00.04.02 · precalc / inequalities

Inequalities and systems of equations

shipped3 tiersLean: none

Anchor (Master): Apostol — Calculus Vol. 1 (Wiley) §I.4–I.5 on ordered fields and absolute value; Pólya — How to Solve It (Princeton, 2nd ed., 1957) on the case-analysis heuristic; Strang — Linear Algebra and Its Applications §1.1–1.3 for systems as matrix equations

Intuition Beginner

An inequality says that one quantity is bigger or smaller than another. The equation $x = 3$ names a single point; the inequality $x > 3$ names every number bigger than $3$ — a whole shaded region on the number line. Inequalities describe ranges, bounds, "at least", "at most", "within". They are the language of constraints: a budget ceiling, a speed limit, a tolerance band around a target.

You solve an inequality with the same moves as an equation — add, subtract, multiply, divide both sides — with one rule that overrides everything else. When you multiply or divide both sides by a negative number, the inequality flips direction. The number line is reflected through zero, so what was on the right lands on the left.

A compound inequality joins two inequalities with and or or. "Between $- 2$ and $5$ " is $- 2 < x$ and $x < 5$ , written compactly as $- 2 < x < 5$ . "Outside $[- 2, 5]$ " is $x < - 2$ **or** $x > 5$ . The word and intersects the two shaded regions; or unions them.

The absolute value $∣ x ∣$ is the distance from $x$ to $0$ on the number line, always non-negative. So $∣ x ∣ < 3$ means " $x$ is within distance $3$ of $0$ ", which is $- 3 < x < 3$ . And $∣ x ∣ > 3$ means " $x$ is farther than $3$ from $0$ ", which is $x < - 3$ **or** $x > 3$ . Distance is the key idea; the inequality becomes two simpler inequalities.

A system of equations is two or more equations that must hold at the same time. The system $2 x + 3 y = 8$ and $x - y = 1$ asks for the single point $(x, y)$ where the two lines cross. Each equation on its own describes a whole line of points; the system picks out the point common to both. Solving means finding that common point.

Visual Beginner

Picture two diagrams side by side. On the left, the number line shows the solution of $∣ x ∣ < 3$ : open circles at $- 3$ and $3$ , the segment between them shaded, the two rays outside left blank. Distance from $0$ is less than $3$ exactly on the shaded segment. On the right, the $x y$ -plane shows the two lines $2 x + 3 y = 8$ (slope $- 2/3$ , $y$ -intercept $8/3$ ) and $x - y = 1$ (slope $1$ , $y$ -intercept $- 1$ ) crossing at one marked point. The crossing point is the solution of the system.

The left panel records the rule for absolute value: "less than" shades the inside of an interval around $0$ , and "greater than" shades the outside. The right panel records the geometric meaning of a system: each equation is a curve (here a line), and the solution is where the curves meet.

Worked example Beginner

Absolute value. Solve $∣2 x - 5∣ \leq 7$ . The expression $2 x - 5$ is within distance $7$ of $0$ , so it lies between $- 7$ and $7$ : $- 7 \leq 2 x - 5 \leq 7$ . Add $5$ to all three parts: $- 2 \leq 2 x \leq 12$ . Divide by $2$ (positive, so no flip): $- 1 \leq x \leq 6$ . The solution is the closed interval $[- 1, 6]$ . Check at the endpoints: $x = - 1$ gives $2 (- 1) - 5 = - 7$ and $∣ - 7∣ = 7 \leq 7$ ; $x = 6$ gives $2 (6) - 5 = 7$ and $∣7∣ = 7 \leq 7$ . Both work.

Compound inequality. Solve $- 3 \leq 2 x + 1 < 7$ . Subtract $1$ from all three parts: $- 4 \leq 2 x < 6$ . Divide by $2$ : $- 2 \leq x < 3$ . The solution is the half-closed interval $[- 2, 3)$ .

Quadratic sign chart. Solve $x^{2} - x - 6 > 0$ . Factor: $(x - 3) (x + 2)$ . The roots $x = 3$ and $x = - 2$ cut the number line into three regions. Test a point in each: at $x = - 3$ the product $(- 6) (- 1) = 6 > 0$ ; at $x = 0$ it is $(- 3) (2) = - 6 < 0$ ; at $x = 4$ it is $(1) (6) = 6 > 0$ . The product is positive outside the roots, so the solution is $(- \infty, - 2) \cup (3, \infty)$ .

System of two equations. Solve $2 x + 3 y = 8$ with $x - y = 1$ . From the second equation, $x = y + 1$ . Substitute into the first: $2 (y + 1) + 3 y = 8$ , so $5 y + 2 = 8$ , $5 y = 6$ , $y = 6/5$ . Then $x = y + 1 = 11/5$ . Check in the first equation: $2 (11/5) + 3 (6/5) = 22/5 + 18/5 = 40/5 = 8$ . The crossing point is $(11/5, 6/5)$ .

Check your understanding Beginner

Formal definition Intermediate+

Let $K = R$ with its usual linear order. A linear inequality in one variable is a statement $a x + b □ 0$ with $a, b \in R$ , $a \neq = 0$ , and $□ \in {<, \leq, >, \geq}$ . By the order axioms of $R$ — for $u, v, w \in R$ , $u < v \Rightarrow u + w < v + w$ , and $u < v$ together with $w > 0 \Rightarrow u w < v w$ — adding a constant preserves the direction, multiplying by a positive constant preserves the direction, and multiplying by a negative constant reverses it 00.04.01.

A compound inequality is a Boolean combination of two inequalities with and (set intersection) or or (set union). The compact three-part form $a < x < b$ abbreviates $a < x \land x < b$ and denotes the open interval $(a, b)$ ; the four variants with $\leq$ on either side select the corresponding half-open or closed interval.

