Functions: Fractional functions
Linear fractional equations
A linear fractional equation is of the form
\[\frac{\blue{A}}{\green{B}}=\frac{\purple{C}}{\orange{D}}\]
where #\blue{A}#, #\green{B}#, #\purple{C}# and #\orange{D}# are linear expressions, which are expressions of the form #ax+b#.
Examples
\[\begin{array}{rcl}\dfrac{\blue{2x+3}}{\green{x+1}}&=&\dfrac{\purple{4x+2}}{\orange{3x+5}} \\ \\ \dfrac{\blue{2x+3}}{\green{3x+6}}&=&\purple{2}\end{array}\]
Solving linear fractional equations by cross multiplication
Example We solve the linear fractional equation of the form #\frac{\blue{A}}{\green{B}}=\frac{\purple{C}}{\orange{D}}#. 
Example #\frac{\blue{2x+4}}{\green{3x+3}}=\frac{\purple{x+2}}{\orange{2x3}}# 

Step 1 
We cross multiply the equation. This gives us: \[\blue{A} \cdot \orange{D}=\purple{C} \cdot \green{B}\] 
#\left(\blue{2x+4}\right) \left(\orange{2x3}\right) = \left(\purple{x+2}\right) \left(\green{3x+3}\right)# 
Step 2  Solve the obtained quadratic equation.  #x=2 \lor x=9# 
Stap 3  Check the found solutions by investigating if they don't make the denominator of the original equation equal to #0#. 
Both solutions are valid. 
\[{{92\cdot x}\over{7x}}=7\]
Give your answer in the form of #x=x_1#, in which #x_1# is a simplified fraction.
#\begin{array}{rcl}
\dfrac{92\cdot x}{7x}&=& 7 \\
&&\phantom{xxx}\blue{\text{original equation }}\\
92\cdot x &=& 7 \cdot \left(7x\right)\\
&&\phantom{xxx}\blue{\text{left and right multiplied by }7x }\\
92\cdot x &=& 7\cdot x49\\
&&\phantom{xxx}\blue{\text{expanded brackets on the right} }\\
9\cdot x &=& 58 \\
&&\phantom{xxx}\blue{\text{terms with }x \text{ to the left and numbers to the right} }\\
x &=&{{58}\over{9}} \\
&&\phantom{xxx}\blue{\text{divided by the coefficient of }x}\\
\end{array}#
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