Pair of Linear Equations — Class 10 IMO

Solutions and Their Graphs

What is a graphical solution of linear equations? A pair of linear equations in two variables (x and y) represents two straight lines on the coordinate plane. The graphical solution is the point(s) where these two lines intersect. This intersection point gives the values of x and y that satisfy both equations simultaneously.

General form: \(a_{1}x + b_{1}y + c_{1}\) = 0 \(a_{2}x + b_{2}y + c_{2}\) = 0

Three possible cases when graphing:

Condition	Relationship	Type of Solution	Graph
\(a_{1}\)/\(a_{2}\) ≠ \(b_{1}\)/\(b_{2}\)	Intersecting lines	Unique solution (one point)	Lines cross at one point
\(a_{1}\)/\(a_{2}\) = \(b_{1}\)/\(b_{2}\) = \(c_{1}\)/\(c_{2}\)	Coincident lines	Infinitely many solutions	Lines overlap completely
\(a_{1}\)/\(a_{2}\) = \(b_{1}\)/\(b_{2}\) ≠ \(c_{1}\)/\(c_{2}\)	Parallel lines	No solution	Lines never meet

Real-life analogy: Imagine two roads on a map. They either:

Cross at one intersection (unique meeting point) → unique solution
Are the same road (every point is common) → infinite solutions
Run parallel (never meet) → no solution

┌─────────────────────────────────────────────────────────────┐
│           GRAPHICAL SOLUTIONS - THREE CASES                  │
└─────────────────────────────────────────────────────────────┘

CASE 1: INTERSECTING LINES (UNIQUE SOLUTION)

        y
        │
        │      /
        │     /   ← Equation 2
        │    /
        │   /★ (x₀,y₀)
        │  /
        │ /  ← Equation 1
        │/
        └──────────► x
        
     Example: x + y = 5 and x - y = 1
     Solution: (3, 2)


CASE 2: COINCIDENT LINES (INFINITE SOLUTIONS)

        y
        │
        │  ════════════
        │  ════════════  ← Both equations
        │  ════════════    represent the
        │  ════════════    same line
        │  ════════════
        │
        └──────────► x
        
     Example: x + y = 3 and 2x + 2y = 6


CASE 3: PARALLEL LINES (NO SOLUTION)

        y
        │
        │  ────────  ← Equation 1
        │  
        │  ────────  ← Equation 2
        │
        │
        └──────────► x
        
     Example: x + y = 4 and x + y = 7


STEP-BY-STEP GRAPHING METHOD:

     Equation 1: x + y = 4
     Equation 2: x - y = 2

     FIND TWO POINTS FOR EACH LINE:

     For x + y = 4:          For x - y = 2:
     x=0 → y=4 → (0,4)       x=0 → y=-2 → (0,-2)
     x=4 → y=0 → (4,0)       x=2 → y=0 → (2,0)

     PLOT AND FIND INTERSECTION:

        (0,4) 
           \
            \  (3,1) ← SOLUTION!
             \
        (0,-2) \  (2,0)
                 \
                  (4,0)
     
     VERIFY: 3+1=4 ✓ and 3-1=2 ✓

Example 1: Solve graphically: x + y = 6 and x − y = 2

Step 1: For x + y = 6: points (0,6) and (6,0)
Step 2: For x − y = 2: points (0,−2) and (2,0)
Step 3: Plot both lines on graph paper
Step 4: Find intersection point: both equations satisfy x = 4, y = 2
Step 5: Verify: 4 + 2 = 6 ✓ and 4 − 2 = 2 ✓

Answer: x = 4, y = 2

Example 2: Determine if the pair 2x + 3y = 6 and 4x + 6y = 12 has a unique solution.

Step 1: Write in standard form: 2x + 3y − 6 = 0 and 4x + 6y − 12 = 0
Step 2: Compare ratios: \(a_{1}\)/\(a_{2}\) = 2/4 = 1/2, \(b_{1}\)/\(b_{2}\) = 3/6 = 1/2, \(c_{1}\)/\(c_{2}\) = −6/−12 = 1/2
Step 3: All ratios equal → lines are coincident

Answer: No unique solution; infinitely many solutions

Example 3: For what value of k will the lines 3x + y = 5 and 6x + 2y = k be parallel?

