Substitution, elimination, 3-variable Gaussian elimination, nonlinear systems, linear programming, partial fractions, and matrices — with fully worked examples and step-by-step solutions.
Chapter 10 of Stewart Precalculus expands the idea of solving equations to solving multiple equations simultaneously. A system of equations is a set of two or more equations that share variables — you are looking for values that satisfy all equations at once.
The chapter progresses from two-variable linear systems, through three-variable systems, into nonlinear systems, graphing inequalities, linear programming, partial fraction decomposition, and the matrix methods that connect everything together.
| Section | Topic | Key Skill |
|---|---|---|
| 10.1 | 2-Variable Linear Systems | Substitution and elimination |
| 10.2 | Systems of Linear Equations in Several Variables | Gaussian elimination, back-substitution |
| 10.3 | Matrices and Systems of Linear Equations | Augmented matrices, row reduction |
| 10.4 | The Algebra of Matrices | Matrix operations, matrix equations |
| 10.5 | Inverses of Matrices and Matrix Equations | Inverse matrix method |
| 10.6 | Determinants and Cramer's Rule | Determinants, Cramer's rule |
| 10.7 | Partial Fractions | Decomposing rational expressions |
| 10.8 | Systems of Inequalities | Graphing feasible regions |
| 10.9 | Linear Programming | Objective function, corner point theorem |
A 2x2 linear system has two equations and two unknowns. Geometrically, each equation is a line, and the solution is the point where the lines intersect.
Solve one equation for one variable, then substitute that expression into the other equation. Best when one variable has a coefficient of 1 or negative 1.
Step-by-Step
Worked Example — Substitution
Solve: x + 2y = 8 and 3x minus y = 3
Step 1: Isolate x in equation 1: x = 8 minus 2y
Step 2: Substitute into equation 2:
3(8 minus 2y) minus y = 3
24 minus 6y minus y = 3
24 minus 7y = 3
minus 7y = minus 21
y = 3
Step 3: Back-substitute: x = 8 minus 2(3) = 8 minus 6 = 2
Step 4: Check in eq 1: 2 + 2(3) = 2 + 6 = 8 ✓
Step 4: Check in eq 2: 3(2) minus 3 = 6 minus 3 = 3 ✓
Solution: (2, 3)
Multiply one or both equations by constants so that one variable has equal and opposite coefficients, then add the equations to eliminate that variable. Best for integer coefficients.
Step-by-Step
Worked Example — Elimination
Solve: 2x + 3y = 12 and 4x minus y = 10
Goal: eliminate y. LCM of coefficients 3 and 1 is 3.
Multiply equation 2 by 3: 12x minus 3y = 30
Add to equation 1:
(2x + 3y) + (12x minus 3y) = 12 + 30
14x = 42
x = 3
Back-substitute into equation 1: 2(3) + 3y = 12
6 + 3y = 12 → 3y = 6 → y = 2
Check eq 2: 4(3) minus 2 = 12 minus 2 = 10 ✓
Solution: (3, 2)
Write each equation in slope-intercept form y = mx + b, graph both lines, and identify their intersection point. Graphing is conceptually important but impractical for exact answers with non-integer solutions.
Three Geometric Outcomes
One Solution
Lines intersect at exactly one point. Different slopes.
Example: y = 2x + 1 and y = minus x + 4
No Solution
Lines are parallel — same slope, different y-intercepts.
Example: y = 2x + 1 and y = 2x minus 3
Infinite Solutions
Lines are identical — same slope and same intercept.
Example: y = 2x + 1 and 2y = 4x + 2
When you apply elimination or substitution, the algebra will reveal which case you are in before you ever graph the lines.
No Solution — Inconsistent System
x + y = 3
x + y = 7
Subtract eq 1 from eq 2:
0 = 4 (contradiction)
The lines x + y = 3 and x + y = 7 are parallel. No point satisfies both.
Infinite Solutions — Dependent System
2x + 4y = 8
x + 2y = 4
Multiply eq 2 by 2:
2x + 4y = 8
Subtract: 0 = 0 (always true)
Both equations describe the same line. Solution: x = t, y = 2 minus (t/2) for any real t.
