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Mathematics AA · HL · Unit 1: Number and Algebra

Unit 1Topic 1.16Key Concepts53 min read

Systems of linear equations

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Meaning of simultaneous and system equations

A system of linear equations is a collection of linear equations that must all be true for the same values of the unknowns. The words simultaneous equations emphasize that the equations are solved together, not independently.

A solution to a two-variable system is an ordered pair x,yx,y. A solution to a three-variable system is an ordered triple x,y,zx,y,z. Substitution into every original equation is the definitive check.

Each equation restricts the possible values. Solving the system means finding the intersection of all restrictions: one common point, no common point, or an entire family of common points.

Linear means every unknown has power one and variables are not multiplied together. Expressions such as xy, x2x^{2} or 1x\frac{1}{x} make an equation nonlinear and require different methods.

Key terms

System
Two or more equations imposed on the same unknowns.
Solution
Values that satisfy every equation in the system simultaneously.
Linear equation
An equation in which each variable appears only to the first power and variables are not multiplied together.
Ordered triple
A three-variable solution written in the fixed order x,y,zx,y,z.

Worked examples

Verify a proposed solution

Does (x,y)=(3,2)(x,y)=(3,2) solve 2x+3y=122x+3y=12 and 5x−3y=95x-3y=9?

  1. 1First equation: 2(3)+3(2)=6+6=122(3)+3(2)=6+6=12.
  2. 2Second equation: 5(3)−3(2)=15−6=95(3)-3(2)=15-6=9.
  3. 3Both equations are true, so the pair is a simultaneous solution.

Answer: Yes, the solution is 3,23,2.

Recognize a nonlinear equation

Is the pair x+y=4x+y=4 and xy=3xy=3 a linear system?

  1. 1The first equation is linear.
  2. 2The product xy in the second equation makes that equation nonlinear.

Answer: No. The system contains a nonlinear equation.

Rules and formulas

General two-variable linear system

a1x+b1y=c1,a2x+b2y=c2a_1x+b_1y=c_1,\qquad a_2x+b_2y=c_2

General three-variable linear equation

ax+by+cz=dax+by+cz=d

Solving systems with two unknowns by substitution

Substitution is efficient when one equation already isolates a variable, or when a variable has coefficient 1 or -1. Rearrange one equation, replace that variable in the other equation, then back-substitute.

Use parentheses around the substituted expression. This protects signs and ensures that every term is multiplied correctly.

If substitution produces a true identity such as 0=00=0, the equations are dependent and further interpretation is needed. If it produces a contradiction such as 0=50=5, the system is inconsistent.

Key terms

Substitution
Replacing a variable by an equal expression from another equation.
Back-substitution
Using a solved variable value in an earlier equation to obtain the remaining variable.

Worked examples

Direct substitution

Solve y=2x−1y=2x-1 and 3x+y=143x+y=14.

  1. 1Substitute y=2x−1y=2x-1 into 3x+y=143x+y=14.
  2. 23x+(2x−1)=143x+(2x-1)=14, so 5x=155x=15 and x=3x=3.
  3. 3Back-substitute: y=2(3)−1=5y=2(3)-1=5.
  4. 4Check: 3(3)+5=143(3)+5=14.

Answer: x,y=(3,5)x,y=(3,5).

Rearrange before substitution

Solve x+y=7x+y=7 and 2x−3y=−12x-3y=-1.

  1. 1From x+y=7x+y=7, write x=7−yx=7-y.
  2. 2Substitute: 2(7−y)−3y=−12(7-y)-3y=-1.
  3. 314−5y=−114-5y=-1, so y=3y=3 and x=4x=4.

Answer: x,y=(4,3)x,y=(4,3).

Rules and formulas

Substitution pattern

x=g(y)⟹a g(y)+by=cx=g(y)\quad\Longrightarrow\quad a\,g(y)+by=c

Solving systems with two unknowns by elimination

Elimination combines equations so that one variable cancels. Multiply one or both equations by nonzero constants to create equal or opposite coefficients, then add or subtract the complete equations.

Write equations in aligned standard form before combining them. This reduces sign errors and makes omitted terms visible.

After finding one variable, substitute into an original equation. A final check is especially important when equations were multiplied or subtracted.

Key terms

Elimination
Combining equations to remove one unknown.
Equivalent equation
An equation with the same solution set, obtained by a valid reversible operation.

Worked examples

Opposite coefficients already present

Solve 2x+3y=122x+3y=12 and 5x−3y=95x-3y=9.

  1. 1Add the equations: 7x=217x=21.
  2. 2Therefore x=3x=3.
  3. 3Substitute into 2x+3y=122x+3y=12: 6+3y=126+3y=12, so y=2y=2.

Answer: x,y=(3,2)x,y=(3,2).

