Unit: Algebra – 1
Chapter: Solving Linear Equations – Algebraic Methods
Reference: – Introduction to Solving Linear Equations in Two Variables, Systems of Linear Equations, Substitution Method, Elimination Method, Cross Multiplication Method, Comparison Method, Consistent vs Inconsistent Systems, Dependent Systems, Solved Examples, Odd-One-Out Problems, Common Mistakes
After studying this chapter, you should be able to understand:
- What is a System of Linear Equations in Two Variables
- Substitution Method
- Elimination Method
- Cross Multiplication Method
- Identifying Consistent, Inconsistent, and Dependent Systems
Introduction to Solving Linear Equations in Two Variables
Definition
When we have two linear equations in two variables (usually x and y), we call it a system of linear equations. Solving the system means finding the ordered pair (x, y) that satisfies both equations simultaneously.
In real life, we often need two equations to solve for two unknowns. For example, if you buy apples and oranges, knowing the total cost and the total number of items gives two equations.
When we solve a system of linear equations, we essentially ask:
"What is the common point (x, y) that lies on both lines?"
Geometrically, the solution is the intersection point of the two lines.
Importance of Solving Systems
- Solves real-world problems involving two unknowns (prices, speeds, mixtures, etc.)
- Foundation for higher algebra, linear programming, and optimization
- Used extensively in economics, engineering, physics, and computer science
Example
System:
x + y = 5
x – y = 1
Solution: Adding the two equations gives 2x = 6 → x = 3, then y = 2. So the solution is (3, 2), which satisfies both equations.
Subtopics
1. System of Linear Equations – Types of Solutions
When we solve two linear equations in two variables, three possibilities exist:
Case 1 – Unique Solution (Consistent System):
The two lines intersect at exactly one point. There is one unique pair (x, y) that satisfies both equations.
Example: x + y = 5 and x – y = 1 → solution (3, 2)
Case 2 – No Solution (Inconsistent System):
The two lines are parallel and never intersect. There is no pair (x, y) that satisfies both equations.
Example: x + y = 5 and x + y = 7 (same left side, different right side)
Case 3 – Infinite Solutions (Dependent System):
The two lines coincide (are the same line). Every point on the line is a solution.
Example: x + y = 5 and 2x + 2y = 10 (second equation is just twice the first)
General Form for Checking: For equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
- Unique solution if a₁/a₂ ≠ b₁/b₂
- No solution if a₁/a₂ = b₁/b₂ ≠ c₁/c₂
- Infinite solutions if a₁/a₂ = b₁/b₂ = c₁/c₂
Method 1: Substitution Method
Definition
In the substitution method, we solve one equation for one variable in terms of the other, and then substitute that expression into the second equation.
Steps of Substitution Method:
- Step 1: From one equation, express one variable (say y) in terms of the other variable (x).
- Step 2: Substitute this expression into the other equation.
- Step 3: Solve the resulting single-variable equation for x.
- Step 4: Substitute x back to find y.
- Step 5: Verify the solution in both original equations.
Example 1: Solve by substitution: x + y = 10 and 2x – y = 5
Solution:
From the first equation: y = 10 – x
Substitute into second equation: 2x – (10 – x) = 5
Simplify: 2x – 10 + x = 5 → 3x – 10 = 5 → 3x = 15 → x = 5
Now find y: y = 10 – 5 = 5
Verification: First: 5+5=10 ✓; Second: 2(5)-5=10-5=5 ✓
Answer: (5, 5)
Example 2: Solve by substitution: 3x + 2y = 12 and y = x + 1
Solution:
Since y = x + 1 is already solved for y, substitute into first equation:
3x + 2(x + 1) = 12 → 3x + 2x + 2 = 12 → 5x + 2 = 12 → 5x = 10 → x = 2
Now find y: y = x + 1 = 2 + 1 = 3
Verification: First: 3(2)+2(3)=6+6=12 ✓; Second: 3=2+1 ✓
Answer: (2, 3)
Example 3: Solve by substitution: 2x + y = 7 and x – y = 2
Solution:
From the second equation: x = y + 2
Substitute into first: 2(y + 2) + y = 7 → 2y + 4 + y = 7 → 3y + 4 = 7 → 3y = 3 → y = 1
Then x = 1 + 2 = 3
Answer: (3, 1)
Method 2: Elimination Method
Definition
In the elimination method, we add or subtract the two equations to eliminate one variable. Sometimes we need to multiply one or both equations by constants to make the coefficients of one variable opposites.
