Key Concepts
- 1Consistent pair: unique solution
- 2Inconsistent pair: no solution
- 3Dependent pair: infinite solutions
- 4Elimination method
- 5Substitution method
- 6Cross-multiplication method
- 7Word problem strategy
- 8Consistent pair: unique solution
- 9Inconsistent pair: no solution
- 10Dependent pair: infinite solutions
Important Formulas & Facts
a₁/a₂ ≠ b₁/b₂ → Intersecting lines → One solution
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → Parallel lines
a₁/a₂ = b₁/b₂ = c₁/c₂ → Coincident lines
Multiply equations to make one variable's coefficient equal, then add/subtract.
Express one variable from one equation, substitute in the other.
x/(b₁c₂-b₂c₁) = y/(c₁a₂-c₂a₁) = 1/(a₁b₂-a₂b₁)
1) Define variables 2) Form equations 3) Solve 4) Verify
a₁/a₂ ≠ b₁/b₂ → Intersecting lines → One solution
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → Parallel lines
a₁/a₂ = b₁/b₂ = c₁/c₂ → Coincident lines
Must-Know Questions
Q1A fraction becomes 9/11 if 2 is added to both numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction.
Let fraction = x/y. (x+2)/(y+2) = 9/11 → 11x - 9y = -4...(i). (x+3)/(y+3) = 5/6 → 6x - 5y = -3...(ii). From (i): x = (-4+9y)/11. Sub in (ii): 6(-4+9y)/11 - 5y = -3 → -24+54y-55y = -33 → y = 9. x = (-4+81)/11 = 7. Fraction = 7/9.
Q2Solve: 2x + y = 5 and 3x - y = 5.
Adding: 5x = 10, x = 2. From eq(1): 2(2) + y = 5, y = 1. Solution: x = 2, y = 1. Verify: 3(2) - 1 = 5 ✓.
Q3For what value of k, the pair kx + 3y = k - 3 and 12x + ky = k has no solution?
No solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂. k/12 = 3/k → k² = 36 → k = ±6. Check k = 6: 6/12 = 3/6 = 1/2, c₁/c₂ = 3/6 = 1/2 (equal → infinite solutions). Check k = -6: -6/12 = 3/(-6) = -1/2, c₁/c₂ = -9/(-6) = 3/2 ≠ -1/2. So k = -6.
Q4The pair x + 2y - 5 = 0 and -4x - 8y + 20 = 0 has:
a₁/a₂ = 1/(-4), b₁/b₂ = 2/(-8) = 1/(-4), c₁/c₂ = (-5)/20 = -1/4. Since a₁/a₂ = b₁/b₂ = c₁/c₂ (all equal -1/4), infinitely many solutions.
Q5Solve by elimination: 3x + 4y = 10, 2x - 2y = 2.
From eq(2): x - y = 1 → x = y + 1. Sub in eq(1): 3(y+1) + 4y = 10 → 7y = 7 → y = 1, x = 2.
Practice Pair of Linear Equations in Two Variables
Reinforce what you just revised with practice questions