Create matching coefficients

Solve 3x+2y=73x+2y=7 and 5x−3y=45x-3y=4.

  1. 1Multiply the first equation by 3: 9x+6y=219x+6y=21.
  2. 2Multiply the second equation by 2: 10x−6y=810x-6y=8.
  3. 3Add: 19x=2919x=29, so x=2919x=\frac{29}{19}.
  4. 4Substitute into 3x+2y=73x+2y=7 to obtain y=2319y=\frac{23}{19}.

Answer: (x,y)=(2919,2319)(x,y)=\left(\frac{29}{19},\frac{23}{19}\right).

Rules and formulas

Elimination combination

λ(a1x+b1y=c1)+μ(a2x+b2y=c2)\lambda(a_1x+b_1y=c_1)+\mu(a_2x+b_2y=c_2)

Systems with three unknowns

For a 3 by 3 system, eliminate the same variable from two different pairs of equations. This creates a 2 by 2 system in the remaining variables.

Label the equations and record each operation. A clear elimination chain earns method marks and makes the work auditable.

Once two variables are known, back-substitute to find the third, then verify the ordered triple in all three original equations.

Key terms

3 by 3 system
A system of three linear equations in three unknowns.
Triangular system
A system arranged so the final equation has one unknown, the preceding equation has two, and so on.

Worked examples

Systematic elimination

Solve x+y+z=6x+y+z=6, 2x−y+z=32x-y+z=3 and x+2y−z=2x+2y-z=2.

  1. 1Subtract the first equation from the second: x−2y=−3x-2y=-3.
  2. 2Subtract the first equation from the third: y−2z=−4y-2z=-4.
  3. 3From the first new equation, x=2y−3x=2y-3. From the second, y=2z−4y=2z-4.
  4. 4Substitute into x+y+z=6x+y+z=6 to obtain 7z=217z=21, so z=3z=3.
  5. 5Then y=2y=2 and x=1x=1. Check all three original equations.

Answer: x,y,z=(1,2,3)x,y,z=(1,2,3).

Choose a convenient elimination order

Why might x be eliminated first when its coefficients are 1, 2 and -1?

  1. 1Small coefficients need little or no scaling.
  2. 2Combining the first equation with each other equation quickly produces a 2 by 2 system.

Answer: Eliminate x first to minimize arithmetic and sign risk.

Rules and formulas

Three-equation model

a1x+b1y+c1z=d1a2x+b2y+c2z=d2a3x+b3y+c3z=d3\begin{aligned}a_1x+b_1y+c_1z&=d_1\\a_2x+b_2y+c_2z&=d_2\\a_3x+b_3y+c_3z&=d_3\end{aligned}

Matrix representation of a linear system

A linear system can be written compactly as Ax=bA\mathbf{x}=\mathbf{b}, where AA is the coefficient matrix, x\mathbf{x} is the column of unknowns and b\mathbf{b} is the constants column.

The augmented matrix places the constants column beside the coefficients. The vertical divider is organizational; row operations act across the complete row, including the augmented entry.

Fix the variable order before building the matrix. A missing variable must be represented by a zero coefficient.

Key terms

Coefficient matrix
The rectangular array containing the coefficients of the unknowns.
Augmented matrix
The coefficient matrix with the constants column appended.
Matrix equation
The compact representation Ax=bA\mathbf{x}=\mathbf{b} of a linear system.

Worked examples

Build a matrix equation

Represent 2x−y=52x-y=5 and x+3y=1x+3y=1 as Ax=bA\mathbf{x}=\mathbf{b}.

  1. 1Use variable order x,y.
  2. 2Read each row of coefficients and its right-hand constant.

Answer: [2−113][xy]=[51]\begin{bmatrix}2&-1\\1&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\1\end{bmatrix}

Include a zero coefficient

Write x+2z=4x+2z=4 and y−z=1y-z=1 as an augmented matrix in x,y,z order.

  1. 1The first equation has coefficient 0 for y.
  2. 2The second equation has coefficient 0 for x.

Answer: [102401−11]\left[\begin{array}{ccc|c}1&0&2&4\\0&1&-1&1\end{array}\right]

Rules and formulas

Matrix form

Ax=bA\mathbf{x}=\mathbf{b}

Augmented matrix

[A∣b][A\mid\mathbf{b}]

Two-variable coefficient matrix

A=[a1b1a2b2]A=\begin{bmatrix}a_1&b_1\\a_2&b_2\end{bmatrix}

Elementary row operations

Three reversible row operations preserve the solution set: interchange two rows, multiply a row by a nonzero constant, or add a multiple of one row to another row.

Row replacement is written as a single instruction, for example R2R_{2} to R2R_{2}-2R12R_{1}. Calculate every new entry from the old rows before moving to the next operation.

