Types of Linear Equations: Systems, Methods, and Solved Examples

Linear equations are the most frequently tested algebra topic across TCS NQT, AMCAT, and eLitmus quantitative sections. Every placement-aptitude exam includes at least a couple of problems that reduce to solving a system of first-degree equations.

This article covers the classification of linear systems, the standard solution methods, and worked problems at placement-exam difficulty.

What Makes an Equation Linear

A linear equation is any equation where every variable appears with an exponent of exactly 1. No squares, no cubes, no roots of variables. The general forms by variable count:

Variables	General form	Example
One	ax + b = 0	3x + 6 = 0 (solution: x = -2)
Two	ax + by = c	2x + 5y = 10
Three	ax + by + cz = d	x + 2y + 3z = 14

Key properties:

• The degree (highest power of any variable) is always 1
• Solving for n unknowns requires n independent equations
• The graph of a two-variable linear equation is a straight line; for three variables, a plane

Types of Linear Systems

For a pair of equations a1x + b1y = c1 and a2x + b2y = c2, the system type depends on coefficient ratios as described in NCERT Class 10 Chapter 3:

Type	Condition	Solutions	Geometric meaning
Consistent (unique)	a1/a2 not equal to b1/b2	Exactly one	Two lines intersecting at one point
Inconsistent	a1/a2 = b1/b2 but not equal to c1/c2	None	Two parallel lines
Dependent	a1/a2 = b1/b2 = c1/c2	Infinite	Same line

This ratio test is the fastest way to classify a system on an aptitude exam without solving it fully.

Consistent System (Unique Solution)

• Equations: x + y = 7 and x + 2y = 10
• Ratio check: 1/1 not equal to 1/2, so consistent.
• Solution: Subtract equation 1 from equation 2: y = 3. Then x = 4.
• Verification: 4 + 3 = 7 (correct), 4 + 6 = 10 (correct).

Inconsistent System (No Solution)

• Equations: 7x + 8y = 15 and 14x + 16y = 20
• Ratio check: 7/14 = 1/2, 8/16 = 1/2, 15/20 = 3/4. Since 1/2 = 1/2 but not 3/4, inconsistent.
• Why: Multiply equation 1 by 2: gives 14x + 16y = 30. But equation 2 says 14x + 16y = 20. Contradiction.

Dependent System (Infinite Solutions)

• Equations: 6x + 21y = 75 and 4x + 14y = 50
• Ratio check: 6/4 = 3/2, 21/14 = 3/2, 75/50 = 3/2. All ratios equal.
• Why infinite: Multiply equation 2 by 3/2: gives 6x + 21y = 75, identical to equation 1.

Methods to Solve Linear Equations

Four standard methods appear in placement-prep contexts. Khan Academy’s systems module covers these in depth.

Substitution Method

Best when one variable already has coefficient 1.

• Problem: x + 2y = 8 and 3x + y = 9
• Step 1: From equation 1, x = 8 - 2y
• Step 2: Substitute into equation 2: 3(8 - 2y) + y = 9
• Step 3: 24 - 6y + y = 9, so -5y = -15, giving y = 3
• Step 4: x = 8 - 2(3) = 2
• Answer: x = 2, y = 3
• Check: 2 + 6 = 8 (correct), 6 + 3 = 9 (correct).

Elimination Method

Fastest method for timed exams when coefficients are small integers.

• Problem: 2x + 3y = 12 and 4x + y = 14
• Step 1: Multiply equation 2 by 3: 12x + 3y = 42
• Step 2: Subtract equation 1: 10x = 30, so x = 3
• Step 3: Substitute back: 4(3) + y = 14, so y = 2
• Answer: x = 3, y = 2
• Check: 6 + 6 = 12 (correct), 12 + 2 = 14 (correct).