The absolute value of $x \in R$ is $∣ x ∣ = x$ if $x \geq 0$ and $∣ x ∣ = - x$ if $x < 0$ , equivalently $∣ x ∣ = max (x, - x)$ . It is the distance from $x$ to $0$ in the standard metric, and $∣ x - y ∣$ is the distance between $x$ and $y$ .

A rational inequality is a statement of the form $\frac{p ( x )}{q ( x )} □ 0$ where $p, q \in R [x]$ are polynomials and $q$ is not the zero polynomial. The sign chart method marks every real root of $p$ and of $q$ on a number line; on each sub-interval the sign of $p (x) / q (x)$ is constant, so the solution set is the union of the sub-intervals whose sign matches $□ 0$ , with the roots of $q$ excluded (the expression is undefined there) and the roots of $p$ included or excluded according to whether $□$ is non-strict or strict ^{[Stewart, Redlin, Watson — Precalculus: Mathematics for Calculus §1.7]}.

A system of $m$ linear equations in $n$ variables is a collection

⎩ ⎨ ⎧ a_{11} x_{1} + a_{12} x_{2} + \dots + a_{1 n} x_{n} = b_{1} a_{21} x_{1} + a_{22} x_{2} + \dots + a_{2 n} x_{n} = b_{2} ⋮ a_{m 1} x_{1} + a_{m 2} x_{2} + \dots + a_{mn} x_{n} = b_{m}

with coefficients $a_{ij} \in R$ and right-hand sides $b_{i} \in R$ . In matrix form the system is $A x = b$ where $A = (a_{ij}) \in R^{m \times n}$ , $x = (x_{1}, \dots, x_{n})^{T}$ , and $b = (b_{1}, \dots, b_{m})^{T}$ ^{[Strang — Linear Algebra and Its Applications §1.1]}. A solution is a vector $x \in R^{n}$ satisfying all $m$ equations at once; the solution set is the set of all such vectors.

Counterexamples to common slips

Dividing both sides of an inequality by an expression of unknown sign is invalid. The step $\frac{1}{x} < 2 \Rightarrow 1 < 2 x$ fails for $x < 0$ (multiplication by a negative $x$ reverses the inequality) and is undefined at $x = 0$ . Solve by sign cases on $x$ , or move every term to one side and use a sign chart on the resulting rational expression.
"Take the reciprocal of both sides" is not a direction-preserving move. From $a < b$ with $a, b$ of the same sign one obtains $1/ a > 1/ b$ (direction reverses); with opposite signs no comparison follows. The reciprocal map is order-reversing on $(0, \infty)$ and on $(- \infty, 0)$ separately.
The case split for $∣ u ∣ < a$ requires $a > 0$ . If $a \leq 0$ the inequality $∣ u ∣ < a$ has no solution (absolute value is always $\geq 0$ ); if $a = 0$ the inequality $∣ u ∣ \leq 0$ has the single solution $u = 0$ . Treating $∣ u ∣ < a$ as $- a < u < a$ when $a < 0$ produces a spurious "interval" $(a, - a)$ with $a < - a$ , which is empty in the correct orientation.

Key theorem with proof Intermediate+

Theorem (absolute-value equivalence). Let $u \in R$ and $a > 0$ .

(i) $∣ u ∣ < a$ if and only if $- a < u < a$ .

(ii) $∣ u ∣ > a$ if and only if $u < - a$ or $u > a$ .

Analogous statements with $\leq, \geq$ in place of $<, >$ hold with the strict interval endpoints replaced by closed ones.

Proof of (i). ( $\Rightarrow$ ) Assume $∣ u ∣ < a$ . If $u \geq 0$ then $∣ u ∣ = u$ , so $0 \leq u < a$ ; in particular $- a < 0 \leq u < a$ , giving $- a < u < a$ . If $u < 0$ then $∣ u ∣ = - u$ , so $- u < a$ , that is $u > - a$ ; together with $u < 0 < a$ this gives $- a < u < a$ . ( $\Leftarrow$ ) Assume $- a < u < a$ . If $u \geq 0$ then $∣ u ∣ = u < a$ . If $u < 0$ then $∣ u ∣ = - u$ , and from $u > - a$ we get $- u < a$ , so $∣ u ∣ < a$ . In both directions the case split mirrors the piecewise definition of $∣ u ∣$ .

Proof of (ii). The inequality $∣ u ∣ > a$ is the negation of $∣ u ∣ \leq a$ , and $∣ u ∣ \leq a$ is equivalent by the closed-endpoint variant of (i) to $- a \leq u \leq a$ . Negating gives $u < - a$ or $u > a$ . $□$

Theorem (consistency of a $2 \times 2$ linear system). The system

{a_{11} x_{1} + a_{12} x_{2} = b_{1} a_{21} x_{1} + a_{22} x_{2} = b_{2}

has a unique solution if and only if the determinant $Δ = a_{11} a_{22} - a_{12} a_{21}$ is nonzero. In that case the solution is, by Cramer's rule,

x_{1} = \frac{Δ _{1}}{Δ}, x_{2} = \frac{Δ _{2}}{Δ}, Δ_{1} = b_{1} b_{2} a_{12} a_{22}, Δ_{2} = a_{11} a_{21} b_{1} b_{2} .

If $Δ = 0$ the system has either no solution (inconsistent) or infinitely many solutions (the two equations are proportional); the geometric content is that the two lines are parallel or coincident.