Multiplying by zero is forbidden because it destroys information and is not reversible.

Key terms

Row interchange
Swapping the positions of two equations.
Row scaling
Multiplying an entire row by a nonzero constant.
Row replacement
Adding a multiple of one row to another row.
Pivot
The leading nonzero entry used to eliminate entries below or above it.

Worked examples

Apply one row replacement

Apply R2→R2−2R1R_2\to R_2-2R_1 to [123258]\left[\begin{array}{cc|c}1&2&3\\2&5&8\end{array}\right].

  1. 1Keep R1R_{1} unchanged.
  2. 2Compute the new second row entry by entry: 2,5,82,5,8-21,2,31,2,3=0,1,20,1,2.

Answer: [123012]\left[\begin{array}{cc|c}1&2&3\\0&1&2\end{array}\right]

Choose a pivot row

If the first entry of row 1 is zero but row 2 has a nonzero first entry, what should be done?

  1. 1Interchange the two rows.
  2. 2Use the new leading entry as the first pivot.

Answer: Use R1↔R2R_1\leftrightarrow R_2.

Rules and formulas

Row interchange

Ri↔RjR_i\leftrightarrow R_j

Nonzero row scaling

Ri→λRi,λ≠0R_i\to\lambda R_i,\qquad\lambda\ne0

Row replacement

Ri→Ri+λRjR_i\to R_i+\lambda R_j

Gaussian elimination and back-substitution

Gaussian elimination uses row operations to reach row-echelon form: pivots move to the right as rows descend, and entries below each pivot are zero.

Once the system is triangular, solve the last equation and back-substitute upward. Gauss-Jordan elimination continues until each pivot is the only nonzero entry in its column, producing reduced row-echelon form.

Exact fractions expose structure. Premature decimal rounding can turn a dependent row into an apparently nonzero row or distort a parameter condition.

Key terms

Row-echelon form
A stepped matrix with zero entries below each pivot.
Reduced row-echelon form
Row-echelon form in which each pivot is 1 and is the only nonzero entry in its column.
Gaussian elimination
Forward elimination to echelon form followed by back-substitution.
Gauss-Jordan elimination
Row reduction continued to reduced row-echelon form.

Worked examples

Gaussian elimination for a 3 by 3 system

Solve x+2y−z=3x+2y-z=3, 2x+5y+z=92x+5y+z=9 and −x+y+2z=2-x+y+2z=2.

  1. 1Start with [12−132519−1122]\left[\begin{array}{ccc|c}1&2&-1&3\\2&5&1&9\\-1&1&2&2\end{array}\right].
  2. 2Use R2→R2−2R1R_2\to R_2-2R_1 and R3→R3+R1R_3\to R_3+R_1 to get [12−1301330315]\left[\begin{array}{ccc|c}1&2&-1&3\\0&1&3&3\\0&3&1&5\end{array}\right].
  3. 3Use R3→R3−3R2R_3\to R_3-3R_2 to obtain −8z=−4-8z=-4, so z=12z=\frac12.
  4. 4Back-substitute: y+3z=3y+3z=3 gives y=32y=\frac32, then x+2y−z=3x+2y-z=3 gives x=12x=\frac12.

Answer: (x,y,z)=(12,32,12)(x,y,z)=\left(\frac12,\frac32,\frac12\right).

Read a triangular system

Interpret [12−1401350026]\left[\begin{array}{ccc|c}1&2&-1&4\\0&1&3&5\\0&0&2&6\end{array}\right].

  1. 1The last row gives 2z=62z=6, so z=3z=3.
  2. 2The second row gives y+9=5y+9=5, so y=−4y=-4.
  3. 3The first row gives x−8−3=4x-8-3=4, so x=15x=15.

Answer: x,y,z=(15,−4,3)x,y,z=(15,-4,3).

Rules and formulas

Row-echelon structure

[∗∗∗0∗∗00∗]\begin{bmatrix}*&*&*\\0&*&*\\0&0&*\end{bmatrix}

Back-substitution direction

z⟶y⟶xz\longrightarrow y\longrightarrow x

Unique solution, no solution and infinitely many solutions

A unique solution occurs when every variable column contains a pivot. There are no free variables, so one ordered tuple satisfies the system.

No solution occurs when row reduction creates a contradiction row such as 0=10=1. The augmented column introduces a pivot that is absent from the coefficient matrix.

Infinitely many solutions occur when the system is consistent but at least one variable is free. A zero row 0=00=0 indicates redundant information, not a contradiction.

Rank is the number of pivots. The system Ax=bA\mathbf{x}=\mathbf{b} is consistent exactly when rank⁡(A)=rank⁡([A∣b])\operatorname{rank}(A)=\operatorname{rank}([A\mid\mathbf{b}]). If that common rank is smaller than the number of unknowns, free variables produce infinitely many solutions.

