Chapter 11 — Systems of Equations and Inequalities
When one equation isn’t enough to pin down the unknowns, you need a system. This chapter covers solving systems of linear equations in 2 and 3 variables, nonlinear systems, partial fraction decomposition, and the powerful matrix methods: Gaussian elimination, matrix algebra, inverse matrices, and Cramer’s Rule.
Table of Contents
1 — Systems of Linear Equations: Two Variables
2 — Systems of Linear Equations: Three Variables
3 — Systems of Nonlinear Equations and Inequalities
4 — Partial Fractions
- 4.1 Proper vs. Improper Fractions
- 4.2 Distinct Linear Factors
- 4.3 Repeated Linear Factors
- 4.4 Irreducible Quadratic Factors
5 — Matrices and Gaussian Elimination
- 5.1 Augmented Matrices
- 5.2 Row Operations
- 5.3 Row-Echelon Form and Gaussian Elimination
- 5.4 Reduced Row-Echelon Form
6 — Matrices and Matrix Operations
7 — Inverses of Matrices
- 7.1 The Identity Matrix
- 7.2 The Inverse of a 2×2 Matrix
- 7.3 Finding Inverses by Row Reduction
- 7.4 Solving Systems with Matrix Inverses
8 — Cramer’s Rule
Glossary
| Term | Definition |
|---|---|
| System of equations | Two or more equations with the same variables |
| Consistent | Has at least one solution |
| Inconsistent | Has no solution (parallel lines/planes) |
| Independent | Exactly one solution |
| Dependent | Infinitely many solutions (same line/plane) |
| Augmented matrix | Matrix combining coefficients and constants $[A | \mathbf{b}]$ |
| Row-echelon form | Upper triangular form with leading 1s (each below and right) |
| Determinant | A scalar value computed from a square matrix; zero ↔ singular |
| Inverse matrix | $A^{-1}$ such that $AA^{-1} = A^{-1}A = I$ |
1 — Systems of Linear Equations: Two Variables
1.1 Graphical Method
Each linear equation in two variables is a line. The solution is the intersection point.
1.2 Substitution Method
- Solve one equation for one variable.
- Substitute into the other equation.
- Solve the resulting equation.
- Back-substitute to find the other variable.
Solve: $y = 3x - 1$ and $2x + y = 9$.
Substitute: $2x + (3x - 1) = 9 \Rightarrow 5x = 10 \Rightarrow x = 2$
$y = 3(2) - 1 = 5$
Solution: $(2, 5)$
1.3 Elimination Method
- Multiply one or both equations so a variable has opposite coefficients.
- Add the equations to eliminate that variable.
- Solve and back-substitute.
Solve: $3x + 2y = 14$ and $5x - 2y = 2$.
Add: $8x = 16 \Rightarrow x = 2$.
$3(2) + 2y = 14 \Rightarrow 2y = 8 \Rightarrow y = 4$
Solution: $(2, 4)$
1.4 Types of Systems
| Type | Graph | Solutions |
|---|---|---|
| Independent | Lines intersect at one point | Exactly one |
| Dependent | Lines coincide | Infinitely many |
| Inconsistent | Lines are parallel | None |
Detection during solving: If you eliminate both variables and get a true statement ($0 = 0$), the system is dependent. A false statement ($0 = 5$) means inconsistent.
2 — Systems of Linear Equations: Three Variables
2.1 Strategy for Three Variables
A system of three equations in three unknowns represents three planes in 3D space. The solution is the point (or line/plane) where all three intersect.
Elimination strategy:
- Use two pairs of equations to eliminate the same variable, producing a 2×2 system.
- Solve the 2×2 system.
- Back-substitute to find the third variable.
2.2 Worked Example
Solve:
\(x + y + z = 6 \tag{1}\) \(2x - y + z = 3 \tag{2}\) \(x + 2y - z = 5 \tag{3}\)
Step 1: Add (1) and (3) to eliminate $z$:
$2x + 3y = 11 \tag{4}$
Step 2: Add (2) and (3) to eliminate $z$:
$3x + y = 8 \tag{5}$
Step 3: Solve (4) and (5):
From (5): $y = 8 - 3x$. Substitute into (4): $2x + 3(8 - 3x) = 11 \Rightarrow 2x + 24 - 9x = 11 \Rightarrow -7x = -13 \Rightarrow x = \dfrac{13}{7}$
$y = 8 - 3!\left(\dfrac{13}{7}\right) = \dfrac{56 - 39}{7} = \dfrac{17}{7}$
$z = 6 - x - y = 6 - \dfrac{13}{7} - \dfrac{17}{7} = \dfrac{42 - 30}{7} = \dfrac{12}{7}$
Solution: $\left(\dfrac{13}{7}, \dfrac{17}{7}, \dfrac{12}{7}\right)$
3 — Systems of Nonlinear Equations and Inequalities
3.1 Solving Nonlinear Systems
A nonlinear system contains at least one equation that is not linear (e.g., quadratic, circle, hyperbola).
