Fractional linear equations

Linear Equations with a Single Unknown: Variations

Fractional linear equations

A fractional linear function is an expression of the form #\dfrac{ax+b}{cx+d}#, where #a#, #b#, #c#, and #d# are fixed numbers and #x# is a variable.

The equation \[\dfrac{ax+b}{cx+d} = w\] with unknown #x# is called a fractional linear equation.

The equation \[\dfrac{ax+b}{cx+d} = w\] with unknown #x# deals with the question for which #x# the function #\frac{ax+b}{cx+d}# with variable #x# attains the value #w#. In the chapter Functions the notion function will be discussed.

The fractional linear function is a quotient of the linear function #ax+b# by the linear function #cx+d#. The function is not defined if #x# satisfies #cx+d=0# (for, then the denominator equals #0#). So you need to solve a linear equation to find out where the function is defined.

The fractional linear equation \[\dfrac{ax+b}{cx+d} = w\] is equivalent to the system \[ax+b=w\cdot (cx+d)\land cx+d\ne0\tiny.\]

This can be seen by observing that the fraction \(c x + d=0\) is not defined if #cx+d=0# and by multiplying both sides by # c x +d#.

In this way the solving of a fractional linear equation is reduced to solving a linear equation and selecting the solution(s) #x# with #cx+d\ne0#.

It may happen that the fractional linear equation has no solution, but also that every #x# is a solution. Below you see the graphs of #y = \dfrac{ax+b}{cx+d}# and #y = w#. Solutions to #\dfrac{ax+b}{cx+d}= w# are the #x#-coordinates of point where the two graphs meet. Move the sliders to get an idea how many solutions of the equation #\dfrac{ax+b}{cx+d} = w# with unknown #x# to expect.

Solve for #x# in the equation #\;\dfrac{8 x + 19}{12 x + 51} = 1#.

#x = -8#

This is done by

first multiplying the left and right hand side by #(12 x + 51)# ;

#\;{8 x + 19} = 12 x + 51#.

then move all terms containing #x# to the left hand side and all constant terms to the right hand side:

#(8-12) x = 51-19#;

next divide by #(8-12)=-4# ;

#\displaystyle x=-8#

and finally check that the solution #x=-8# will not be zero in #12 x + 51# (the denominator in the left hand side of the equation #\;\dfrac{8 x + 19}{12 x + 51} = 1# cannot be equal to zero).

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