Key terms

Rank
The number of pivot positions in a matrix.
Free variable
A non-pivot variable allowed to vary over its domain.
Contradiction row
A row representing 0=c0=c for nonzero c, proving the system inconsistent.

Worked examples

Classify three echelon forms

Classify systems ending in ii 0=40=4, (ii) 0=00=0 with a free variable, and (iii) one pivot per variable. In (ii) and (iii), assume there is no contradictory row anywhere in the echelon form.

  1. 1Case ii is contradictory.
  2. 2Case iiii is consistent but underdetermined.
  3. 3Case iiiiii fixes every variable.

Answer: ii no solution; iiii infinitely many solutions; iiiiii one unique solution.

Communicate an infinite family

Solve x+2y=3x+2y=3 and 2x+4y=62x+4y=6.

  1. 1The second equation is twice the first, so one equation is redundant.
  2. 2Let y=ty=t for any real t.
  3. 3Then x=3−2tx=3-2t.

Answer: (x,y)=(3−2t,t),t∈R(x,y)=(3-2t,t),\qquad t\in\mathbb{R}.

Rules and formulas

Consistency criterion

rank⁡(A)=rank⁡([A∣b])\operatorname{rank}(A)=\operatorname{rank}([A\mid\mathbf b])

Unique-solution rank condition

rank⁡(A)=rank⁡([A∣b])=n\operatorname{rank}(A)=\operatorname{rank}([A\mid\mathbf b])=n

Infinite-solution rank condition

rank⁡(A)=rank⁡([A∣b])<n\operatorname{rank}(A)=\operatorname{rank}([A\mid\mathbf b])<n

No-solution rank condition

rank⁡(A)<rank⁡([A∣b])\operatorname{rank}(A)<\operatorname{rank}([A\mid\mathbf b])

Geometric interpretation of lines and planes

In two dimensions, each linear equation represents a line. Two nonparallel lines meet once, distinct parallel lines never meet, and coincident lines share every point.

In three dimensions, a linear equation represents a plane. Three planes may meet at one point, share a line or a plane, or have no common point. Pairwise intersections alone do not guarantee one point common to all three planes.

The algebraic and geometric classifications agree: unique means one common point, inconsistent means no common point, and dependent means a common family of points.

Three coordinate panels showing intersecting, parallel and coincident line pairs.Intersecting, parallel and coincident lines correspond to one solution, no solution and infinitely many solutions.

Open graph full size (opens in a new tab)

Key terms

Intersecting lines
Nonparallel lines with exactly one common point.
Parallel lines
Distinct coplanar lines with equal gradient and no common point.
Coincident lines
Different equation forms representing the same line.
Plane
The three-dimensional graph of a linear equation in x, y and z.

Worked examples

Read a 2D system geometrically

Classify y=2x+1y=2x+1 and y=2x−3y=2x-3.

  1. 1Both lines have gradient 2.
  2. 2Their y-intercepts differ, so they are distinct parallel lines.

Answer: The system has no solution.

Recognize coincident equations

Compare x+2y=3x+2y=3 and 3x+6y=93x+6y=9.

  1. 1Every coefficient and the constant in the second equation is three times the first.
  2. 2The equations describe the same line.

Answer: The system has infinitely many solutions on x+2y=3x+2y=3.

Rules and formulas

Two-line determinant

D=a1b2−a2b1D=a_1b_2-a_2b_1

Nonparallel-line condition

D≠0⟹one intersectionD\ne0\quad\Longrightarrow\quad\text{one intersection}

Consistency, independence and dependency

A consistent system has at least one solution. An inconsistent system has no solution.

Independent equations contribute distinct constraints. Dependent equations contain redundant information because one equation is a linear combination of others.

Dependency does not automatically mean infinitely many solutions. A larger system may contain one redundant equation while the remaining independent equations still determine a unique solution.

For a square 2 by 2 system, a nonzero determinant guarantees a unique solution. A zero determinant tells you only that the system is nonunique; compare constants or row-reduce to distinguish no solution from infinitely many solutions.

Key terms

Consistent system
A system with at least one solution.
Inconsistent system
A system with no solution.
Independent equations
Equations that supply nonredundant constraints.
Dependent equations
Equations for which at least one is a linear combination of the others.

Worked examples

Dependent pair

Classify 2x−4y=62x-4y=6 and −x+2y=−3-x+2y=-3.

  1. 1Multiplying the second equation by -2 gives the first.
  2. 2The equations describe the same line.

Answer: The system is consistent and dependent, with infinitely many solutions.

Zero determinant needs another test

What can be concluded immediately when D=0D=0 for a 2 by 2 system?

  1. 1The coefficient rows are dependent.
  2. 2The constants may preserve the same ratio or break it.