Methods: Substitution is usually the best approach — solve the simpler equation for one variable and substitute into the other.
Solve: $x^2 + y^2 = 25$ and $y = x + 1$.
Substitute: $x^2 + (x + 1)^2 = 25$
$x^2 + x^2 + 2x + 1 = 25 \Rightarrow 2x^2 + 2x - 24 = 0 \Rightarrow x^2 + x - 12 = 0$
$(x + 4)(x - 3) = 0 \Rightarrow x = -4$ or $x = 3$
$x = -4$: $y = -3$. $x = 3$: $y = 4$.
Solutions: $(-4, -3)$ and $(3, 4)$
3.2 Systems of Inequalities
To solve a system of inequalities, graph each inequality (shade the appropriate half-plane) and find the region of overlap — the feasible region.
Each boundary is graphed as:
- Solid line for $\le$ or $\ge$
- Dashed line for $<$ or $>$
4 — Partial Fractions
4.1 Proper vs. Improper Fractions
- Proper: degree of numerator < degree of denominator → apply partial fractions directly
- Improper: degree of numerator ≥ degree of denominator → perform polynomial long division first
4.2 Distinct Linear Factors
\(\frac{P(x)}{(a_1 x + b_1)(a_2 x + b_2)\cdots(a_n x + b_n)} = \frac{A_1}{a_1 x + b_1} + \frac{A_2}{a_2 x + b_2} + \cdots + \frac{A_n}{a_n x + b_n}\)
Decompose: $\dfrac{3x + 11}{(x + 3)(x - 1)}$
\[\frac{3x + 11}{(x + 3)(x - 1)} = \frac{A}{x + 3} + \frac{B}{x - 1}\]Multiply both sides by $(x + 3)(x - 1)$:
$3x + 11 = A(x - 1) + B(x + 3)$
$x = 1$: $14 = 4B \Rightarrow B = \dfrac{7}{2}$
$x = -3$: $2 = -4A \Rightarrow A = -\dfrac{1}{2}$
\(\frac{3x + 11}{(x + 3)(x - 1)} = \frac{-1/2}{x + 3} + \frac{7/2}{x - 1}\)
4.3 Repeated Linear Factors
If a factor $(ax + b)$ appears $m$ times, include terms for each power:
\(\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_m}{(ax + b)^m}\)
Decompose: $\dfrac{5x + 1}{(x - 2)^2}$
\[\frac{5x + 1}{(x - 2)^2} = \frac{A}{x - 2} + \frac{B}{(x - 2)^2}\]$5x + 1 = A(x - 2) + B$
$x = 2$: $11 = B$
Expand: $5x + 1 = Ax - 2A + 11$. Compare $x$: $A = 5$.
\(\frac{5}{x - 2} + \frac{11}{(x - 2)^2}\)
4.4 Irreducible Quadratic Factors
For each irreducible quadratic factor $ax^2 + bx + c$ (discriminant $< 0$):
\[\frac{Ax + B}{ax^2 + bx + c}\]If the quadratic is repeated $m$ times, include terms up to $(ax^2 + bx + c)^m$.
5 — Matrices and Gaussian Elimination
5.1 Augmented Matrices
A system like:
\(2x + 3y = 8\) \(x - y = 1\)
can be written as the augmented matrix:
\(\left[\begin{array}{cc|c} 2 & 3 & 8 \\ 1 & -1 & 1 \end{array}\right]\)
5.2 Row Operations
Three elementary row operations (produce equivalent systems):
- Swap two rows: $R_i \leftrightarrow R_j$
- Multiply a row by a nonzero scalar: $kR_i \to R_i$
- Add a multiple of one row to another: $R_i + kR_j \to R_i$
5.3 Row-Echelon Form and Gaussian Elimination
Row-Echelon Form (REF):
- All zero rows are at the bottom.
- Each leading entry (first nonzero entry in a row) is to the right of the leading entry above it.
- All entries below a leading entry are zero.
Gaussian Elimination: Use row operations to reduce to REF, then back-substitute.