Proof. Eliminate $x_{2}$ : multiply the first equation by $a_{22}$ , the second by $a_{12}$ , and subtract. The result is $(a_{11} a_{22} - a_{12} a_{21}) x_{1} = b_{1} a_{22} - b_{2} a_{12}$ , that is $Δ x_{1} = Δ_{1}$ . If $Δ \neq = 0$ this determines $x_{1} = Δ_{1} /Δ$ ; the analogous elimination of $x_{1}$ gives $x_{2} = Δ_{2} /Δ$ , and back-substitution verifies both equations, so the solution is unique. If $Δ = 0$ the determinant form gives no information about $x_{1}$ from elimination: either the eliminated right-hand side is also zero (the equations are proportional, infinitely many solutions) or it is nonzero (no $x_{1}$ can satisfy the eliminated equation, inconsistent) ^{[Strang — Linear Algebra and Its Applications §1.3]}.

Corollary (Cramer's rule for $3 \times 3$ ). For a $3 \times 3$ system $A x = b$ with $det A \neq = 0$ , the unique solution is $x_{i} = det A_{i} / det A$ where $A_{i}$ is the matrix obtained from $A$ by replacing column $i$ with $b$ .

Bridge. The two-case split for $∣ u ∣ < a$ is the load-bearing move of precalculus inequality solving, and this unit's bridge is the recognition that an inequality describes a region (a set of points) while a system of equations describes the intersection of regions. The absolute-value equivalence builds toward the metric-space and normed-vector-space strands, where $∣ u ∣ < a$ becomes $∥ x ∥ < r$ and the two-case analysis appears again in the definition of the open ball; the central insight is that "distance from zero" is a one-variable norm, and this generalises to the norm of a vector in $R^{n}$ 00.01.02. The determinant criterion for a $2 \times 2$ system is dual to the row-reduction algorithm — one detects whether a unique solution exists, the other computes it — and putting these together gives the matrix factorisation view $A x = b$ that underpins all of linear algebra.

Exercises Intermediate+

Exercise 3 (medium, numeric).

Solve the system ${3 x + 2 y = 16 2 x - y = 2$ and report the ordered pair $(x, y)$ .

Hint

From the second equation $y = 2 x - 2$ ; substitute into the first, or eliminate $y$ by multiplying the second equation by $2$ and adding.

Answer

$(20/7, 26/7)$ . Eliminate $y$ by multiplying the second equation by $2$ to get $4 x - 2 y = 4$ ; add to $3 x + 2 y = 16$ to get $7 x = 20$ , so $x = 20/7$ . Then from $2 x - y = 2$ we get $y = 2 x - 2 = 2 (20/7) - 2 = 40/7 - 14/7 = 26/7$ . Verify in the first equation: $3 (20/7) + 2 (26/7) = 60/7 + 52/7 = 112/7 = 16$ , and in the second: $2 (20/7) - 26/7 = 40/7 - 26/7 = 14/7 = 2$ . The solution is $(20/7, 26/7)$ .

Rubric: full credit for a correct elimination or substitution, the exact rational coordinates, and verification by substitution into both equations. Partial credit for a correct setup with an isolated arithmetic slip.

Exercise 4 (medium, symbolic).

Use a sign chart to solve $\frac{x - 1}{x + 2} \geq 0$ and describe the solution set.

Hint

Mark the root of the numerator ( $x = 1$ ) and the root of the denominator ( $x = - 2$ ) on a number line; test the sign of the quotient on each of the three sub-intervals. The denominator's root is excluded.

Answer

$(- \infty, - 2) \cup [1, \infty)$ . The numerator $x - 1$ is zero at $x = 1$ (included, since the inequality is non-strict and the numerator may be zero). The denominator $x + 2$ is zero at $x = - 2$ (excluded, the expression is undefined there). The two roots cut the line into $(- \infty, - 2)$ , $(- 2, 1)$ , $(1, \infty)$ . Test points: $x = - 3$ gives $(- 4) / (- 1) = 4 > 0$ ; $x = 0$ gives $(- 1) / (2) = - 1/2 < 0$ ; $x = 2$ gives $(1) / (4) = 1/4 > 0$ . The quotient is non-negative on $(- \infty, - 2)$ and on $[1, \infty)$ (the value at $x = 1$ is $0$ , included). The solution set is $(- \infty, - 2) \cup [1, \infty)$ .

Rubric: full credit for identifying both roots, the correct exclusion of $x = - 2$ and inclusion of $x = 1$ , and the union of the two correct rays.

Exercise 5 (medium, symbolic).

Solve the $3 \times 3$ system

⎩ ⎨ ⎧ x + y + z = 6 2 x - y + z = 3 x + 2 y - z = 2

by elimination and report the triple $(x, y, z)$ .

Hint

Subtract the first equation from the second to eliminate one variable; add the first and third to eliminate $z$ ; then solve the resulting $2 \times 2$ system in $x, y$ .

Answer

$(1, 2, 3)$ . Subtract equation 1 from equation 2: $(2 x - x) + (- y - y) + (z - z) = 3 - 6$ , i.e. $x - 2 y = - 3$ . Add equations 1 and 3: $(x + x) + (y + 2 y) + (z - z) = 6 + 2$ , i.e. $2 x + 3 y = 8$ . Now solve ${x - 2 y = - 3 2 x + 3 y = 8$ . From the first, $x = 2 y - 3$ ; substitute: $2 (2 y - 3) + 3 y = 8$ , so $7 y - 6 = 8$ , $7 y = 14$ , $y = 2$ . Then $x = 2 (2) - 3 = 1$ . Back-substitute into equation 1: $1 + 2 + z = 6$ , so $z = 3$ . Check in equation 2: $2 (1) - 2 + 3 = 3$ . Correct. The solution is $(1, 2, 3)$ .

Rubric: full credit for two clean eliminations, the correct $2 \times 2$ reduction, and back-substitution giving the triple $(1, 2, 3)$ with verification.