Three equations, three unknowns. Each equation is a plane in 3D space. A unique solution is a point where all three planes meet. Use systematic elimination to reduce to simpler subsystems.
The goal is to transform the system into triangular form — where equation 3 has one variable, equation 2 has two variables, and equation 1 has all three — then back-substitute from the bottom up.
Allowable Row Operations
| Operation | Symbol | Example |
|---|---|---|
| Interchange two equations | Eq1 ↔ Eq2 | Swap equation 1 and equation 2 |
| Multiply equation by nonzero constant | k · Eq1 | Multiply equation 1 by minus 1/2 |
| Add multiple of one equation to another | Eq2 + k · Eq1 | Replace Eq2 with Eq2 minus 3 times Eq1 |
Fully Worked 3x3 Example
Given system:
Eq 1: x + y + z = 6
Eq 2: 2x minus y + z = 3
Eq 3: x + 2y minus z = 2
Step 1 — Eliminate x from Eq 2 and Eq 3:
New Eq 2 = Eq 2 minus 2 times Eq 1:
(2x minus y + z) minus 2(x + y + z) = 3 minus 12
minus 3y minus z = minus 9 → Eq 2'
New Eq 3 = Eq 3 minus Eq 1:
(x + 2y minus z) minus (x + y + z) = 2 minus 6
y minus 2z = minus 4 → Eq 3'
Reduced system:
Eq 1: x + y + z = 6
Eq 2': minus 3y minus z = minus 9
Eq 3': y minus 2z = minus 4
Step 2 — Eliminate y from Eq 3':
Multiply Eq 3' by 3 to align y-coefficients: 3y minus 6z = minus 12
Add to Eq 2': (minus 3y minus z) + (3y minus 6z) = minus 9 + (minus 12)
minus 7z = minus 21
z = 3
Step 3 — Back-substitute:
From Eq 3': y minus 2(3) = minus 4 → y = minus 4 + 6 = 2
From Eq 1: x + 2 + 3 = 6 → x = 1
Solution: (x, y, z) = (1, 2, 3)
Verify: 1+2+3=6 ✓ 2(1) minus 2 + 3 = 3 ✓ 1+2(2) minus 3 = 2 ✓
Special Cases in 3-Variable Systems
Unique Solution
Back-substitution yields specific values for x, y, z. Three planes meet at one point.
No Solution
Elimination produces a contradiction like 0 = 5. Planes do not all share a common point.
Infinite Solutions
Elimination produces 0 = 0. Planes intersect along a line. Express solution with a parameter t.
When at least one equation in a system is nonlinear (quadratic, circle, parabola, etc.), graphically the curves can intersect in 0, 1, 2, or more points. Substitution is almost always the best strategy.
The most common nonlinear system in precalculus is a parabola intersected with a line. Substitute the linear equation into the quadratic to get a quadratic equation in one variable.
Worked Example
Solve: y = x squared and y = x + 2
Step 1: Both equations express y, so set equal:
x squared = x + 2
Step 2: Rearrange: x squared minus x minus 2 = 0
Step 3: Factor: (x minus 2)(x + 1) = 0
Step 4: x = 2 or x = minus 1
Step 5: Find y-values:
When x = 2: y = 2 squared = 4 → point (2, 4)
When x = minus 1: y = (minus 1) squared = 1 → point (minus 1, 1)
Step 6: Check both in y = x + 2:
4 = 2 + 2 = 4 ✓ and 1 = minus 1 + 2 = 1 ✓
Solutions: (2, 4) and (minus 1, 1)
When both equations are conics, use substitution or strategic elimination. This type appears on exams and requires careful algebra.
