Solve absolute value inequations of the form |ax + b| < k, and inequations with the unknown in the denominator, by multiplying through by the square of the denominator

What this dot point is asking

NESA wants you to solve two kinds of inequation that the linear and quadratic methods from earlier do not handle cleanly:

• absolute value inequations of the form $|ax + b| < k$ (and $\le$, $>$, $\ge$), and
• inequations with the unknown in the denominator, such as $\dfrac{x}{x - 2} \ge 3$.

The trap in both is the same: an absolute value hides two cases, and a denominator can be positive or negative, so the naive moves (drop the bars, or multiply both sides by the denominator) silently break the logic. This page gives the safe methods, the reasoning behind each, and the degenerate cases that catch people out.

A quick note on language: we solve an inequation such as $|2x + 1| > 3$, and we prove an inequality such as $(x - 2)^2 \ge 0$. NESA uses the word "inequality" for both, so do not be thrown if a question says "solve the inequality".

The answer

Absolute value inequations: three equivalent methods

The single idea behind every method is that $|x - a|$ is the distance from $x$ to $a$ on the number line, and $|x|$ is the distance from $x$ to $0$. There are three standard ways to solve $|ax + b| < k$, and a good candidate can move between them.

Method 1, graph it. Solve the equation $|ax + b| = k$ first, draw $y = |ax + b|$ (a V shape) and the horizontal line $y = k$ on one set of axes, and read the solution off the picture. Where the V is below the line solves $<$; where it is above solves $>$.

Method 2, distance on the number line. Rewrite the inequation so the coefficient of $x$ is $1$, then read it as a distance. The two preparatory steps are: force $a$ to be positive (for example write $|{-2x} + 3|$ as $|2x - 3|$, since $|-u| = |u|$), then divide through by the now-positive $a$.

Method 3, the algebraic rewrite. Strip the bars by splitting into the two cases the absolute value stands for:

• $|f(x)| = k$ becomes $f(x) = k$ or $f(x) = -k$;
• $|f(x)| < k$ becomes $-k < f(x) < k$ (a single sandwich);
• $|f(x)| > k$ becomes $f(x) < -k$ or $f(x) > k$ (two separate pieces).

All three give the same answer; pick the one that suits the question. The distance method is fastest for a number such as $|2x - 3| < 5$; the algebraic rewrite is essential once the inside is more complicated than $ax + b$.

The degenerate cases (where $k \le 0$)

An absolute value can never be negative. That single fact resolves the cases that have no "interval" answer:

• $|f(x)| = k$ and $|f(x)| < k$ have no solution when $k < 0$ (a distance cannot be negative).
• $|f(x)| > k$ and $|f(x)| \ge k$ are true for every $x$ in the domain when $k < 0$ (a distance always beats a negative number); the inequation is then an identity.
• $|f(x)| < 0$ has no solution, while $|f(x)| \ge 0$ is true for all $x$; and $|f(x)| \le 0$ holds only where $f(x) = 0$.

Always glance at the sign of the right-hand side before starting work. A question such as "solve $|3x + 2| < -7$" is testing exactly this: the answer is "no solutions", and it takes one line.

Inequations with the unknown in the denominator

Consider $\dfrac{5}{x - 4} \ge 1$. The tempting move is to multiply both sides by $x - 4$, but $x - 4$ is positive for $x > 4$ and negative for $x < 4$, and multiplying an inequality by a negative number reverses it. You would have to split into cases.

The clean way to avoid cases is to multiply through by the square of the denominator, which is positive (or zero) for every value of $x$, so the inequality sign never flips.

Worked schematically on $\dfrac{5}{x - 4} \ge 1$, with $x \ne 4$:

which gives $4 \le x \le 9$; discarding the undefined $x = 4$ leaves $4 < x \le 9$.

There is a second route. You can instead move everything to one side over a common denominator and analyse the sign of the resulting single fraction across its critical values (a sign table). That method is developed on the companion page the sign of a function, and applied to harder rational inequations in the Year 12 page polynomial and rational inequalities. The multiply-by-the-square method on this page is the most direct for a single fraction against a constant.

Solve a rational inequation, stage by stage

Once you have multiplied by the square of the denominator and reached a factored quadratic inequation, a sign diagram is the clearest way to finish and to show your working. Take $\dfrac{x}{x - 2} \ge 3$ again. Multiplying by $(x - 2)^2$ and collecting terms (as in the worked example below) turns it into $2(x - 2)(x - 3) \le 0$, equivalently $(x - 2)(x - 3) \le 0$, with the constraint $x \ne 2$ carried over from the original denominator. The critical values are $x = 2$ and $x = 3$.