Steps of Elimination Method:
- Step 1: Write both equations in standard form (ax + by = c).
- Step 2: Make the coefficients of one variable (say x) equal in magnitude (opposite signs for addition).
- Step 3: Add or subtract the equations to eliminate that variable.
- Step 4: Solve for the remaining variable.
- Step 5: Substitute back to find the other variable.
- Step 6: Verify the solution.
Example 1: Solve by elimination: x + y = 10 and x – y = 4
Solution:
Add the two equations: (x + y) + (x – y) = 10 + 4 → 2x = 14 → x = 7
Substitute x = 7 into first equation: 7 + y = 10 → y = 3
Verification: Second: 7 – 3 = 4 ✓
Answer: (7, 3)
Example 2: Solve by elimination: 2x + 3y = 13 and 4x – 3y = 11
Solution:
Notice y coefficients are +3 and -3 (already opposites). Add the equations:
(2x + 3y) + (4x – 3y) = 13 + 11 → 6x = 24 → x = 4
Substitute x = 4 into first equation: 2(4) + 3y = 13 → 8 + 3y = 13 → 3y = 5 → y = 5/3
Answer: (4, 5/3)
Example 3: Solve by elimination: 3x + 2y = 8 and 2x + 3y = 7
Solution:
To eliminate x, multiply first equation by 2 and second equation by 3:
First ×2: 6x + 4y = 16
Second ×3: 6x + 9y = 21
Subtract second from first: (6x + 4y) – (6x + 9y) = 16 – 21 → -5y = -5 → y = 1
Substitute y = 1 into first original: 3x + 2(1) = 8 → 3x + 2 = 8 → 3x = 6 → x = 2
Answer: (2, 1)
Example 4: Solve by elimination: 5x – 3y = 11 and 2x + 4y = 6
Solution:
To eliminate x, multiply first equation by 2 and second equation by 5:
First ×2: 10x – 6y = 22
Second ×5: 10x + 20y = 30
Subtract: (10x – 6y) – (10x + 20y) = 22 – 30 → -26y = -8 → y = 8/26 = 4/13
Substitute y = 4/13 into second original: 2x + 4(4/13) = 6 → 2x + 16/13 = 6 → Multiply by 13: 26x + 16 = 78 → 26x = 62 → x = 62/26 = 31/13
Answer: (31/13, 4/13)
Method 3: Cross Multiplication Method
Definition
The cross-multiplication method is a formula-based approach for solving systems of the form a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.
Formula:
x / (b₁c₂ – b₂c₁) = y / (c₁a₂ – c₂a₁) = 1 / (a₁b₂ – a₂b₁)
From this, we get:
x = (b₁c₂ – b₂c₁) / (a₁b₂ – a₂b₁)
y = (c₁a₂ – c₂a₁) / (a₁b₂ – a₂b₁)
Note: This method is efficient when the coefficients are large or when rote memorization saves time.
Example: Solve using cross multiplication: 2x + 3y – 8 = 0 and 3x + 2y – 7 = 0
Solution:
Here a₁=2, b₁=3, c₁=-8; a₂=3, b₂=2, c₂=-7
First compute the denominator (a₁b₂ – a₂b₁) = (2×2 – 3×3) = 4 – 9 = -5
Now find x: x = (b₁c₂ – b₂c₁) / (a₁b₂ – a₂b₁) = (3×(-7) – 2×(-8)) / (-5) = (-21 + 16) / (-5) = (-5)/(-5) = 1
Now find y: y = (c₁a₂ – c₂a₁) / (a₁b₂ – a₂b₁) = ((-8)×3 – (-7)×2) / (-5) = (-24 + 14) / (-5) = (-10)/(-5) = 2
Answer: (1, 2)
Comparison of Methods
Substitution Method: Best when one equation is already solved for a variable (like y = 2x + 3) or can be easily solved. Works well for all systems but can get messy with fractions.
