=\frac{2 \green{y}5}{3 \green{y}2}# 
Step 6 
Swap the #\blue x# into a #\green y# and the #\green y# into a #\blue x# to get the inverse function. 
#\green y=\frac{2 \blue{x}5}{3 \blue{x}2}# 
Isolate #x# in
\[y={{5\cdot x+9}\over{5\cdot x2}}\]
\[y={{5\cdot x+9}\over{5\cdot x2}}\]
#x={{2\cdot y9}\over{5\cdot y+5}}#
#\begin{array}{rcl}
y&=&{{5\cdot x+9}\over{5\cdot x2}} \\ &&\phantom{xxx}\blue{\text{the original function }}\\
y \cdot \left(5\cdot x2\right)&=& 5\cdot x+9 \\ &&\phantom{xxx}\blue{\text{both sides divided by }5\cdot x2}\\
5\cdot x\cdot y2\cdot y&=&5\cdot x+9 \\ &&\phantom{xxx}\blue{\text{brackets expanded}}\\
5\cdot x\cdot y5\cdot x &=&2\cdot y+9 \\&&\phantom{xxx}\blue{\text{terms with } x \text{ to the left hand side, terms without }x \text{ to the right hand side }}\\
x\cdot \left(5\cdot y5\right) &=& 2\cdot y+9 \\ &&\phantom{xxx}\blue{x \text{ moved outside brackets}}\\
x&=&{{2\cdot y9}\over{5\cdot y+5}} \\ &&\phantom{xxx}\blue{\text{divided by }5\cdot y5}\\
\end{array}#
#\begin{array}{rcl}
y&=&{{5\cdot x+9}\over{5\cdot x2}} \\ &&\phantom{xxx}\blue{\text{the original function }}\\
y \cdot \left(5\cdot x2\right)&=& 5\cdot x+9 \\ &&\phantom{xxx}\blue{\text{both sides divided by }5\cdot x2}\\
5\cdot x\cdot y2\cdot y&=&5\cdot x+9 \\ &&\phantom{xxx}\blue{\text{brackets expanded}}\\
5\cdot x\cdot y5\cdot x &=&2\cdot y+9 \\&&\phantom{xxx}\blue{\text{terms with } x \text{ to the left hand side, terms without }x \text{ to the right hand side }}\\
x\cdot \left(5\cdot y5\right) &=& 2\cdot y+9 \\ &&\phantom{xxx}\blue{x \text{ moved outside brackets}}\\
x&=&{{2\cdot y9}\over{5\cdot y+5}} \\ &&\phantom{xxx}\blue{\text{divided by }5\cdot y5}\\
\end{array}#
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