Worked Example
Solve: x squared + y squared = 5 and y = x squared minus 1
Step 1: Substitute y = x squared minus 1 into the circle equation:
x squared + (x squared minus 1) squared = 5
x squared + x to the 4th minus 2x squared + 1 = 5
x to the 4th minus x squared minus 4 = 0
Step 2: Let u = x squared: u squared minus u minus 4 = 0
u = (1 plus or minus sqrt(1 + 16)) / 2 = (1 plus or minus sqrt(17)) / 2
Since u = x squared must be non-negative, take u = (1 + sqrt(17)) / 2 approximately 2.56
x = plus or minus sqrt(2.56) approximately plus or minus 1.60
y = x squared minus 1 approximately 1.56
Approximately (1.60, 1.56) and (minus 1.60, 1.56)
How to Count Solutions Using the Discriminant
When a nonlinear system reduces to a quadratic equation ax squared + bx + c = 0, the discriminant b squared minus 4ac tells you the number of real intersection points:
b² minus 4ac greater than 0
Two distinct real solutions — 2 intersection points
b² minus 4ac = 0
One repeated solution — curves are tangent
b² minus 4ac less than 0
No real solutions — curves do not intersect
A system of inequalities asks for the set of all points (x, y) that satisfy every inequality simultaneously. The solution is a region in the plane, not a single point.
Step-by-Step
Worked Example
Graph the system:
x + y less than or equal to 4
x greater than or equal to 0
y greater than or equal to 0
Boundary 1: x + y = 4 (solid line — passes through (4,0) and (0,4))
Test (0,0): 0 + 0 = 0 less than or equal to 4 ✓ — shade below and left of this line
Boundary 2: x = 0 (the y-axis, solid) — shade to the right
Boundary 3: y = 0 (the x-axis, solid) — shade above
Feasible region: the triangle with vertices (0,0), (4,0), and (0,4)
Linear programming finds the maximum or minimum of a linear objective function subject to constraints that define a feasible region. It is one of the most important real-world applications of systems.
| Term | Definition |
|---|---|
| Objective function | The linear expression you want to maximize or minimize (e.g., P = 3x + 2y) |
| Constraints | The system of inequalities that limit the variables |
| Feasible region | The set of all points satisfying all constraints simultaneously |
| Corner points (vertices) | The intersection points of boundary lines at the edges of the feasible region |
| Corner Point Theorem | The maximum and minimum of the objective function occur at a corner point |
| Bounded region | A feasible region enclosed on all sides — both max and min exist |
| Unbounded region | A feasible region open on one side — only max or only min may exist |
If the feasible region is bounded and nonempty, the objective function P = ax + by attains its maximum and minimum values at corner points of the feasible region. To find them: evaluate P at every corner point and compare.
Procedure — Solve a Linear Programming Problem
Fully Worked Linear Programming Example
Problem:
Maximize P = 5x + 4y
Subject to:
x + y less than or equal to 6
2x + y less than or equal to 10
x greater than or equal to 0
y greater than or equal to 0
Step 1 — Find corner points:
Intersection of x + y = 6 and x-axis (y = 0): (6, 0)
Intersection of 2x + y = 10 and y-axis (x = 0): (0, 10) — but y less than or equal to 6 from eq 1,
so intersect x = 0 with x + y = 6: (0, 6)
Intersection of x + y = 6 and 2x + y = 10:
Subtract: x = 4, then y = 2 → (4, 2)
Origin: (0, 0)
Step 2 — Evaluate P = 5x + 4y at each corner:
At (0, 0): P = 0
At (6, 0): P = 5(6) + 4(0) = 30
At (4, 2): P = 5(4) + 4(2) = 20 + 8 = 28
At (0, 6): P = 5(0) + 4(6) = 24
Maximum P = 30 at corner point (6, 0)
Minimum P = 0 at corner point (0, 0)
Partial fraction decomposition reverses the process of adding fractions — it breaks a complex rational expression into simpler pieces. This technique is essential both in precalculus and in calculus integration.
The fraction must be a proper rational expression: the degree of the numerator must be strictly less than the degree of the denominator. If not, do polynomial long division first to extract the polynomial part, then decompose the remainder.
