Linear fractional equations

Functions: Fractional functions

Linear fractional equations

Linear fractional equation

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{2x-3}}#
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{2x-3}\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.

Explanation step 3

To check #x=-2#, we substitute #x=-2# in both denominators to check if the denominator does not become equal to #0#. Which gives us:

#\green{3 \cdot -2+3}=-3 \ne 0#

#\orange{2 \cdot -2-3}=-7 \ne 0#

Hence, #x=-2# is valid.

In the same way we check if #x=9# is valid.

#\green{3 \cdot 9+3}=30 \ne 0#

#\orange{2 \cdot 9-3}=15 \ne 0#

Hence, #x=9# is valid.

Reduce equations

Linear fractional equations are not always of the form \[\frac{\blue{A}}{\green{B}}=\frac{\purple{C}}{\orange{D}}\] If they are not, we can use the normal reduction rules to get them into that form.

Example

\[\begin{array}{rcl}\dfrac{3x-2}{2x-6}+4&=&0 \\ \dfrac{3x-2}{2x-6}&=&-4 \end{array}\]

Solve:
\[{{9-2\cdot x}\over{7-x}}=-7\]
Give your answer in the form of #x=x_1#, in which #x_1# is a simplified fraction.

# x={{58}\over{9}} #

#\begin{array}{rcl}
\dfrac{9-2\cdot x}{7-x}&=& -7 \\
&&\phantom{xxx}\blue{\text{original equation }}\\
9-2\cdot x &=& -7 \cdot \left(7-x\right)\\
&&\phantom{xxx}\blue{\text{left and right multiplied by }7-x }\\
9-2\cdot x &=& 7\cdot x-49\\
&&\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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