Answer: There is not a unique solution; row-reduce or compare ratios to decide between none and infinitely many.

Rules and formulas

Two-by-two determinant

det⁡A=∣a1b1a2b2∣=a1b2−a2b1\det A=\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix}=a_1b_2-a_2b_1

Unique 2 by 2 system

det⁡A≠0\det A\ne0

Parameter-based systems

A parameter changes the coefficients or constants, so the solution type can change at special values. Identify the expression that controls a pivot or determinant, then split into cases.

Never divide by an expression involving a parameter until the zero case has been handled separately. Dividing by k-1 silently discards the case k=1k=1.

For a square system, determinant not equal to zero gives a unique solution. At determinant zero, inspect the augmented equations to decide whether the system is dependent or inconsistent.

Key terms

Parameter
A fixed but unspecified quantity whose value may change the behavior of a system.
Case split
Separate analysis for parameter values that make a controlling expression zero or nonzero.

Worked examples

A parameter controlling uniqueness

For x+y=2x+y=2 and kx+y=3+y=3, classify and solve the system.

  1. 1Subtract the first equation from the second: (k−1)x=1(k-1)x=1.
  2. 2If k≠1k\ne1, x=1k−1x=\frac{1}{k-1} and y=2−1k−1=2k−3k−1y=2-\frac{1}{k-1}=\frac{2k-3}{k-1}.
  3. 3If k=1k=1, the equations become x+y=2x+y=2 and x+y=3x+y=3, a contradiction.

Answer: k≠1: (x,y)=(1k−1,2k−3k−1);k=1: no solutionk\ne1:\ (x,y)=\left(\frac1{k-1},\frac{2k-3}{k-1}\right);\qquad k=1:\ \text{no solution}.

Parameter in the augmented column

Classify x+2y=3x+2y=3 and 2x+4y=m2x+4y=m.

  1. 1The second coefficient row is twice the first.
  2. 2If m=6m=6, the full equation is twice the first and the system is dependent.
  3. 3If m is not 6, row reduction gives 0=m−60=m-6, a contradiction.

Answer: m=6: infinitely many solutions;m≠6: no solutionm=6:\ \text{infinitely many solutions};\qquad m\ne6:\ \text{no solution}.

Rules and formulas

Parameter case trigger

p(k)=0orp(k)≠0p(k)=0\quad\text{or}\quad p(k)\ne0

Determinant case split

det⁡A(k)≠0⇒unique;det⁡A(k)=0⇒inspect [A∣b]\det A(k)\ne0\Rightarrow\text{unique};\qquad \det A(k)=0\Rightarrow\text{inspect }[A\mid\mathbf b]

Applications and mathematical modeling

A modeling problem begins by defining unknowns with units. Translate each independent condition into a linear equation, solve, then interpret the tuple in the original context.

Check feasibility as well as algebra: counts may need to be nonnegative integers, masses and times should be nonnegative, and a monetary total should match the stated revenue.

If the equations are inconsistent, the data or assumptions cannot all hold simultaneously. If they are dependent, the information is insufficient for one unique answer.

Key terms

Constraint
A condition translated into an equation or inequality.
Feasible solution
A mathematical solution that also satisfies contextual restrictions.
Model validation
Checking the result against the original data, units and assumptions.

Worked examples

Ticket revenue model

A theater sells 120 tickets. Adult tickets cost 12 dollars and student tickets cost 7 dollars. Revenue is 1090 dollars. Find each number.

  1. 1Let a be adult tickets and s be student tickets.
  2. 2Write a+s=120a+s=120 and 12a+7s=109012a+7s=1090.
  3. 3Use s=120−as=120-a: 12a+7(120−a)=109012a+7(120-a)=1090.
  4. 4Then 5a=2505a=250, so a=50a=50 and s=70s=70.
  5. 5Check the revenue: 12(50)+7(70)=109012(50)+7(70)=1090.

Answer: 50 adult tickets and 70 student tickets.

Mixture model

How many liters of 20 percent and 50 percent solution make 30 liters of 40 percent solution?

  1. 1Let x and y be the respective volumes.
  2. 2Write x+y=30x+y=30 and 0.20x+0.50y=0.40(30)=120.20x+0.50y=0.40(30)=12.
  3. 3Solving gives y=20y=20 and x=10x=10.
  4. 4Both volumes are nonnegative and sum to 30.

Answer: 10 L of 20 percent solution and 20 L of 50 percent solution.

Rules and formulas

Two-category total

x+y=Tx+y=T

Weighted-total model

p1x+p2y=Pp_1x+p_2y=P

Calculator and GDC methods

A GDC may provide a simultaneous-equation solver, matrix inverse command, row-reduced echelon form command, or numerical solve feature. Enter the coefficients in a fixed variable order and include zeros for missing variables.