Proper vs. Improper Fractions
Proper (ready to decompose)
(3x + 1) / (x squared minus 1)
Numerator degree 1 less than denominator degree 2
Improper (divide first)
(x cubed + 2x) / (x squared minus 1)
Numerator degree 3 is not less than denominator degree 2
Case 1 — Distinct Linear Factors
Denominator factors completely into distinct linear factors (x minus a)(x minus b) etc.
P(x) / [(x minus a)(x minus b)] = A / (x minus a) + B / (x minus b)
Example: (3x + 5) / [(x + 1)(x minus 2)] = A/(x+1) + B/(x minus 2)
Case 2 — Repeated Linear Factors
Denominator contains a repeated linear factor (x minus a) to the power n. Include one term for each power from 1 up to n.
P(x) / (x minus a) squared = A / (x minus a) + B / (x minus a) squared
For (x minus a) cubed: add C / (x minus a) cubed as a third term
Case 3 — Irreducible Quadratic Factors
Denominator contains ax squared + bx + c where b squared minus 4ac is less than 0. The numerator of the partial fraction must be linear.
P(x) / [(x minus a)(x squared + bx + c)] = A / (x minus a) + (Bx + C) / (x squared + bx + c)
Example: 2x / [(x+1)(x squared + 4)] = A/(x+1) + (Bx+C)/(x squared + 4)
Case 4 — Repeated Irreducible Quadratic Factors
Denominator contains (ax squared + bx + c) to the power n. Include one linear-over-quadratic term for each power.
P(x) / (x squared + 1) squared = (Ax + B) / (x squared + 1) + (Cx + D) / (x squared + 1) squared
Worked Example — Case 1 (Distinct Linear Factors)
Decompose: (5x + 1) / [(x + 1)(x minus 3)]
Step 1: Set up: (5x + 1) / [(x+1)(x minus 3)] = A/(x+1) + B/(x minus 3)
Step 2: Multiply both sides by (x+1)(x minus 3):
5x + 1 = A(x minus 3) + B(x + 1)
Step 3: Plug in x = minus 1 (eliminates B):
5(minus 1) + 1 = A(minus 1 minus 3) + 0
minus 4 = minus 4A → A = 1
Step 4: Plug in x = 3 (eliminates A):
5(3) + 1 = 0 + B(3 + 1)
16 = 4B → B = 4
Answer: 1/(x+1) + 4/(x minus 3)
Worked Example — Case 3 (Irreducible Quadratic)
Decompose: (x squared + 3) / [x(x squared + 1)]
Step 1: Set up: A/x + (Bx + C)/(x squared + 1)
Step 2: Multiply through: x squared + 3 = A(x squared + 1) + (Bx + C)(x)
= Ax squared + A + Bx squared + Cx
Step 3: Equate coefficients:
x squared: 1 = A + B
x: 0 = C
constant: 3 = A
So A = 3, C = 0, B = 1 minus 3 = minus 2
Answer: 3/x + (minus 2x)/(x squared + 1) = 3/x minus 2x/(x squared + 1)
A matrix is a rectangular array of numbers. In precalculus, matrices are primarily used to solve linear systems via row reduction and to perform matrix arithmetic.
Writing a linear system as an augmented matrix organizes the coefficients and constants into rows so that row operations correspond exactly to the elimination steps you would do algebraically. The goal is to reach reduced row echelon form (RREF).
Row Echelon Form Requirements
Example — Augmented Matrix Setup
System: 2x + 3y = 7 and x minus y = 1
Augmented matrix:
[ 2 3 | 7 ]
[ 1 -1 | 1 ]
R1 ↔ R2 (put the easier row first):
[ 1 -1 | 1 ]
[ 2 3 | 7 ]
R2 minus 2 R1:
[ 1 -1 | 1 ]
[ 0 5 | 5 ]
(1/5) R2:
[ 1 -1 | 1 ]
[ 0 1 | 1 ]
R1 + R2:
[ 1 0 | 2 ] → x = 2
[ 0 1 | 1 ] → y = 1
Solution: (2, 1)
| Operation | Requirement | How To |
|---|---|---|
| Addition (A + B) | Same dimensions | Add corresponding entries |
| Scalar multiplication (kA) | Any matrix | Multiply every entry by k |
| Matrix multiplication (AB) | Columns of A = Rows of B | Dot product of A-rows with B-columns |
| Transpose (A^T) | Any matrix | Reflect entries across the main diagonal |
| Inverse (A^-1) | Square matrix with nonzero determinant | Row-reduce [A | I] to get [I | A^-1] |
Matrix Multiplication Is Not Commutative
In general AB does not equal BA. Always check whether the dimensions are compatible: an m by n matrix multiplied by an n by p matrix produces an m by p result. If the inner dimensions do not match, the product is undefined.