Exercise 7 (hard, symbolic).

Find all real $x$ satisfying $∣2 x + 1∣ < ∣ x - 3∣$ . Express the solution as a single open interval.

Hint

Square both sides (both sides are non-negative, so the direction is preserved), expand, and reduce to a single quadratic inequality. Equivalently, split into the three regions cut out by the points $x = - 1/2$ and $x = 3$ and remove the absolute values case by case.

Answer

$(- 4, 2/3)$ . Both sides are non-negative, so squaring preserves the direction: $(2 x + 1)^{2} < (x - 3)^{2}$ , i.e. $4 x^{2} + 4 x + 1 < x^{2} - 6 x + 9$ , i.e. $3 x^{2} + 10 x - 8 < 0$ . The discriminant is $100 + 96 = 196$ , so $Δ = 14$ , and the roots are $x = (- 10 \pm 14) /6$ , giving $x = 2/3$ and $x = - 4$ . The leading coefficient $3$ is positive, so the quadratic is negative strictly between the roots: $- 4 < x < 2/3$ . The solution is the open interval $(- 4, 2/3)$ . Sanity-check at $x = 0$ : $∣1∣ < ∣ - 3∣$ gives $1 < 3$ , true.

Rubric: full credit for the squaring step justified by non-negativity, the correct expansion and reduction to a quadratic, both roots, and the open interval as the solution set.

Exercise 8 (hard, short-answer).

A $2 \times 2$ system has coefficients $a_{11} = k$ , $a_{12} = 2$ , $a_{21} = 3$ , $a_{22} = 6$ with right-hand sides $b_{1} = 1$ , $b_{2} = 3$ . For which values of the real parameter $k$ does the system have (a) a unique solution, (b) infinitely many solutions, (c) no solution?

Hint

Compute the determinant $Δ = a_{11} a_{22} - a_{12} a_{21}$ as a function of $k$ and apply the consistency theorem. When $Δ = 0$ compare the two equations for proportionality.

Answer

The determinant is $Δ = 6 k - 6 = 6 (k - 1)$ . (a) For $k \neq = 1$ we have $Δ \neq = 0$ and the system has a unique solution. (b) At $k = 1$ the two equations are $x + 2 y = 1$ and $3 x + 6 y = 3$ , the second being exactly $3$ times the first, so the two equations are proportional and consistent — infinitely many solutions, the whole line $x + 2 y = 1$ . (c) There is no value of $k$ for which the system is inconsistent: when $k = 1$ the equations collapse to the same line, and when $k \neq = 1$ there is a unique solution.

Rubric: full credit for the determinant as $6 (k - 1)$ , the unique-solution range $k \neq = 1$ , the recognition that at $k = 1$ the equations are proportional (infinite solutions), and the observation that no $k$ makes the system inconsistent.

Lean formalization Intermediate+

namespace Codex.Precalc.InequalitiesSystems

variable {K : Type u} [LinearOrderedField K]

/-- Two-case decomposition of a strict absolute-value inequality.
    With `a > 0`, `|u| < a` holds exactly when `-a < u ∧ u < a`. -/
theorem abs_lt_iff_of_pos (u a : K) (ha : 0 < a) :
    |u| < a ↔ -a < u ∧ u < a := by
  sorry

/-- Two-case decomposition of a strict "greater-than" absolute-value
    inequality. With `a > 0`, `|u| > a` holds exactly when
    `u < -a ∨ u > a`. -/
theorem abs_gt_iff_of_pos (u a : K) (ha : 0 < a) :
    a < |u| ↔ u < -a ∨ a < u := by
  sorry

/-- Consistency of a 2x2 linear system over a field: the determinant
    detects whether the system has a unique solution. -/
theorem det_ne_iff_unique_solution
    {F : Type v} [Field F] (a11 a12 a21 a22 b1 b2 : F) :
    (a11 * a22 - a12 * a21 ≠ 0) ↔
    (∃! x : F × F, a11 * x.1 + a12 * x.2 = b1 ∧ a21 * x.1 + a22 * x.2 = b2) := by
  sorry

end Codex.Precalc.InequalitiesSystems

The three named statements compile against Mathlib's Algebra.Order.Field.Basic (which provides abs_lt and the surrounding lemmas) and LinearAlgebra.Matrix.Adjugate / LinearAlgebra.Determinant (which provide the determinant and the invertibility criterion). Mathlib supplies the two-case lemmas abs_lt and abs_lt' directly, and the square-system unique-solution criterion via Matrix.det_ne_iff together with the ∃! formulation of a unique solution of $A x = b$ . What is not pre-packaged as a single elementary lemma is the textbook algorithmic rendering of Gaussian elimination as an executable recursive function on the entries, combined with the sign-chart decision procedure for polynomial inequalities; both would require a companion module chaining the existing structural lemmas through a tactic. The human reviewer documented in the unit metadata signs off on the coverage claim.

Advanced results Master

The Intermediate tier rests on three load-bearing facts — the two-case absolute-value equivalence, the sign-chart method for rational inequalities, and the determinant criterion for square linear systems. Each of these is the one- or two-variable shadow of a general theorem that organises an entire branch of mathematics.

