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 $\frac{7x+20}{7x-5}\lt -3$ .

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.

$x\gt -{{5}\over{28}} \land x\lt {{5}\over{7}}$

As we cannot divide by zero, $7x-5 \ne 0$ , and hence $x= {{5}\over{7}}$ is not a solution to the inequality. To solve the inequality, we distinguish according to the sign of $7x-5$ , that is, between $x\gt {{5}\over{7}}$ and $x\lt {{5}\over{7}}$ . Multilplying both sides of $\frac{7x+20}{7x-5}\lt -3$ by $7x-5$ in each case, we see that the given inequality is equivalent to $(x\gt {{5}\over{7}}\land 7x+20 \lt -3\cdot \left(7\cdot x-5\right)) \lor (x\lt {{5}\over{7}}\land 7x+20 \gt -3\cdot \left(7\cdot x-5\right))$ . Now the usual Rules of calculation for inequalities imply that this is equivalent to $(x\gt {{5}\over{7}}\land x \lt -{{5}\over{28}}) \lor (x\lt {{5}\over{7}}\land x \gt -{{5}\over{28}})$ . As ${{5}\over{7}} \ge -{{5}\over{28}}$ , we find $x\gt -{{5}\over{28}} \land x\lt {{5}\over{7}}$ .

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Linear Inequalities: Variations

Fractional linear inequalities

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