Systems of equations model countless real-world problems. The key is identifying two or more unknown quantities and writing two or more equations that relate them.
Mixture problems involve combining two substances at different concentrations, prices, or strengths to get a target mixture. Always write a total-amount equation and a total-value equation.
Worked Example — Acid Solution
How many liters of 20% acid and 50% acid make 12 liters of 30% acid?
Let x = liters of 20% solution, y = liters of 50% solution.
Equation 1 (total volume): x + y = 12
Equation 2 (total acid): 0.20x + 0.50y = 0.30(12) = 3.6
From Eq 1: x = 12 minus y
Substitute: 0.20(12 minus y) + 0.50y = 3.6
2.4 minus 0.20y + 0.50y = 3.6
0.30y = 1.2
y = 4
Then x = 12 minus 4 = 8
Answer: 8 liters of 20% and 4 liters of 50%
Investment problems combine a principal equation (amounts sum to total invested) with an interest equation (simple interest = principal times rate times time).
Worked Example
$10,000 split between 4% and 6% accounts earns $480 total interest per year. How much is in each?
Let x = amount at 4%, y = amount at 6%.
Eq 1 (principal): x + y = 10000
Eq 2 (interest): 0.04x + 0.06y = 480
Multiply Eq 1 by 0.04: 0.04x + 0.04y = 400
Subtract from Eq 2: 0.02y = 80 → y = 4000
Then x = 10000 minus 4000 = 6000
Answer: $6,000 at 4% and $4,000 at 6%
A business breaks even when total revenue equals total cost. Set the revenue function equal to the cost function and solve for the quantity.
Worked Example
A product sells for $15. Fixed costs are $600. Variable cost is $9 per unit. Find the break-even quantity.
Revenue: R = 15q
Cost: C = 600 + 9q
Set R = C: 15q = 600 + 9q
6q = 600
q = 100 units
At q = 100: R = 15(100) = $1,500 and C = 600 + 9(100) = $1,500 ✓
Break-even point: 100 units, $1,500
Above 100 units the business is profitable; below 100 units it operates at a loss.
Mistake: Forgetting to check both solutions
Always substitute back into BOTH original equations — one equation is not enough.
Mistake: Declaring a system has no solution when it has infinite solutions
Getting 0 = 0 means infinite solutions (dependent), not no solution. Getting 0 = k (k nonzero) means no solution.
Mistake: Using the wrong partial fraction form
Check whether each factor is linear, repeated linear, or irreducible quadratic before setting up the form. Mistakes in setup guarantee a wrong answer.
Mistake: Sign errors in elimination
When subtracting an equation, distribute the negative to EVERY term. Write each step clearly.
Mistake: Evaluating the objective function at the wrong points
Evaluate only at corner points of the feasible region — interior points and edge midpoints are irrelevant.
Mistake: Not performing long division before partial fractions
If the numerator degree is greater than or equal to the denominator degree, divide first. Skipping this step gives an incorrect decomposition.
Mistake: Row operation errors in Gaussian elimination
Write out every step explicitly. Abbreviate as R2 minus 2R1 before computing. A single arithmetic error propagates through all subsequent steps.