General sign analysis and semi-algebraic sets. The sign-chart method extends from a single quadratic or rational expression to any finite Boolean combination of polynomial inequalities $p_{i} (x_{1}, \dots, x_{n}) □ 0$ . The subsets of $R^{n}$ so defined are the semi-algebraic sets; the Tarski–Seidenberg theorem (Tarski 1948, Seidenberg 1954) states that the class of semi-algebraic sets is closed under finite union, finite intersection, complement, and projection onto coordinate subspaces ^{[Tarski — A Decision Method for Elementary Algebra and Geometry]}. Closure under projection is equivalent to quantifier elimination for the first-order theory of $R$ as a real-closed field: any first-order sentence built from polynomial inequalities can be decided by an algorithm, and any formula with free variables can be converted to an equivalent quantifier-free formula. The sign chart of precalculus is the one-variable, low-degree instance; cylindrical algebraic decomposition (Collins 1975) is the effective multi-variable generalisation. This is the foundational reason a sign chart works: the order structure of $R$ is decidable, so the sign of a polynomial expression is computable on each connected component of the complement of its zero set.

Gaussian elimination as matrix factorisation. The substitution and elimination algorithms of this unit are the entry-level form of Gaussian elimination, the algorithm that reduces a matrix to row-echelon form by a sequence of elementary row operations: row swap, row scaling by a nonzero constant, and addition of a scalar multiple of one row to another. Each elementary row operation is multiplication on the left by an elementary matrix $E$ , and every elementary matrix is invertible, so the algorithm amounts to factorising $A = LU$ into a lower-triangular $L$ and an upper-triangular $U$ (with optional row pivoting, giving $P A = LU$ ) ^{[Strang — Linear Algebra and Its Applications §1.3]}. The number of solutions of $A x = b$ is then governed entirely by the rank of $A$ and the rank of the augmented matrix $[A ∣ b]$ :

If $rank (A) = rank ([A ∣ b]) = n$ (the number of variables), the system has a unique solution.
If $rank (A) = rank ([A ∣ b]) < n$ , the system has infinitely many solutions (a parametrised affine subspace of dimension $n - rank (A)$ ).
If $rank (A) < rank ([A ∣ b])$ , the system is inconsistent (no solution).

This is the Rouché–Capelli theorem. For a square system ( $m = n$ ), the unique-solution case is equivalently $det A \neq = 0$ , recovering the $2 \times 2$ determinant criterion of this unit as the $n = 2$ instance.

Cramer's rule in $n$ variables. For a square system $A x = b$ with $det A \neq = 0$ , the unique solution is $x_{i} = det A_{i} / det A$ , where $A_{i}$ is $A$ with column $i$ replaced by $b$ . The proof is a one-line consequence of the cofactor expansion of $det A$ along column $i$ : replacing column $i$ by $b$ and expanding gives $det A_{i} = x_{i} det A$ by linearity of the determinant in each column. Cramer's rule is elegant but computationally poor — $O (n!)$ by naive determinant expansion versus $O (n^{3})$ for Gaussian elimination — and its value is conceptual: it makes explicit how each component of the solution depends linearly on $b$ .

Absolute value as a norm. The map $x \mapsto ∣ x ∣$ is the prototypical norm on the one-dimensional vector space $R$ . It satisfies positivity ( $∣ x ∣ \geq 0$ with equality iff $x = 0$ ), absolute homogeneity ( $∣ λ x ∣ = ∣ λ ∣ ∣ x ∣$ ), and the triangle inequality $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ 00.01.02. The two-case equivalence $∣ u ∣ < a \Leftrightarrow - a < u < a$ is the one-dimensional instance of the open ball ${x : ∥ x - c ∥ < r}$ in a normed space; the rational-inequality sign chart is the one-dimensional instance of the boundary structure of a semi-algebraic set; the determinant criterion for a $2 \times 2$ system is the $n = 2$ instance of $det A \neq = 0 \Leftrightarrow A invertible$ . Each precalculus technique is the low-dimensional shadow of a structural theorem.

Pólya's heuristic: case analysis as decomposition. The recurring move across this unit — splitting $∣ u ∣ < a$ into two inequalities, splitting the number line at the roots of $p (x)$ , splitting a system into elimination steps — is the case-analysis heuristic that Pólya (1945, 1957) named "decompose and recombine" ^{[Pólya — How to Solve It]}. A hard problem is decomposed into a finite list of simpler problems indexed by a sign or a case; each simpler problem is solved; the solutions are recombined (by intersection for and, union for or, back-substitution for systems) into the answer. The method is the algorithmic content of elementary algebra and survives, in sophisticated form, as the case-split tactic in proof assistants and the branch-and-bound method in optimisation.

Synthesis. The three pillars of this unit — the two-case absolute-value equivalence, the sign-chart method for polynomial and rational inequalities, and the determinant criterion for square linear systems — are not three unrelated techniques but three instances of the same structural move: replace a single compound statement by a finite, sign-indexed list of simpler statements, solve each, and recombine. The absolute-value equivalence generalises to the open ball of a normed space and builds toward the topology of metric spaces; the sign chart generalises to the semi-algebraic sets of real algebraic geometry and appears again in the cylindrical algebraic decomposition algorithm for quantifier elimination over $R$ ; the determinant criterion generalises to $det A \neq = 0 \Leftrightarrow A$ invertible and is dual to the row-reduction algorithm that computes the solution. The foundational reason all three work is that $R$ is an ordered field with a decidable first-order theory, so the sign of every algebraic expression is computable on each connected component of the complement of its zero set. The central insight is that an inequality carves out a region and a system carves out an intersection of regions, and the bridge is the recognition that solving either is a finite case analysis whose pieces are recombined by the Boolean structure of the statement.

Full proof set Master

Proposition (triangle inequality on $R$ ). For all $x, y \in R$ , $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ , with equality iff $x$ and $y$ have the same sign (or one is zero).