Elimination Method: Best when coefficients line up nicely or can be easily matched. Very systematic and avoids fractions until the end. Often the fastest method for most problems.
Cross Multiplication Method: Best when you need a quick formula-based solution, especially for large coefficients. Less intuitive but very efficient once memorized.
Special Cases – Consistent, Inconsistent, Dependent
Case 1 – Unique Solution (Consistent System):
The lines intersect at one point. The algebraic condition is a₁/a₂ ≠ b₁/b₂.
Example: 2x + y = 5 and x – y = 1
Here 2/1 ≠ 1/(-1) → 2 ≠ -1 → Unique solution (2, 1)
Case 2 – No Solution (Inconsistent System):
The lines are parallel and distinct. Condition: a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
Example: x + y = 5 and x + y = 8
Here 1/1 = 1/1 ≠ 5/8 → No solution
Case 3 – Infinite Solutions (Dependent System):
The lines are coincident (same line). Condition: a₁/a₂ = b₁/b₂ = c₁/c₂.
Example: x + y = 5 and 2x + 2y = 10
Here 1/2 = 1/2 = 5/10 → Infinite solutions
Solved Examples
Example 1 – Substitution Method: Solve 2x + 3y = 8 and x – y = -1
Solution: From second equation: x = y – 1
Substitute into first: 2(y – 1) + 3y = 8 → 2y – 2 + 3y = 8 → 5y – 2 = 8 → 5y = 10 → y = 2
Then x = 2 – 1 = 1
Answer: (1, 2)
Example 2 – Elimination Method: Solve 4x – 5y = 3 and 3x + 2y = 8
Solution: Multiply first by 2 and second by 5 to eliminate y:
First ×2: 8x – 10y = 6
Second ×5: 15x + 10y = 40
Add: 23x = 46 → x = 2
Substitute into second original: 3(2) + 2y = 8 → 6 + 2y = 8 → 2y = 2 → y = 1
Answer: (2, 1)
Example 3 – No Solution Case: Solve x + 2y = 5 and 2x + 4y = 12
Solution: Divide second equation by 2: x + 2y = 6
Now we have x + 2y = 5 and x + 2y = 6. These are contradictory.
Answer: No solution (Inconsistent system)
Example 4 – Infinite Solutions Case: Solve 3x – y = 4 and 6x – 2y = 8
Solution: Divide second equation by 2: 3x – y = 4, which is exactly the same as the first equation.
Answer: Infinite solutions (Dependent system)
Common Mistakes to Avoid
Mistake 1 – Forgetting to multiply the entire equation
When using elimination, if you multiply an equation by a number, multiply EVERY term, not just one.
Correct approach: 2(x + y = 5) → 2x + 2y = 10 (both terms multiplied)
Mistake 2 – Making sign errors when subtracting equations
When subtracting, be careful with negative signs.
Correct approach: (2x + 3y) – (2x – y) = 2x + 3y – 2x + y = 4y (not 2y)
Mistake 3 – Substituting back into the same equation
After finding x, substitute into the OTHER equation to verify or find y.
Correct approach: Use an equation you haven't used yet for verification.
Mistake 4 – Mixing up the condition for no solution vs infinite solutions
Remember: a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → no solution; a₁/a₂ = b₁/b₂ = c₁/c₂ → infinite solutions.
Correct approach: Check all three ratios carefully.
Mistake 5 – Giving only one value when there are infinite solutions
For dependent systems, the answer is "infinite solutions" or "all points on the line," not a single pair.
Correct approach: Recognize the condition and state appropriately.
Quick Reference Summary
Substitution Method:
Solve for one variable → substitute into other equation → solve → back-substitute
Elimination Method:
Make coefficients opposites → add/subtract to eliminate → solve → back-substitute
Cross Multiplication Method:
Use formula x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)
Types of Systems:
Unique solution → a₁/a₂ ≠ b₁/b₂
No solution → a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Infinite solutions → a₁/a₂ = b₁/b₂ = c₁/c₂