For a nonsingular square matrix, x=A−1bx=A^{-1}b is valid. If A is singular, the inverse does not exist; use row reduction to distinguish no solution from infinitely many solutions.

GDC output must still be interpreted. A decimal tuple should be checked in the original equations, and parameter-based classification requires algebraic reasoning rather than sampling a few values.

On examinations, record the equations or augmented matrix and state the solution. Technology is a tool, not a replacement for mathematical communication.

Key terms

Singular matrix
A square matrix with determinant zero and no inverse.
RREF command
A calculator command that returns reduced row-echelon form.
Residual
The difference between a left-hand side evaluated at the proposed solution and its right-hand constant.

Worked examples

Use inverse-matrix mode safely

What should be checked before calculating A inverse times b?

  1. 1Confirm A is square.
  2. 2Confirm detAA is not zero.
  3. 3Enter b in the same equation order and variables in the same column order.

Answer: Use x=A−1b\mathbf{x}=A^{-1}\mathbf{b} only when det⁡A≠0\det A\ne0.

Interpret a calculator error

A GDC reports ‘singular matrix’. Does this prove there is no solution?

  1. 1A singular coefficient matrix proves only that a unique solution is impossible.
  2. 2Row-reduce the augmented matrix to test consistency.

Answer: No. The system may have no solution or infinitely many solutions.

Rules and formulas

Inverse-matrix solution

x=A−1b,det⁡A≠0\mathbf{x}=A^{-1}\mathbf{b},\qquad\det A\ne0

Residual vector

r=Ax^−b\mathbf r=A\widehat{\mathbf x}-\mathbf b

Common mistakes and how to prevent them

Sign errors often enter when subtracting equations. Put brackets around the full equation being subtracted or add a clearly written negative multiple instead.

A row operation must change every entry in the row, including the augmented constant. Applying it only to selected coefficients changes the system.

Do not confuse a zero row with a contradiction row, dependent equations with inconsistent equations, or a singular matrix with proof of no solution.

In parameter questions, dividing by an expression before treating its zero case loses a possible classification. In contextual problems, failing to interpret units or feasibility leaves the solution incomplete.

Premature decimal rounding can hide exact dependency. Keep exact values during elimination and check the final tuple in every original equation.

Worked examples

Spot an illegal row operation

A student uses R2→0R2R_2\to0R_2. Why is this invalid?

  1. 1The original second equation is replaced by 0=00=0.
  2. 2The operation cannot be reversed and can destroy a necessary constraint. It is not an allowed elementary row operation.

Answer: Multiplying a row by zero does not in general preserve the solution set. A particular redundant row may be removed without changing solutions, but that does not make zero row-scaling a valid elementary operation.

Spot a lost parameter case

A student divides (k−2)x=3(k-2)x=3 by k-2 without comment.

  1. 1The division assumes k is not 2.
  2. 2The case k=2k=2 must be tested in the original system.

Answer: Split into k=2k=2 and k not equal to 2 before dividing.

IB HL exam strategy and communication

Start by deciding whether substitution, elimination, matrices or technology is most efficient. For small exact systems, algebra usually communicates the structure best; for larger numerical systems, a GDC can reduce arithmetic.

Show enough structure for method marks: define variables, align equations, label row operations, identify parameter cases and state the final solution type.

If the system has infinitely many solutions, introduce free parameters and specify their domain. If it has no solution, show the contradiction. If it has one solution, verify the tuple.

For a contextual response, finish with a sentence containing units and a reasonableness check. For approximate GDC results, use the requested accuracy and avoid rounding intermediate values.

Worked examples

Select a method

Choose a method for y=4x−7y=4x-7 and 3x+2y=83x+2y=8.

  1. 1One variable is already isolated.
  2. 2Substitution needs no preliminary scaling.

Answer: Use substitution.

State a complete classification

Row reduction ends with [102301−140000]\left[\begin{array}{ccc|c}1&0&2&3\\0&1&-1&4\\0&0&0&0\end{array}\right].

  1. 1There are pivots in x and y columns but not z.
  2. 2Let z=tz=t.
  3. 3Then x=3−2tx=3-2t and y=4+ty=4+t.

Answer: The system is consistent with infinitely many solutions: (x,y,z)=(3−2t,4+t,t), t∈R(x,y,z)=(3-2t,4+t,t),\ t\in\mathbb R.

Rules and formulas

Solution verification

Ax^=bA\widehat{\mathbf x}=\mathbf b

IB HL-style practice with worked solutions

Attempt each problem without the solution first. Use exact arithmetic unless an approximation is requested, and verify numerical solutions in the original equations.

The set progresses from two-variable techniques through row reduction, classification, parameters, modeling and three-variable reasoning.

Worked examples

Practice 1: substitution

Solve y=2x−1y=2x-1 and 3x+y=143x+y=14.