Solve using Gaussian elimination:
\[\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & -1 & 1 & 3 \\ 1 & 2 & -1 & 5 \end{array}\right]\]| $R_2 - 2R_1 \to R_2$: $\left[\begin{array}{ccc | c} 1 & 1 & 1 & 6 \ 0 & -3 & -1 & -9 \ 1 & 2 & -1 & 5 \end{array}\right]$ |
| $R_3 - R_1 \to R_3$: $\left[\begin{array}{ccc | c} 1 & 1 & 1 & 6 \ 0 & -3 & -1 & -9 \ 0 & 1 & -2 & -1 \end{array}\right]$ |
| $R_2 \leftrightarrow R_3$: $\left[\begin{array}{ccc | c} 1 & 1 & 1 & 6 \ 0 & 1 & -2 & -1 \ 0 & -3 & -1 & -9 \end{array}\right]$ |
| $R_3 + 3R_2 \to R_3$: $\left[\begin{array}{ccc | c} 1 & 1 & 1 & 6 \ 0 & 1 & -2 & -1 \ 0 & 0 & -7 & -12 \end{array}\right]$ |
Back-substitute: $z = \dfrac{12}{7}$, $y = -1 + 2!\left(\dfrac{12}{7}\right) = \dfrac{17}{7}$, $x = 6 - \dfrac{17}{7} - \dfrac{12}{7} = \dfrac{13}{7}$
5.4 Reduced Row-Echelon Form
Reduced Row-Echelon Form (RREF): row-echelon form with:
- Each leading entry is $1$.
- Each leading $1$ is the only nonzero entry in its column.
RREF gives the solution directly without back-substitution (Gauss-Jordan elimination).
6 — Matrices and Matrix Operations
6.1 Matrix Notation
A matrix is a rectangular array of numbers. An $m \times n$ matrix has $m$ rows and $n$ columns.
Entry in row $i$, column $j$: $a_{ij}$
Two matrices are equal if they have the same dimensions and all corresponding entries are equal.
6.2 Addition, Subtraction, and Scalar Multiplication
For $m \times n$ matrices $A$ and $B$ and scalar $c$:
\((A + B)_{ij} = a_{ij} + b_{ij}\) \((cA)_{ij} = c \cdot a_{ij}\)
Matrices must have the same dimensions to add/subtract.
6.3 Matrix Multiplication
If $A$ is $m \times n$ and $B$ is $n \times p$, then $AB$ is $m \times p$:
\[(AB)_{ij} = \sum_{k=1}^{n} a_{ik} b_{kj}\]Row $i$ of $A$ “dot” column $j$ of $B$.
Key properties:
- Not commutative: $AB \ne BA$ in general
- Associative: $(AB)C = A(BC)$
- Distributive: $A(B + C) = AB + AC$
For $AB$ to exist, the number of columns of $A$ must equal the number of rows of $B$.
7 — Inverses of Matrices
7.1 The Identity Matrix
The $n \times n$ identity matrix $I_n$ has $1$s on the main diagonal and $0$s elsewhere.
\(AI = IA = A\)
7.2 The Inverse of a 2×2 Matrix
For $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$:
\[A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\]provided $\det(A) = ad - bc \ne 0$.
If $\det(A) = 0$, the matrix is singular (no inverse exists).
7.3 Finding Inverses by Row Reduction
Augment $A$ with the identity: $[A \mid I]$. Row-reduce until the left side becomes $I$. The right side is $A^{-1}$:
\([A \mid I] \xrightarrow{\text{row ops}} [I \mid A^{-1}]\)
7.4 Solving Systems with Matrix Inverses
A system $A\mathbf{x} = \mathbf{b}$ has solution:
\[\mathbf{x} = A^{-1}\mathbf{b}\](when $A$ is invertible).
8 — Cramer’s Rule
8.1 Determinants
2×2 determinant:
\[\det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc\]3×3 determinant (cofactor expansion along row 1):
\(\det\begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} = a_1(b_2 c_3 - b_3 c_2) - b_1(a_2 c_3 - a_3 c_2) + c_1(a_2 b_3 - a_3 b_2)\)
8.2 Cramer’s Rule for 2×2 Systems
For the system $a_1 x + b_1 y = c_1$, $a_2 x + b_2 y = c_2$:
\[x = \frac{D_x}{D}, \qquad y = \frac{D_y}{D}\]where:
\[D = \det\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}, \quad D_x = \det\begin{bmatrix} c_1 & b_1 \\ c_2 & b_2 \end{bmatrix}, \quad D_y = \det\begin{bmatrix} a_1 & c_1 \\ a_2 & c_2 \end{bmatrix}\]$D_x$: replace the $x$-column with the constants. $D_y$: replace the $y$-column.
8.3 Cramer’s Rule for 3×3 Systems
Same principle: $x = D_x/D$, $y = D_y/D$, $z = D_z/D$, replacing one column at a time with the constants. Only works when $D \ne 0$.
Solve using Cramer’s Rule: $2x + y = 5$, $3x - 2y = -4$.
