Fractional linear inequalities

Linear Inequalities: Variations

Fractional linear inequalities

Just as in the case of equality in the chapter Linear equations with one unknown, inequalities in which quotients of linear expressions #x# occur, can be solved by reduction to linear inequalities. Below a few examples are given.

A fraction in the form #\dfrac{ax+b}{cx+d}#, where #a#, #b#, #c#, and #d# are fixed numbers with at least one of #c# and #d# distinct from zero, is called a fractional linear expression in #x#.

The case in which #c# and #d# are both zero is excluded because in that case the denominator of the fraction would be equal to zero.

The case in which #c=0# corresponds to a linear expression in #x#, namely, #\frac{a}{d}x+\frac{b}{d}#.

We recall that a set of inequalities (with operators like '#\lor#' and '#\land#', which stand for "or" and "and") is equivalent to another set if the two composite inequalities have the same solution set.

Let #\dfrac{ax+b}{cx+d}# be a fractional linear expression in #x#. Inequality #\dfrac{ax+b}{cx+d}\ge0# is equivalent to \[\left(ax+b\ge0\land cx+d\gt0\right)\lor\left(ax+b\le0\land cx+d\lt0\right)\tiny.\]

fractional linear function has a denominator equal to #0# and #cx+d=0#. Therefore #cx+d\gt0\lor cx+d\lt0# is satisfied.

Suppose #cx+d\gt0#. According to the theorem Ordering of rational numbers #\dfrac{ax+b}{cx+d}\ge0# is true if and only if #{ax+b}\ge0#. In this case, therefore, we have #ax+b\ge0\land cx+d\gt0#.

Now assume #cx+d\lt0#. According to the last part of the above-mentioned theorem #\dfrac{ax+b}{cx+d}\ge0# holds if and only if #{ax+b}\le0#. In this case, therefore, we have #ax+b\le0\land cx+d\lt0#.

We conclude that #\left(ax+b\ge0\land cx+d\gt0\right)\lor\left(ax+b\le0\land cx+d\lt0\right)#.

Determine all #x# for which #\frac{8x-27}{9x-9}\lt -6#.

Provide your answer in the form #x \lt a\lor x\gt a# or #\left(x\lt a\right)\lor\left(x\gt b\right)# or # \left(x \gt a\right)\land \left(x \lt b\right)#, where #a\lt b#, or similar.

#x\gt 1 \land x\lt {{81}\over{62}}#

As we cannot divide by zero, #9x-9 \ne 0#, and hence #x= 1# is not a solution to the inequality. To solve the inequality, we distinguish according to the sign of #9x-9#, that is, between #x\gt 1# and # x\lt 1#. Multilplying both sides of #\frac{8x-27}{9x-9}\lt -6# by #9x-9# in each case, we see that the given inequality is equivalent to #(x\gt 1\land 8x-27 \lt -6\cdot \left(9\cdot x-9\right)) \lor (x\lt 1\land 8x-27 \gt -6\cdot \left(9\cdot x-9\right)) #. Now the usual Rules of calculation for inequalities imply that this is equivalent to # (x\gt 1\land x \lt {{81}\over{62}}) \lor (x\lt 1\land x \gt {{81}\over{62}}) #. As # 1 \le {{81}\over{62}}#, we find #x\gt 1 \land x\lt {{81}\over{62}}#.

New example