  1. 1Substitute y into the second equation: 3x+2x−1=143x+2x-1=14.
  2. 2Thus x=3x=3 and y=5y=5.
  3. 3Both equations check.

Answer: x,y=(3,5)x,y=(3,5).

Practice 2: elimination

Solve 2x+3y=122x+3y=12 and 5x−3y=95x-3y=9.

  1. 1Add the equations to get 7x=217x=21.
  2. 2Hence x=3x=3.
  3. 3Substitute to get y=2y=2.

Answer: x,y=(3,2)x,y=(3,2).

Practice 3: exact fractional solution

Solve 3x+2y=73x+2y=7 and 5x−3y=45x-3y=4.

  1. 1Use 9x+6y=219x+6y=21 and 10x−6y=810x-6y=8.
  2. 2Add to obtain 19x=2919x=29.
  3. 3Then x=2919x=\frac{29}{19} and y=2319y=\frac{23}{19}.

Answer: (x,y)=(2919,2319)(x,y)=\left(\frac{29}{19},\frac{23}{19}\right).

Practice 4: three unknowns

Solve x+y+z=4x+y+z=4, x−y+z=2x-y+z=2 and 2x+y−z=12x+y-z=1.

  1. 1Subtract the second equation from the first: 2y=22y=2, so y=1y=1.
  2. 2Then x+z=3x+z=3.
  3. 3The third equation becomes 2x−z=02x-z=0, so z=2xz=2x.
  4. 4Hence 3x=33x=3, giving x=1x=1 and z=2z=2.

Answer: x,y,z=(1,1,2)x,y,z=(1,1,2).

Practice 5: Gaussian elimination

Solve x+2y−z=3x+2y-z=3, 2x+5y+z=92x+5y+z=9 and −x+y+2z=2-x+y+2z=2.

  1. 1Use R2→R2−2R1R_2\to R_2-2R_1 and R3→R3+R1R_3\to R_3+R_1.
  2. 2Then use R3→R3−3R2R_3\to R_3-3R_2 to get −8z=−4-8z=-4.
  3. 3Back-substitution gives z=12z=\frac12, y=32y=\frac32 and x=12x=\frac12.

Answer: (x,y,z)=(12,32,12)(x,y,z)=\left(\frac12,\frac32,\frac12\right).

Practice 6: inconsistent system

Classify x+2y=3x+2y=3 and 2x+4y=72x+4y=7.

  1. 1Subtract twice the first equation from the second.
  2. 2This gives 0=10=1, a contradiction.

Answer: The system is inconsistent and has no solution.

Practice 7: dependent system

Solve x+2y=3x+2y=3 and 2x+4y=62x+4y=6.

  1. 1The second equation is twice the first.
  2. 2Let y=ty=t.
  3. 3Then x=3−2tx=3-2t.

Answer: (x,y)=(3−2t,t),t∈R(x,y)=(3-2t,t),\quad t\in\mathbb R.

Practice 8: parameter controlling uniqueness

Classify x+y=2x+y=2 and kx+y=3+y=3 for all real k.

  1. 1Subtract to get (k−1)x=1(k-1)x=1.
  2. 2For k≠1k\ne1, x=1k−1x=\frac1{k-1} and y=2k−3k−1y=\frac{2k-3}{k-1}.
  3. 3For k=1k=1, the equations contradict each other.

Answer: k≠1: unique solution;k=1: no solutionk\ne1:\ \text{unique solution};\qquad k=1:\ \text{no solution}.

Practice 9: parameter in the constants

Classify x+2y=3x+2y=3 and 2x+4y=m2x+4y=m.

  1. 1The coefficient rows are proportional.
  2. 2For m=6m=6, the whole second equation is twice the first.
  3. 3For m not equal to 6, elimination gives a contradiction.

Answer: m=6: infinitely many;m≠6: nonem=6:\ \text{infinitely many};\qquad m\ne6:\ \text{none}.

Practice 10: ticket model

A theater sells 120 adult and student tickets for 1090 dollars. Adult tickets cost 12 dollars and student tickets 7 dollars. Find the numbers.

  1. 1Let a+s=120a+s=120 and 12a+7s=109012a+7s=1090.
  2. 2Substitute s=120−as=120-a to get 5a=2505a=250.
  3. 3Thus a=50a=50 and s=70s=70.
  4. 4The total and revenue both check.

Answer: 50 adult tickets and 70 student tickets.

Practice 11: interpret RREF

Interpret [10−1201210000]\left[\begin{array}{ccc|c}1&0&-1&2\\0&1&2&1\\0&0&0&0\end{array}\right].

  1. 1The z column has no pivot, so let z=tz=t.
  2. 2The first row gives x−z=2x-z=2, so x=2+tx=2+t.
  3. 3The second gives y+2z=1y+2z=1, so y=1−2ty=1-2t.