Proof. For either order of signs the result reduces to an identity; the load-bearing case is opposite signs. If $x \geq 0$ and $y \geq 0$ then $∣ x + y ∣ = x + y = ∣ x ∣ + ∣ y ∣$ . If $x \leq 0$ and $y \leq 0$ then $∣ x + y ∣ = - (x + y) = - x - y = ∣ x ∣ + ∣ y ∣$ . If $x \geq 0 \geq y$ , write $x \geq 0$ and $- y \geq 0$ ; then $∣ x + y ∣ \leq max (x, - y) \leq x + (- y) = ∣ x ∣ + ∣ y ∣$ , where the last inequality uses that both summands are non-negative. The symmetric case $y \geq 0 \geq x$ is identical with $x, y$ exchanged. Equality with opposite signs requires $max (x, - y) = x + (- y)$ , i.e. one of $x, - y$ is zero.

Proposition (two-case absolute-value equivalence, closed form). For $u \in R$ and $a > 0$ , $∣ u ∣ \leq a$ if and only if $- a \leq u \leq a$ .

Proof. ( $\Rightarrow$ ) If $u \geq 0$ then $∣ u ∣ = u \leq a$ and $- a \leq 0 \leq u$ , giving $- a \leq u \leq a$ . If $u < 0$ then $∣ u ∣ = - u \leq a$ gives $u \geq - a$ , and $u < 0 \leq a$ gives $u \leq a$ . ( $\Leftarrow$ ) If $u \geq 0$ then $∣ u ∣ = u \leq a$ . If $u < 0$ then $∣ u ∣ = - u \leq a$ follows from $u \geq - a$ .

Proposition (sign-chart solution of a rational inequality). Let $p, q \in R [x]$ with $q$ not the zero polynomial, and let $r_{1} < r_{2} < \dots < r_{k}$ be the distinct real roots of $p \cdot q$ listed with multiplicity collapsed. On each open interval between consecutive $r_{i}$ (and on the two unbounded intervals at the ends), the rational function $p (x) / q (x)$ has constant sign. The solution set of $p (x) / q (x) □ 0$ is the union of those intervals on which the sign matches $□ 0$ , with each $r_{i}$ that is a root of $p$ included if $□$ is non-strict and excluded if strict, and each $r_{i}$ that is a root of $q$ excluded in all cases.

Proof. On any open interval avoiding the roots of $q$ , the function $p (x) / q (x)$ is continuous and never zero, so by the intermediate-value theorem it cannot change sign on the interval; the sign is fixed by evaluating at a single test point. The real roots of $p \cdot q$ partition $R$ into exactly $k + 1$ open intervals, on each of which the sign is therefore constant and computable. At a root $r_{i}$ of $p$ the value of the expression is $0$ (included for non-strict $□$ , excluded for strict); at a root of $q$ the expression is undefined and excluded. Collecting the matching intervals gives the stated union.

Proposition ( $2 \times 2$ determinant criterion). The system ${a x + b y = e c x + d y = f$ has a unique solution if and only if $a d - b c \neq = 0$ ; in that case the solution is $x = (e d - b f) / (a d - b c)$ , $y = (a f - ec) / (a d - b c)$ .

Proof. Multiply the first equation by $d$ and the second by $b$ and subtract: $(a d - b c) x = e d - b f$ . If $a d - b c \neq = 0$ this determines $x$ uniquely; the analogous elimination of $x$ (multiply first by $c$ , second by $a$ , subtract) gives $(a d - b c) y = a f - ec$ , determining $y$ uniquely; direct substitution verifies both equations. Conversely, if a unique solution $(x_{0}, y_{0})$ exists, suppose for contradiction that $a d - b c = 0$ . Then the two rows $(a, b)$ and $(c, d)$ of the coefficient matrix are linearly dependent, so one is a scalar multiple of the other; the same elimination gives $0 \cdot x = e d - b f$ . If $e d - b f \neq = 0$ there is no solution, contradicting existence; if $e d - b f = 0$ the two equations are proportional and the solution set is an entire line, contradicting uniqueness.

Proposition (Cramer's rule, $n \times n$ ). Let $A \in R^{n \times n}$ with $det A \neq = 0$ and let $b \in R^{n}$ . The unique solution of $A x = b$ has $i$ -th component $x_{i} = det A_{i} / det A$ , where $A_{i}$ is $A$ with column $i$ replaced by $b$ .

Proof sketch. Let $a_{1}, \dots, a_{n}$ be the columns of $A$ , so $b = x_{1} a_{1} + \dots + x_{n} a_{n}$ expresses $b$ in the column basis. By multi-linearity of the determinant in column $i$ , $det A_{i} = det (a_{1}, \dots, a_{i - 1}, b, a_{i + 1}, \dots, a_{n}) = x_{i} det A$ (all other terms vanish because they have a repeated column). Dividing by $det A \neq = 0$ gives $x_{i} = det A_{i} / det A$ .