$D = \det\begin{bmatrix} 2 & 1 \ 3 & -2 \end{bmatrix} = -4 - 3 = -7$
$D_x = \det\begin{bmatrix} 5 & 1 \ -4 & -2 \end{bmatrix} = -10 + 4 = -6$
$D_y = \det\begin{bmatrix} 2 & 5 \ 3 & -4 \end{bmatrix} = -8 - 15 = -23$
$x = \dfrac{-6}{-7} = \dfrac{6}{7}$, $y = \dfrac{-23}{-7} = \dfrac{23}{7}$
Key Takeaways
- Three methods for 2-variable linear systems: graphing, substitution, elimination.
- Systems are consistent (independent or dependent) or inconsistent — check for $0 = 0$ vs $0 = c$.
- Three-variable systems reduce to two-variable systems by eliminating one variable.
- Nonlinear systems: substitution works best; expect multiple solutions (intersection of curves).
- Partial fractions decompose rational expressions by factor type: distinct linear, repeated linear, irreducible quadratic.
- Gaussian elimination row-reduces the augmented matrix to echelon form for back-substitution.
- Matrix multiplication is row-by-column; dimensions must be compatible; not commutative.
- Inverse matrix: $A^{-1}$ exists iff $\det(A) \ne 0$; gives solution $\mathbf{x} = A^{-1}\mathbf{b}$.
- Cramer’s Rule: elegant for small systems; replace each column of the coefficient matrix with the constants.
Practice Questions
Q1. Solve by elimination: $4x - 3y = 10$ and $2x + 5y = 8$.
Answer:
Multiply the second equation by $-2$: $-4x - 10y = -16$.
Add: $-13y = -6 \Rightarrow y = \dfrac{6}{13}$
$2x + 5!\left(\dfrac{6}{13}\right) = 8 \Rightarrow 2x = 8 - \dfrac{30}{13} = \dfrac{74}{13} \Rightarrow x = \dfrac{37}{13}$
Solution: $\left(\dfrac{37}{13}, \dfrac{6}{13}\right)$
Q2. Decompose $\dfrac{2x + 3}{(x - 1)(x + 2)}$ into partial fractions.
Answer:
$\dfrac{2x + 3}{(x - 1)(x + 2)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 2}$
$2x + 3 = A(x + 2) + B(x - 1)$
$x = 1$: $5 = 3A \Rightarrow A = \dfrac{5}{3}$
$x = -2$: $-1 = -3B \Rightarrow B = \dfrac{1}{3}$
\(\frac{5/3}{x - 1} + \frac{1/3}{x + 2}\)
Q3. Find the determinant: $\det\begin{bmatrix} 2 & -1 & 3 \ 0 & 4 & -2 \ 1 & 0 & 5 \end{bmatrix}$.
Answer:
$= 2(4 \cdot 5 - (-2) \cdot 0) - (-1)(0 \cdot 5 - (-2) \cdot 1) + 3(0 \cdot 0 - 4 \cdot 1)$
$= 2(20) + 1(2) + 3(-4) = 40 + 2 - 12 = 30$
Q4. Find the inverse of $A = \begin{bmatrix} 3 & 1 \ 5 & 2 \end{bmatrix}$.
Answer:
$\det(A) = 6 - 5 = 1$
$A^{-1} = \dfrac{1}{1}\begin{bmatrix} 2 & -1 \ -5 & 3 \end{bmatrix} = \begin{bmatrix} 2 & -1 \ -5 & 3 \end{bmatrix}$
Q5. Solve the nonlinear system: $x^2 + y^2 = 13$ and $x + y = 5$.
Answer:
$y = 5 - x$. Substitute: $x^2 + (5 - x)^2 = 13$
$x^2 + 25 - 10x + x^2 = 13 \Rightarrow 2x^2 - 10x + 12 = 0 \Rightarrow x^2 - 5x + 6 = 0$
$(x - 2)(x - 3) = 0 \Rightarrow x = 2$ or $x = 3$
Solutions: $(2, 3)$ and $(3, 2)$
Q6. Use Cramer’s Rule to solve: $x + 2y + z = 7$, $2x + y - z = 0$, $3x - y + 2z = 11$.
Answer:
$D = \det\begin{bmatrix} 1 & 2 & 1 \ 2 & 1 & -1 \ 3 & -1 & 2 \end{bmatrix} = 1(2 - 1) - 2(4 + 3) + 1(-2 - 3) = 1 - 14 - 5 = -18$
$D_x = \det\begin{bmatrix} 7 & 2 & 1 \ 0 & 1 & -1 \ 11 & -1 & 2 \end{bmatrix} = 7(2 - 1) - 2(0 + 11) + 1(0 - 11) = 7 - 22 - 11 = -26$
$x = \dfrac{-26}{-18} = \dfrac{13}{9}$
(Similarly compute $D_y$ and $D_z$ for the full solution.)