Answer: (x,y,z)=(2+t,1−2t,t),t∈R(x,y,z)=(2+t,1-2t,t),\quad t\in\mathbb R.

Practice 12: a three-equation parameter system

Classify x+y+z=1x+y+z=1, x+ay+z=2x+ay+z=2 and x+y+az=3x+y+az=3 for all real aa.

  1. 1Subtract the first equation from the second: (a−1)y=1(a-1)y=1.
  2. 2Subtract the first from the third: (a−1)z=2(a-1)z=2.
  3. 3For a≠1a\ne1, y=1a−1y=\frac1{a-1}, z=2a−1z=\frac2{a-1} and x=1−3a−1x=1-\frac3{a-1}.
  4. 4For a=1a=1, the first two equations become x+y+z=1x+y+z=1 and x+y+z=2x+y+z=2, a contradiction.

Answer: a≠1: (x,y,z)=(1−3a−1,1a−1,2a−1);a=1: no solutiona\ne1:\ (x,y,z)=\left(1-\frac3{a-1},\frac1{a-1},\frac2{a-1}\right);\qquad a=1:\ \text{no solution}.

Topic summary and cheat sheet

A solution satisfies every equation simultaneously. Use substitution for an isolated variable, elimination for convenient matching coefficients, and Gaussian elimination for systematic matrix work.

One pivot per variable gives a unique solution. A contradiction row gives no solution. A consistent system with free variables has infinitely many solutions.

In two dimensions these cases correspond to intersecting, parallel and coincident lines. In three dimensions, interpret equations as planes and focus on their common intersection.

For parameters, split the zero and nonzero pivot or determinant cases before dividing. With a GDC, record the setup, interpret singular output and verify the result.

Rules and formulas

System form

Ax=bA\mathbf x=\mathbf b

Allowed row replacement

Ri→Ri+λRjR_i\to R_i+\lambda R_j

Consistency test

rank⁡(A)=rank⁡([A∣b])\operatorname{rank}(A)=\operatorname{rank}([A\mid\mathbf b])

Unique solution test

rank⁡(A)=rank⁡([A∣b])=n\operatorname{rank}(A)=\operatorname{rank}([A\mid\mathbf b])=n

Infinite solutions test

rank⁡(A)=rank⁡([A∣b])<n\operatorname{rank}(A)=\operatorname{rank}([A\mid\mathbf b])<n

No solution test

rank⁡(A)<rank⁡([A∣b])\operatorname{rank}(A)<\operatorname{rank}([A\mid\mathbf b])

Two-by-two determinant test

det⁡A=a1b2−a2b1\det A=a_1b_2-a_2b_1

Inverse method condition

x=A−1b,det⁡A≠0\mathbf x=A^{-1}\mathbf b,\qquad\det A\ne0

Topic summary

  • A system requires the same values to satisfy every equation simultaneously.
  • Solutions are ordered pairs for two variables and ordered triples for three variables.
  • Substitution is efficient when a variable is already isolated or easy to isolate.
  • Elimination combines equivalent equations to cancel an unknown.
  • For three unknowns, eliminate the same variable twice to create a 2 by 2 system.
  • The matrix equation Ax=bA\mathbf{x}=\mathbf{b} separates coefficients, unknowns and constants.
  • An augmented matrix must preserve a fixed variable order and include zero coefficients.
  • Valid row operations are row interchange, nonzero scaling and row replacement.
  • Gaussian elimination reaches echelon form; back-substitution then recovers the variables.
  • Gauss-Jordan elimination continues to reduced row-echelon form.
  • A pivot in every variable column gives a unique solution.
  • A contradiction row 0=c0=c with c nonzero proves there is no solution.
  • A zero row 0=00=0 indicates redundancy, not inconsistency.
  • A consistent system with free variables has infinitely many solutions.
  • Consistency requires rank⁡(A)=rank⁡([A∣b])\operatorname{rank}(A)=\operatorname{rank}([A\mid\mathbf{b}]).
  • In two dimensions, unique, none and infinitely many correspond to intersecting, parallel and coincident lines.
  • In three dimensions, linear equations represent planes and the solution is their common intersection.
  • Consistent versus inconsistent and independent versus dependent describe different properties.
  • A zero determinant rules out uniqueness but does not distinguish none from infinitely many solutions.
  • Parameter systems require separate zero and nonzero cases before division.
  • A singular GDC result means no unique solution; row reduction must determine the remaining classification.
  • Exact arithmetic protects dependency and parameter structure from rounding errors.
  • Contextual solutions must satisfy units, nonnegativity, integrality and other feasibility conditions.
  • Finish by checking the solution in every original equation and stating the classification clearly.