Cross-Multiplication Method

For equations in the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0:

• Formula: x / (b1*c2 - b2*c1) = y / (c1*a2 - c2*a1) = 1 / (a1*b2 - a2*b1)

• Problem: 2x + 3y - 12 = 0 and 4x + y - 14 = 0

• a1=2, b1=3, c1=-12, a2=4, b2=1, c2=-14

• Denominator: 2(1) - 4(3) = -10

• x-numerator: 3(-14) - 1(-12) = -42 + 12 = -30

• y-numerator: (-12)(4) - (-14)(2) = -48 + 28 = -20

• Result: x = -30/-10 = 3, y = -20/-10 = 2

Matrix Method (Cramer’s Rule)

For ax + by = e and cx + dy = f, the solution is x = (ed - bf)/(ad - bc) and y = (af - ce)/(ad - bc).

• Problem: 2x + 3y = 12 and 4x + y = 14
• D = 2(1) - 4(3) = -10
• x = (12*1 - 3*14)/-10 = (12 - 42)/-10 = 3
• y = (2*14 - 4*12)/-10 = (28 - 48)/-10 = 2

Placement-Aptitude Worked Problems

Problem 1: Age-Based

• Q: The sum of ages of A and B is 48. Six years ago, A was twice as old as B. Find their present ages.
• Let A’s age = x, B’s age = y
• Equation 1: x + y = 48
• Equation 2: x - 6 = 2(y - 6) simplifies to x - 2y = -6
• From equation 1: x = 48 - y. Substituting: 48 - y - 2y = -6
• Solving: 48 - 3y = -6, so 3y = 54, giving y = 18
• Back-substitute: x = 48 - 18 = 30
• Answer: A is 30, B is 18.
• Check: Sum = 48. Six years ago: A was 24, B was 12. Is 24 = 2(12)? Yes.

Problem 2: Cost-Based

• Q: 3 notebooks and 2 pens cost 120 rupees. 2 notebooks and 3 pens cost 110 rupees. Find the cost of each item.
• Let notebook = x, pen = y
• Equation 1: 3x + 2y = 120
• Equation 2: 2x + 3y = 110
• Multiply eq 1 by 3, eq 2 by 2: 9x + 6y = 360, 4x + 6y = 220
• Subtract: 5x = 140, so x = 28
• Substitute: 2(28) + 3y = 110, so 3y = 54, giving y = 18
• Answer: Notebook = 28 rupees, Pen = 18 rupees.
• Check: 3(28) + 2(18) = 84 + 36 = 120 (correct). 2(28) + 3(18) = 56 + 54 = 110 (correct).

Problem 3: Speed-Distance

• Q: A boat goes 30 km upstream in 3 hours and 44 km downstream in 2 hours. Find the speed of the boat in still water and the speed of the stream.
• Let boat speed = x km/h, stream speed = y km/h
• Upstream: x - y = 30/3 = 10
• Downstream: x + y = 44/2 = 22
• Add both: 2x = 32, so x = 16
• Subtract: 2y = 12, so y = 6
• Answer: Boat speed = 16 km/h, stream speed = 6 km/h.
• Check: Upstream = 10 km/h, covers 30 km in 3 hours (correct). Downstream = 22 km/h, covers 44 km in 2 hours (correct).

When to Use Which Method

Method	Best for	Avoid when
Elimination	Small integer coefficients; timed exams	Large prime coefficients
Substitution	One variable has coefficient 1	Both have multi-digit coefficients
Cross-multiplication	Standard-form competitive exam problems	Quick mental math scenarios
Cramer’s rule	Three-variable systems; formula-based approach	Simple two-variable systems

If you’re also preparing for calendar aptitude problems or clock-based questions, the same approach applies: classify first, pick a method second. Coding and decoding patterns follow analogous logic.

From Ratios to Matrices

The coefficient-ratio test from the systems section is a simplified version of checking whether a matrix determinant equals zero. That single concept connects school algebra to linear algebra, which underlies every gradient-descent step in machine learning. If Cramer’s rule and determinant calculations from this article felt approachable, TinkerLLM’s interactive modules show how those same matrix operations scale into real ML computations, starting at 299 rupees.