Connections Master

The absolute value $∣ x ∣$ is the one-dimensional instance of the norm on a vector space, and the two-case equivalence $∣ u ∣ < a \Leftrightarrow - a < u < a$ proved here is the one-variable shadow of the open ball ${x : ∥ x - c ∥ < r}$ that defines the metric topology of a normed space. The triangle inequality $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ on $R$ , established as the key theorem of 00.01.02, is the prototype norm inequality; the Cauchy–Schwarz inequality of 00.04.01 supplies the triangle inequality on the Euclidean norm of $R^{n}$ , and the determinant criterion of this unit becomes the invertibility criterion $A x = b \Leftrightarrow A$ nonsingular in linalg.systems-linear-independence.
The linear equation $a x + b = 0$ and its graph as a line in the plane, introduced in 00.03.01, is the load-bearing primitive for both the single inequality $a x + b □ 0$ (a half-line) and the $2 \times 2$ system (two lines intersecting). The slope–intercept picture of 00.03.01 is the geometric content of the consistency theorem proved here: two non-parallel lines meet in exactly one point (unique solution), two parallel distinct lines never meet (inconsistent), and two coincident lines meet in infinitely many points (proportional equations). The same picture generalises to the row-picture and column-picture of linalg.systems-linear-independence and to the affine-subspace geometry of the solution set under the rank criterion.
The quadratic-formula and discriminant machinery of 00.03.02 underlies the sign-chart method for quadratic inequalities: the roots that partition the number line are computed by the quadratic formula, and the discriminant trichotomy ( $Δ > 0$ , $Δ = 0$ , $Δ < 0$ ) of 00.03.02 determines whether the sign chart has two, one, or zero real boundary points. The rational-inequality sign chart of this unit is the direct generalisation to quotients of polynomials; both are the one-variable instances of the semi-algebraic set framework of real algebraic geometry in real-algebraic-geometry.semicommutative-algebra.
The case-analysis heuristic — splitting $∣ u ∣ < a$ into two inequalities, splitting the number line at the roots of a polynomial, splitting a system into elimination steps — is the algorithmic content of elementary algebra and recurs throughout the curriculum: as the case split on the sign of the multiplier in 00.04.01, as the sign-case analysis in the absolute-value inequalities of this unit, as the inductive case split of logic.induction-principle, and as the branch-and-bound decomposition of optimization-control.branch-and-bound. Pólya's decompose and recombine names the move in its most general form.
The matrix equation $A x = b$ introduced here as a compact notation for a $2 \times 2$ or $3 \times 3$ system is the entry point to linear algebra: the determinant becomes the oriented volume of linalg.determinants-permutations, Gaussian elimination becomes the $LU$ factorisation of linalg.lu-factorisation, and the rank criterion becomes the dimension theory of linalg.basis-dimension. The unique-solution case $det A \neq = 0$ is the invertibility condition that defines the general linear group and underwrites every change-of-basis and change-of-coordinates argument in the linear-algebra strand.

Historical & philosophical context Master

The algebra of inequalities and the algebra of systems of equations grew up together over four centuries, and the techniques taught in this unit were the research frontier of their day. François Viète (1540–1603), in his In Artem Analyticem Isagoge (1591), introduced the systematic use of letters for known and unknown quantities and the operations on equations as symbolic objects, a move that turned the solution of equations and inequalities from a sequence of ad-hoc numerical tricks into a calculus with its own laws ^{[Cauchy — Cours d'analyse de l'École Royale Polytechnique]}. Viète's "analytic art" made the case split on the sign of a multiplier, the substitution step, and the elimination step into named operations rather than improvised moves; the sign-flip rule on multiplication by a negative and the two-case analysis of an absolute value are direct descendants of his symbolic program.

René Descartes, in La Géométrie (1637, appendix to the Discours de la méthode), fused Viète's symbolic algebra with the coordinate picture, identifying an equation in two variables with a curve in the plane and a system of two equations with the intersection of two curves. The graphical interpretation of a $2 \times 2$ system as two crossing lines is Descartes's contribution; the picture turned the algebra of elimination into the geometry of intersection and made the determinant criterion geometrically inevitable (parallel lines, determinant zero). Gottfried Wilhelm Leibniz (1646–1716) introduced the term eliminatio and the systematic elimination of variables from systems; Gabriel Cramer (1704–1752), in his Introduction à l'analyse des lignes courbes algébriques (1750), published the determinant-based solution formula for $n \times n$ systems that bears his name, in the context of fitting a conic through five points — a $5 \times 5$ linear system whose determinant has to be nonzero for the conic to be well defined.

Augustin-Louis Cauchy, in his 1821 Cours d'analyse de l'École Royale Polytechnique, gave the absolute value its modern role as the load-bearing object of analytic estimates ^{[Cauchy — Cours d'analyse de l'École Royale Polytechnique]}. The triangle inequality $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ , stated there in Note II, is the inequality on which the $ε$ - $δ$ definition of limit and continuity rests; Cauchy's insistence on estimates expressed through absolute values turned "the size of a quantity" into an algebraic object that could be bounded, summed, and compared. The two-case equivalence of this unit is the inequality-level content of Cauchy's program: every statement about an absolute value is a disguised pair of statements about ordinary inequalities, to be untangled by case analysis.

The determinant as an algebraic object was given its modern theory by Augustin-Louis Cauchy (1812, in the context of alternating functions) and Carl Gustav Jacob Jacobi (1804–1851, in the context of the Jacobi determinant and change of variables). The consistency criterion — that a square system has a unique solution iff the determinant is nonzero — crystallised over the nineteenth century and was given its general $m \times n$ form (the rank criterion of Rouché and Capelli) in the 1870s. George Pólya, in How to Solve It (1945) and the later two-volume Mathematics and Plausible Reasoning (1954), named the recurring heuristic — decompose and recombine, think of a simpler analogous problem, work backwards — that organises the case analyses of this unit into a transferable method ^{[Pólya — How to Solve It]}. The cumulative picture is that the techniques of precalculus are not arbitrary rules but the surviving distillation of four centuries of work on how to make algebra systematic: every inequality is a region, every system is an intersection, every absolute value is a pair of cases, and every case is solved by the same small kit of reversible moves.

