Revise Solving equations and inequalities; modulus functions for IB Mathematics: Analysis and Approaches (2.8)

To solve an inequality such as $x^2 - x - 6 > 0$, never guess from a couple of test points. Instead, rearrange so one side is zero, find the critical values where the expression equals zero, and use them to split the number line into intervals. Across each interval the sign of the expression cannot change, so testing a single point in each interval tells you the sign everywhere in it. This picture is called a sign diagram.

A rational inequality involves a fraction, such as $\frac{x-1}{x+2}\ge 0$. The dangerous temptation is to multiply through by the denominator — but its sign is unknown, so you cannot tell whether the inequality flips. The reliable method keeps the fraction intact: move everything to one side so you are comparing a single fraction with zero, then find the critical values from both the numerator (where the fraction is zero) and the denominator (where it is undefined). Build a sign diagram exactly as before, remembering that a value making the denominator zero can never be part of the solution.

The modulus (or absolute value) of a real number $x$, written $|x|$, is its size regardless of sign: $|5|=5$ and $|-5|=5$. Formally it is defined in two pieces.

$|x| = \begin{cases} x, & \text{if } x \ge 0 \\ -x, & \text{if } x < 0 \end{cases}$

The graph of $y=|x|$ is a 'V' with its vertex at the origin, its right arm the line $y=x$ and its left arm the line $y=-x$. Two transformations follow from this. To sketch $y=|f(x)|$, draw $y=f(x)$ and reflect any part below the $x$-axis up into it — the graph can then never be negative. To sketch $y=f(|x|)$, keep only the part of $y=f(x)$ with $x\ge0$ and reflect it in the $y$-axis, producing a graph symmetric about the $y$-axis.

Two standard methods solve modulus equations. The case method uses the definition directly: $|A|=c$ (with $c\ge0$) means $A=c$ or $A=-c$. The squaring method uses $|A|^2=A^2$, so $|A|=|B|$ becomes $A^2=B^2$; this is tidy for equations of the form $|A|=|B|$ but can introduce extraneous solutions elsewhere, so check answers when you square. Always watch for equations like $|A|=-5$, which have no solution because a modulus is never negative.

On Paper 1 the marks are analytic: each is tied to a specific line of your working. A method mark (M) rewards a correct approach — a valid equation, the right rearrangement — even if the arithmetic later slips. An accuracy mark (A) rewards a correct result and is dependent on the method mark it follows: you cannot earn the A without the M being present. Two further conventions protect you: follow-through (FT) lets a correct step after an earlier error still score, and 'ISW' (ignore subsequent working) means that once you have shown the right answer, a clumsy final restatement will not cost you. The engine also accepts equivalent forms — an inequality or the same set in interval notation score equally. Study how each mark below is pinned to one line.

Modulus inequalities split by direction. $|X|<c$ means the distance of $X$ from zero is less than $c$, so $X$ lies between $-c$ and $c$: one connected interval. $|X|>c$ means the distance exceeds $c$, so $X$ is beyond $c$ or beyond $-c$: two separate pieces. Getting these two shapes the right way round — 'and' for less-than, 'or' for greater-than — is the single most examined point in this subtopic.

$\text{For a positive constant } c:\\ |X| < c \iff -c < X < c\\ |X| > c \iff X < -c \ \text{ or }\ X > c$

Every equation $f(x)=g(x)$ can be read off a graph: the solutions are the $x$-coordinates of the points where $y=f(x)$ and $y=g(x)$ cross. Inequalities follow the same picture — $f(x)>g(x)$ holds exactly where the $f$ graph sits above the $g$ graph. This is especially handy for modulus problems such as $|x-3|>|x+1|$: sketch both V-shapes, find where they meet, and read the interval where the first is higher. When a question says 'sketch, hence solve', it is inviting you to use exactly this method.

A system of equations is two or more equations that must hold at once; its solution is every point common to all of them. Graphically, that is the intersection of the graphs. Two straight lines usually meet at one point (solved quickly by substitution or elimination); a line and a curve may meet at two points, one, or none. For example, solving $y=x^2$ together with $y=x+2$ by substitution gives $x^2=x+2$, so $x^2-x-2=0$, $(x-2)(x+1)=0$, giving the two intersection points $(2,4)$ and $(-1,1)$ — precisely where the line cuts the parabola.

Sign diagrams reappear whenever calculus asks where a derivative is positive or negative — that is, where a function increases or decreases — so the critical-value habit you built here transfers straight into optimisation. Modulus reasoning underpins later work on distance, bounds and error, and the graphical view of intersections is the foundation for solving systems and for reading solutions off any pair of curves. Master the routine — rearrange to compare with zero, find the critical values, test the signs, and match your brackets to the inequality — and the harder equation-solving later in the course becomes variations on a method you already own.