Mistake: Confusing the feasible region boundary type
Strict inequalities (less than, greater than) use dashed boundaries. Non-strict inequalities (less than or equal, greater than or equal) use solid boundaries.
| Situation | Best Method | Why |
|---|---|---|
| 2 equations, one variable has coefficient 1 | Substitution | Easy to isolate and substitute |
| 2 equations, integer coefficients | Elimination | Clean arithmetic, no fractions |
| 2 equations, need to understand geometry | Graphing | Visualizes the solution type |
| 3 or more equations and variables | Gaussian elimination (matrices) | Systematic, works for any size |
| System includes a nonlinear equation | Substitution into the nonlinear equation | Reduces to a single-variable polynomial |
| Need max or min of a linear expression | Linear programming | Corner point theorem gives exact answer |
| Rational expression to simplify or integrate | Partial fractions | Breaks complex fraction into simple pieces |
Solve one equation for one variable, then substitute that expression into the other equation. Example: solve x + 2y = 8 and 3x minus y = 3.
Multiply equations so one variable has equal and opposite coefficients, then add to eliminate that variable. Best when coefficients are integers and no variable has coefficient 1.
Example: 2x + 3y = 12 and 4x minus y = 10. Multiply eq 2 by 3: 12x minus 3y = 30. Add: 14x = 42, so x = 3. Then y = 2. Solution: (3, 2).
Three outcomes occur when you solve:
Express an infinite solution set using a parameter: for example, x = t, y = 4 minus t for all real t.
Use Gaussian elimination:
Use substitution. If y = x squared and y = x + 2, substitute: x squared = x + 2. Rearrange to x squared minus x minus 2 = 0 and factor: (x minus 2)(x + 1) = 0. Solutions x = 2 and x = minus 1 give points (2, 4) and (minus 1, 1).
The discriminant of the resulting quadratic tells you whether there are 2, 1, or 0 real intersection points.
Graph each boundary line (solid for less-than-or-equal / greater-than-or-equal, dashed for strict). Use a test point to determine which side to shade. The solution region is the overlap of all shaded areas.
Example for x + y less than or equal to 4, x greater than or equal to 0, y greater than or equal to 0: the feasible region is the triangle with vertices (0,0), (4,0), and (0,4).
The maximum or minimum value of a linear objective function over a bounded feasible region always occurs at a corner point (vertex) of that region.
Procedure: find all corner points by solving pairs of boundary equations, evaluate the objective function at each, and select the largest (for max) or smallest (for min).
Always confirm the fraction is proper (numerator degree less than denominator degree) before decomposing. If not, do polynomial long division first.
An augmented matrix stores the coefficients and constants of a linear system in a rectangular array. For 2x + 3y = 7 and x minus y = 1:
[ 2 3 | 7 ]
[ 1 -1 | 1 ]
Each row is one equation. The separator bar marks the constants column. Row operations on the matrix equal algebraic operations on the original system.
Define two variables (amounts of each component), write a total-amount equation and a total-value/concentration equation, then solve by substitution or elimination.
Example: 8 liters of 20% acid + 4 liters of 50% acid = 12 liters of 30% acid. Equation 1: x + y = 12. Equation 2: 0.20x + 0.50y = 3.6.
Write a principal equation (amounts add to total invested) and an interest equation (0.04x + 0.06y = total interest). Solve the system with elimination.
For $10,000 at 4% and 6% earning $480: multiply the principal equation by 0.04 and subtract from the interest equation to isolate y, giving y = $4,000 at 6% and x = $6,000 at 4%.
The break-even point is where Revenue = Cost. Write R = price times q and C = fixed cost + variable cost times q, then set R = C and solve for q.
At break-even, profit is zero. Above the break-even quantity, revenue exceeds cost — the business is profitable. Below it, the business loses money.
Precalculus Study Guide
Full chapter-by-chapter guide to Stewart Precalculus
Systems of Equations
Core concepts and method reference for 2-variable systems
Matrices
Matrix operations, determinants, and inverses in depth
Partial Fractions
All four cases of partial fraction decomposition with worked examples
Solving Inequalities
Linear and polynomial inequality techniques
Matrix Operations
Addition, multiplication, inverse, and determinant computations
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