Stage 1, mark the critical values. Each factor gives one zero: $x - 2 = 0$ at $x = 2$ and $x - 3 = 0$ at $x = 3$. Plot them; they cut the line into three intervals to test. Mark $x = 2$ as already excluded, because it is the value that made the original denominator zero, so it can never be a solution no matter which way the sign works out.

Stage 2, find the sign of each factor in every interval. Pick a convenient test point inside each interval, here $x = \tfrac{3}{2}, \tfrac{5}{2}, 4$, and record whether each factor is positive or negative. A linear factor $(x - r)$ is negative to the left of its own zero $r$ and positive to the right, so each row is a clean run of minuses then pluses.

Stage 3, multiply the columns and read off the solution. In each interval, multiply the two factor signs: $(-)(-) = +$, then $(+)(-) = -$, then $(+)(+) = +$, so the product row runs $+, -, +$. We need the product $\le 0$, which is the middle interval $[2, 3]$. Now fix the endpoints: drop $x = 2$ (undefined in the original), and keep $x = 3$ (where $\tfrac{3}{3 - 2} = 3$ satisfies $\ge 3$). The solution is $2 < x \le 3$.

How exam questions ask about these inequations

The wording tells you which tool to reach for:

• "Solve $|ax + b| < k$" (or $\le, >, \ge$). Use the distance method: make $a$ positive, divide by $a$, then read off one interval for $<$ or two pieces for $>$. If asked to "show on a number line", draw it; if asked to solve "graphically", sketch the V and the line $y = k$.
• "Solve $|ax + b| = k$, hence solve $|ax + b| > k$." The "hence" means: the two solutions of the equation are the boundary points, and the inequation is the region outside (for $>$) or inside (for $<$) them. Do not start the inequation from scratch.
• "Solve $\dfrac{ax + b}{cx + d} \ge m$" (a fraction against a constant). State the excluded value $cx + d \ne 0$, multiply by $(cx + d)^2$, bring everything to one side, factor without expanding, solve the resulting quadratic inequation, then fix the endpoints.
• "Simplify first, then solve", e.g. $1 + \dfrac{1}{x - 3} > -3$. Collect the constant onto the fraction or onto the other side first so there is a single fraction, then multiply by the square of the denominator.
• "Solve $|f(x)| < k$ where $k < 0$", or $|f(x)| \ge k$ with $k < 0$. This is the degenerate-case test: write "no solutions" or "all real $x$" with a one-line reason. Do not grind through algebra that produces an empty or universal set.
• "Solve $|ax + b| \ge cx + d$" (a pronumeral on both sides). The distance shortcut fails because the right side is not constant. Split into the two cases for the sign of $ax + b$ (or sketch both graphs and compare).

Method 2, stage by stage (the distance method)

The distance method is worth drilling because it is the fastest and the least error-prone. Take $|2x - 3| < 5$. The plan is: make the coefficient of $x$ equal to $1$, read the result as a distance, then mark the interval.

Stage 1, force the coefficient of $x$ to be $1$. The coefficient here is already positive, so no sign fix is needed. Factor the $2$ out of the absolute value using $|2u| = 2|u|$: since $2x - 3 = 2\left(x - \tfrac{3}{2}\right)$, we have $|2x - 3| = 2\left|x - \tfrac{3}{2}\right|$. Dividing both sides of $2\left|x - \tfrac{3}{2}\right| < 5$ by the positive number $2$ gives $\left|x - \tfrac{3}{2}\right| < \tfrac{5}{2}$.

Stage 2, read it as a distance and mark the interval. Now $\left|x - \tfrac{3}{2}\right| < \tfrac{5}{2}$ says the distance from $x$ to $\tfrac{3}{2}$ is less than $\tfrac{5}{2}$. So $x$ lies strictly within $\tfrac{5}{2}$ of $\tfrac{3}{2}$, one unbroken interval. The endpoints are $\tfrac{3}{2} - \tfrac{5}{2} = -1$ and $\tfrac{3}{2} + \tfrac{5}{2} = 4$, both excluded because the inequality is strict. The solution is $-1 < x < 4$.

If the inequality had been $\ge$ instead of $<$, the same boundaries $-1$ and $4$ would apply, but the solution would be the two outside pieces $x \le -1$ or $x \ge 4$, with the endpoints now included (closed circles).