Graphical & algebraic methods, consistency conditions, and word problems
A linear equation in two variables is of the form ax + by + c = 0, where a, b, c are real numbers and a, b ≠ 0 both. A pair of linear equations is a system of two such equations that must be satisfied simultaneously.
a₁x + b₁y + c₁ = 0 ...(i)
a₂x + b₂y + c₂ = 0 ...(ii)
Where a₁, a₂, b₁, b₂, c₁, c₂ are real numbers; a₁² + b₁² ≠ 0 and a₂² + b₂² ≠ 0
You buy 2 pens and 3 notebooks for ₹25, and 4 pens and 1 notebook for ₹30. To find price of each, you set up two equations — exactly a pair of linear equations! The price that satisfies both conditions is your solution.
Every linear equation in two variables represents a straight line. A pair of equations gives two straight lines, and their geometric relationship determines the type of solution.
| Lines | Solutions | Type | Condition |
|---|---|---|---|
| Intersecting | Unique solution | Consistent | a₁/a₂ ≠ b₁/b₂ |
| Coincident (same) | Infinitely many | Consistent (Dependent) | a₁/a₂ = b₁/b₂ = c₁/c₂ |
| Parallel | No solution | Inconsistent | a₁/a₂ = b₁/b₂ ≠ c₁/c₂ |
a₁/a₂ = 2/4 = 1/2 | b₁/b₂ = 3/6 = 1/2 | c₁/c₂ = 8/7
Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → Lines are parallel → No solution (Inconsistent)
Step 1: Express one variable in terms of the other from one equation.
Step 2: Substitute in the second equation → solve for one variable.
Step 3: Substitute back to find the other variable.
From (i): x = 5 – y
Substituting in (ii): 2(5 – y) – 3y = 4 → 10 – 5y = 4 → y = 6/5
x = 5 – 6/5 = 19/5
Answer: x = 19/5, y = 6/5
Step 1: Multiply equations by constants to make coefficients of one variable equal.
Step 2: Add or subtract to eliminate that variable.
Step 3: Solve, then substitute back.
Multiply (ii) by 2: 4x – 4y = 4
Add with (i): 7x = 14 → x = 2
Substitute: 2(2) – 2y = 2 → y = 1
Answer: x = 2, y = 1
For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:
x/(b₁c₂ – b₂c₁) = y/(c₁a₂ – c₂a₁) = 1/(a₁b₂ – a₂b₁)
a₁=2, b₁=1, c₁=–5 | a₂=3, b₂=2, c₂=–8
x = [(1)(–8)–(2)(–5)] / [(2)(2)–(3)(1)] = (–8+10)/(4–3) = 2
y = [(–5)(3)–(–8)(2)] / (1) = (–15+16)/1 = 1
Some non-linear equations can be reduced to linear form by substitution. Replace 1/x with p and 1/y with q (or similar).
Let p = 1/x, q = 1/y → 2p + 3q = 13 and 5p – 4q = –2
Elimination: Multiply (i)×4 and (ii)×3 → 8p + 12q = 52 and 15p – 12q = –6
Adding: 23p = 46 → p = 2 → x = 1/2
From (i): 4 + 3q = 13 → q = 3 → y = 1/3
Answer: x = 1/2, y = 1/3
1. Read carefully and identify the unknowns.
2. Assign variables: Let x = ___ and y = ___
3. Form two equations from the given conditions.
4. Solve using any method (substitution/elimination/cross-multiplication).
5. Verify answers and write conclusion.
A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it goes 40 km upstream and 55 km downstream. Find speed of boat and stream.
Let boat speed = x, stream speed = y → Upstream = (x–y), Downstream = (x+y)
30/(x–y) + 44/(x+y) = 10 and 40/(x–y) + 55/(x+y) = 13
Let u = 1/(x–y), v = 1/(x+y):
30u + 44v = 10 and 40u + 55v = 13 → Solving: u = 1/5, v = 1/11
x – y = 5 and x + y = 11 → x = 8 km/h, y = 3 km/h
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
Unique: a₁/a₂ ≠ b₁/b₂
Infinite: a₁/a₂ = b₁/b₂ = c₁/c₂
None: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
x/(b₁c₂–b₂c₁) = y/(c₁a₂–c₂a₁) = 1/(a₁b₂–a₂b₁)