Step 1: Condition for parallel lines: \(a_{1}\)/\(a_{2}\) = \(b_{1}\)/\(b_{2}\) ≠ \(c_{1}\)/\(c_{2}
Step\) 2: \(a_{1}\)/\(a_{2}\) = 3/6 = 1/2, \(b_{1}\)/\(b_{2}\) = 1/2 = 1/2 ✓
Step 3: For parallel, \(c_{1}\)/\(c_{2}\) should NOT equal 1/2
Step 4: \(c_{1}\) = −5, \(c_{2}\) = −k, so (−5)/(−k) = 5/k
Step 5: Set 5/k ≠ 1/2 → cross multiply: 10 ≠ k

Answer: k ≠ 10 (any value except 10)

Quick recap

Graphical solution = intersection point of two lines
Intersecting lines → unique solution (one point)
Coincident lines → infinitely many solutions
Parallel lines → no solution
To graph a line, find any two points satisfying the equation
Always verify the solution in both original equations

✓ Quick check

The number of solutions of the pair x + 2y = 3 and 2x + 4y = 6 is:

All three ratios equal 1/2, so the lines coincide and solutions are infinite.

The graph of the equation x = 4 is:

x = 4 is a vertical line, parallel to the y-axis.

Conditions on the Coefficients

For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0: a unique solution needs a₁/a₂ ≠ b₁/b₂; no solution needs a₁/a₂ = b₁/b₂ ≠ c₁/c₂; infinitely many needs a₁/a₂ = b₁/b₂ = c₁/c₂.

Example 1: 2x + 3y = 7 and 4x + 6y = 14?

All ratios equal 1/2 → infinitely many solutions.

Example 2: x + 2y = 4 and 2x + 4y = 12?

1/2 = 2/4 but 4/12 differs → no solution.

Quick recap

Unique: a₁/a₂ ≠ b₁/b₂.
None: ratios of a, b equal but c differs; infinite: all equal.

✓ Quick check

The pair 3x + 2y = 5 and 6x + 4y = 10 represents:

3/6 = 2/4 = 5/10, so the equations describe the same (coincident) line.

If 3x + 4y = 10 and 2x − 2y = 2, then x equals:

From 2x − 2y = 2, x = y + 1; then 3(y+1) + 4y = 10 gives 7y = 7, y = 1 and x = 2.

Algebraic Methods and Word Problems

What are algebraic methods for solving linear equations? Algebraic methods allow us to find exact solutions without drawing graphs. The three main methods are:

1. Substitution Method: Solve one equation for one variable, then substitute into the other equation.

2. Elimination Method: Multiply equations to make coefficients of one variable equal, then add or subtract to eliminate that variable.

3. Cross-Multiplication Method: For equations \(a_{1}x + b_{1}y + c_{1}\) = 0 and \(a_{2}x + b_{2}y + c_{2}\) = 0:

x / (\(b_{1}c_{2} - b_{2}c_{1}\)) = y / (\(c_{1}a_{2} - c_{2}a_{1}\)) = 1 / (\(a_{1}b_{2} - a_{2}b_{1}\))

Formula recall: x = (\(b_{1}c_{2} - b_{2}c_{1}\)) / (\(a_{1}b_{2} - a_{2}b_{1}\)) y = (\(c_{1}a_{2} - c_{2}a_{1}\)) / (\(a_{1}b_{2} - a_{2}b_{1}\))

┌─────────────────────────────────────────────────────────────┐
│        ALGEBRAIC METHODS - COMPARISON FLOWCHART              │
└─────────────────────────────────────────────────────────────┘