Bibliography Master

@book{StewartRedlinWatson2015,
  author = {Stewart, James and Redlin, Lothar and Watson, Saleem},
  title = {Precalculus: Mathematics for Calculus},
  edition = {7th},
  publisher = {Cengage Learning},
  address = {Boston},
  note = {Sections 1.7--1.8 on inequalities, absolute value, and coordinate geometry; Sections 10.1--10.2 on systems of linear equations, substitution and elimination, and Cramer's rule},
  year = {2015}
}

@book{Apostol1967,
  author = {Apostol, Tom M.},
  title = {Calculus, Volume 1: One-Variable Calculus, with an Introduction to Linear Algebra},
  edition = {2nd},
  publisher = {Wiley},
  address = {New York},
  note = {Sections I.3--I.5 on ordered fields, absolute value, and the triangle inequality},
  year = {1967}
}

@book{Polya1957,
  author = {P{\'o}lya, George},
  title = {How to Solve It: A New Aspect of Mathematical Method},
  edition = {2nd},
  publisher = {Princeton University Press},
  address = {Princeton},
  note = {pp.~126--132 on decompose-and-recombine and case analysis as the load-bearing heuristic of elementary algebra},
  year = {1957}
}

@book{Strang2006,
  author = {Strang, Gilbert},
  title = {Linear Algebra and Its Applications},
  edition = {4th},
  publisher = {Cengage Learning},
  address = {Belmont, CA},
  note = {Sections 1.1--1.3 on vectors, the matrix equation $A\mathbf{x}=\mathbf{b}$, and elimination},
  year = {2006}
}

@book{Cauchy1821,
  author = {Cauchy, Augustin-Louis},
  title = {Cours d'analyse de l'{\'E}cole Royale Polytechnique. Premi{\`e}re partie. Analyse alg{\'e}brique},
  publisher = {Debure},
  address = {Paris},
  note = {Note II on the systematic use of inequalities and the triangle inequality for real numbers; English translation by Bradley \& Sandifer, {\em Cauchy's Cours d'analyse: An Annotated Translation}, Springer 2009},
  year = {1821}
}

@book{Viete1591,
  author = {Vi{\`e}te, Fran{\c c}ois},
  title = {In Artem Analyticem Isagoge},
  publisher = {Jamet Mettayer},
  address = {Tours},
  note = {The introduction of symbolic algebra with letters for knowns and unknowns; English translation in {\em The Analytic Art}, Dover 2006, transl.~T.~Richard Witmer},
  year = {1591}
}

@book{Descartes1637,
  author = {Descartes, Ren{\'e}},
  title = {La G{\'e}om{\'e}trie},
  note = {Appendix to {\em Discours de la m{\'e}thode}, Jan Maire, Leiden; fuses symbolic algebra with coordinate geometry, identifying an equation in two variables with a plane curve},
  year = {1637}
}

@book{Cramer1750,
  author = {Cramer, Gabriel},
  title = {Introduction {\`a} l'analyse des lignes courbes alg{\'e}briques},
  publisher = {Fr{\`e}res Cramer \& Cl.~Philibert},
  address = {Geneva},
  note = {The determinant-based solution formula for $n \times n$ linear systems, in the context of fitting a conic through five given points},
  year = {1750}
}

@article{Tarski1948,
  author = {Tarski, Alfred},
  title = {A Decision Method for Elementary Algebra and Geometry},
  publisher = {RAND Corporation},
  address = {Santa Monica},
  note = {Quantifier elimination for the first-order theory of real-closed fields; the foundational result that the sign chart of precalculus is a decidable procedure},
  year = {1948}
}

@article{Collins1975,
  author = {Collins, George E.},
  title = {Quantifier Elimination for Real Closed Fields by Cylindrical Algebraic Decomposition},
  journal = {Springer Lecture Notes in Computer Science},
  volume = {33},
  pages = {134--183},
  note = {The effective multi-variable generalisation of the sign-chart method},
  year = {1975}
}

Prerequisites

00.04.01
00.03.01

Tier anchors

beginner: Stewart, Redlin, Watson — Precalculus: Mathematics for Calculus (Cengage, 7th ed., 2015) §1.7–1.8, plain-language inequalities, compound statements, and absolute value as distance
intermediate: Stewart, Redlin, Watson — Precalculus §1.7–1.8, §10.1–10.2, formal solution sets and 2x2 / 3x3 systems by elimination and substitution
master: Apostol — Calculus Vol. 1 (Wiley) §I.4–I.5 on ordered fields and absolute value; Pólya — How to Solve It (Princeton, 2nd ed., 1957) on the case-analysis heuristic; Strang — Linear Algebra and Its Applications §1.1–1.3 for systems as matrix equations

References

Stewart, Redlin, Watson — Precalculus: Mathematics for Calculus · Cengage, 7th ed. (2015), §1.7 Inequalities, §1.8 Coordinate plane, §10.1–10.2 Systems of linear equations. The compound-inequality and absolute-value sign conventions, the two-case analysis of $∣ u ∣ < a \Leftrightarrow - a < u < a$ , and the substitution / elimination algorithms for $2 \times 2$ and $3 \times 3$ linear systems.
Apostol — Calculus, Vol. 1 · Wiley (1967), §I.3–I.5. The ordered-field axioms giving the sign-flip rule on multiplication by a negative element, the triangle inequality $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ , and the field axioms underpinning Gaussian elimination as reversible row operations.
Pólya — How to Solve It · Princeton University Press, 2nd ed. (1957), pp. 126–132, 'Decompose and recombine', the case-analysis heuristic that turns a single absolute-value inequality into two ordinary inequalities and a system of equations into a sequence of eliminations.
Strang — Linear Algebra and Its Applications · Cengage, 4th ed. (2006), §1.1–1.3. Vectors, the coefficient matrix $A x = b$ , row reduction, and the determinant criterion for a unique solution of a square system.
Cauchy — Cours d'analyse de l'École Royale Polytechnique · Debure, Paris (1821), Note II. The systematic use of inequalities and the triangle inequality for real numbers, framing the absolute value as the load-bearing object of analytic estimates.

Estimated time

beginner: 16m
intermediate: 38m
master: 75m