START: Two equations in x and y

        │
        ├─────────────────────────────────────┐
        ▼                                     ▼
  SUBSTITUTION METHOD                  ELIMINATION METHOD
        │                                     │
  1. Solve one eqn for x           1. Make coefficients of
     (or y) in terms of               x (or y) equal
     the other variable              
        │                                     │
  2. Substitute into                2. Add or subtract to
     second equation                   eliminate one variable
        │                                     │
  3. Solve for one                  3. Solve for remaining
     variable                          variable
        │                                     │
  4. Back-substitute to             4. Substitute back to
     find other variable               find other variable
        │                                     │
        └──────────────┬──────────────────────┘
                       ▼
                CROSS-MULTIPLICATION
                       │
             x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)
                       │
                       ▼
                  FINAL SOLUTION


EXAMPLE WITH ALL THREE METHODS:

Equations: 2x + y = 7 ...(1)
           x - y = 2  ...(2)

SUBSTITUTION:
From (2): x = y + 2
Into (1): 2(y+2) + y = 7 → 2y+4+y=7 → 3y=3 → y=1, x=3

ELIMINATION:
Add (1) and (2): (2x+y)+(x-y)=7+2 → 3x=9 → x=3
Then from (2): 3-y=2 → y=1

CROSS-MULTIPLICATION:
a₁=2, b₁=1, c₁=-7; a₂=1, b₂=-1, c₂=-2
x/(1×-2 - (-1)×-7) = x/(-2-7) = x/-9
y/( -7×1 - (-2)×2) = y/(-7+4) = y/-3
1/(2×-1 - 1×1) = 1/(-2-1) = 1/-3
So x/-9 = 1/-3 → x=3; y/-3=1/-3 → y=1

Example 1: Solve: 3x + 2y = 12 and x − y = −1

Step 1: From second equation: x = y − 1
Step 2: Substitute into first: 3(y − 1) + 2y = 12
Step 3: Simplify: 3y − 3 + 2y = 12 → 5y − 3 = 12 → 5y = 15 → y = 3
Step 4: Substitute back: x = 3 − 1 = 2

Answer: x = 2, y = 3

Example 2: Solve: 4x + 3y = 18 and 5x − 2y = 11

Step 1: Multiply first eqn by 2: 8x + 6y = 36
Step 2: Multiply second eqn by 3: 15x − 6y = 33
Step 3: Add the two new equations: (8x+6y)+(15x-6y)=36+33 → 23x = 69 → x = 3
Step 4: Substitute x=3 into 4x+3y=18: 12+3y=18 → 3y=6 → y=2

Answer: x = 3, y = 2

Example 3: Solve: 2x + 5y = 1 and 3x + 2y = 7

Step 1: Write as \(a_{1}x + b_{1}y + c_{1}\) = 0: 2x+5y-1=0 → \(a_{1}\)=2, \(b_{1}\)=5, \(c_{1}\)=-1
3x+2y-7=0 → \(a_{2}\)=3, \(b_{2}\)=2, \(c_{2}\)=-7
Step 2: Calculate denominator: \(a_{1}b_{2} - a_{2}b_{1}\) = (2×2) − (3×5) = 4 − 15 = −11
Step 3: Calculate x numerator: \(b_{1}c_{2} - b_{2}c_{1}\) = (5×-7) − (2×-1) = -35 + 2 = -33
Step 4: Calculate y numerator: \(c_{1}a_{2} - c_{2}a_{1}\) = (-1×3) − (-7×2) = -3 + 14 = 11
Step 5: x = (-33)/(-11) = 3, y = (11)/(-11) = -1

Answer: x = 3, y = -1

Quick recap

Substitution: Isolate one variable, substitute into the other equation
Elimination: Make coefficients equal, add/subtract to eliminate a variable
Cross-multiplication: Direct formula when equations are in standard form
All three methods give the same answer — choose the easiest for each problem
Cross-multiplication fails when \(a_{1}b_{2} - a_{2}b_{1}\) = 0 (parallel or coincident lines)
Always check your answer in both original equations

✓ Quick check

A two-digit number is 4 times the sum of its digits. If the digits sum to 9, the number is:

Number = 4 × 9 = 36, and 3 + 6 = 9 fits.

The sum of two numbers is 50 and their difference is 10. The larger number is:

x + y = 50, x − y = 10 give x = 30 and y = 20.