Introduction — Nature of Solutions

Nature of solutions based on ratios of corresponding coefficients.

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Chapter 4 · Pair of Linear Equations in Two Variables

Introduction — Pair of Linear Equations

Understand what a pair of linear equations is, how to find solutions graphically, the three types of line relationships, the ratio conditions for each case, and how to use them — with fully worked Try This problems.

Class 10 Maths CBSE Telangana Board Andhra Pradesh Board Chapter 4

Recap — Linear Equations in One and Two Variables

A linear equation is an equation whose graph is a straight line. Before working with pairs, let us distinguish between equations in one variable and two variables.

Linear Equation in One Variable — only one unknown:
2x + 1 = 0
3y = 1/2
2p + 1/2 = 7p

Each has exactly one solution (a single number).

Linear Equation in Two Variables — two unknowns:
2x + 3y = 12
5m + 6n − 14 = 0
x/3 + y/6 − 3 = 0

Has infinitely many solutions (a whole line of points).

📐 General form: The standard form of a linear equation in two variables is ax + by + c = 0, where a, b, c are real numbers and at least one of a or b is not zero.

Try This — Identifying Linear Equations (MCQ Practice)

The lesson includes five multiple-choice questions to check understanding of what makes an equation linear, how many solutions a two-variable linear equation has, and how to verify a solution.

#QuestionCorrect OptionReason
1 Which is NOT a linear equation? (c) 3 − x = y² + 4 Has y² — a squared term makes it non-linear
2 Which is a linear equation in ONE variable? (b) 2t − 1 = t + 5 Only one variable (t); (a) has x,y; (c)(d) have x²
3 Solution of 2(x + 3) = 18? (b) 6 2(6+3)=18 ✓   Verify: 2×9=18
4 x satisfying 2x − 4 − x = 5 − x (a) 4.5 Simplify: 2x − 4 − x = 5 − x → 2x = 9 → x = 4.5
5 Solutions of x − 4y = 5? (d) Infinitely many One linear equation in two variables always has ∞ solutions
💡 Key rule: A single linear equation in two variables represents a straight line and has infinitely many solutions — any point on that line is a solution. A pair of such equations may have one, none, or infinitely many common solutions.

Graphical Method of Finding Solutions

To solve a pair of linear equations graphically, we plot both equations as straight lines on the same coordinate plane and look at how the two lines relate to each other. The point of intersection (if any) gives the solution.

For each equation, we build a table of (x, y) values by choosing at least 3 values of x, computing the corresponding y, and plotting the points. Connecting them gives the line.

How to build a points table

For ax + by = c, rearrange to: y = (c − ax) / b
Then substitute x = 0, 1, 2 (or any convenient values)

Case 1 — Intersecting Lines (Unique Solution)

Equations: 3x + y = 7   and   2x − y = 3

Eq (1): y = 7 − 3x
xy = 7−3xPoint
14(1, 4)
3−2(3, −2)
07(0, 7)
Eq (2): y = 2x − 3
xy = 2x−3Point
0−3(0, −3)
−1−5(−1, −5)
1−1(1, −1)

When both lines are plotted, they cross at (2, 1). This is the unique solution.

x y -3 -2 -1 1 2 3 4 2 -2 -4 (2,1)
Lines intersect at (2, 1) → Unique solution: x = 2, y = 1
Result: Lines intersect at (2, 1). So x = 2 and y = 1 is the unique solution. The system is consistent and independent.

Case 2 — Parallel Lines (No Solution)

Equations: 3x + y = 10   and   6x + 2y = 24

Notice that the second equation (6x + 2y = 24) simplifies to 3x + y = 12 — it has the same left-hand side coefficients as the first equation but a different constant. These lines have the same slope but different y-intercepts, so they never meet.

Eq (1): y = 10 − 3x
xyPoint
17(1, 7)
24(2, 4)
31(3, 1)
Eq (2): y = 12 − 3x
xyPoint
26(2, 6)
40(4, 0)
5−3(5, −3)
Result: The two lines are parallel — they never intersect. There is no common solution. The system is inconsistent.

Case 3 — Coincident Lines (Infinitely Many Solutions)

Equations: 2x + y = 3   and   6x + 3y = 9

The second equation is simply 3 times the first: 3 × (2x + y) = 3 × 3 = 9. They represent the same line. Every point that lies on one line also lies on the other.

Eq (1): y = 3 − 2x
xyPoint
11(1, 1)
2−1(2, −1)
3−3(3, −3)
Eq (2): y = 3 − 2x (same!)
xyPoint
−15(−1, 5)
03(0, 3)
4−5(4, −5)
Result: The two lines are coincident (overlap completely). Every point on the line is a solution → infinitely many solutions. The system is consistent and dependent.
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The Three Cases — Visual Comparison

✂️ Intersecting Lines
3x + y = 7
2x − y = 3
a₁/a₂ = 3/2
b₁/b₂ = 1/(−1) = −1
a₁/a₂ ≠ b₁/b₂
✅ Unique solution
Consistent & Independent
(2,1)
⟺ Parallel Lines
3x + y = 10
6x + 2y = 24
a₁/a₂ = 1/2
b₁/b₂ = 1/2
c₁/c₂ = 5/12
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
❌ No solution
Inconsistent
= Coincident Lines
2x + y = 3
6x + 3y = 9
a₁/a₂ = 1/3
b₁/b₂ = 1/3
c₁/c₂ = 1/3
a₁/a₂ = b₁/b₂ = c₁/c₂
♾ Infinitely many solutions
Consistent & Dependent

The Master Ratio Conditions Table

For any pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the nature of the solution depends entirely on the ratios of corresponding coefficients. This table is the most important thing to memorise in Chapter 4.

Ratio ConditionGraphSolutionsAlgebraic Name
a₁/a₂ ≠ b₁/b₂ Lines intersect Unique (one) solution Consistent & Independent
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ Lines are parallel No solution Inconsistent
a₁/a₂ = b₁/b₂ = c₁/c₂ Lines coincide Infinitely many solutions Consistent & Dependent
📌 Board exam shortcut: You do NOT need to draw the graph to answer "unique/no/infinitely many solutions." Just compare the three ratios a₁/a₂, b₁/b₂ and c₁/c₂ algebraically using the table above.

Try This — Worked Problems on Ratio Conditions

Try This 1
For what value of p does 2x + py = −5 and 3x + 3y = −6 have a unique solution?

For a unique solution, the lines must intersect, so: a₁/a₂ ≠ b₁/b₂

Rewrite: 2x + py + 5 = 0 → a₁=2, b₁=p, c₁=5 3x + 3y + 6 = 0 → a₂=3, b₂=3, c₂=6 Condition: a₁/a₂ ≠ b₁/b₂ 2/3 ≠ p/3 p ≠ 2
Answer: For all real values of p ≠ 2, the equations have a unique solution.
📐 When p = 2, both equations become 2x + 2y = −5 and 2x + 2y = −4 — parallel lines (no solution). So p = 2 must be excluded.
Try This 2
Find k such that 2x − ky + 3 = 0 and 4x + 6y − 5 = 0 represent parallel lines.

Parallel lines → a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Use only the first equality to find k.

a₁=2, b₁=−k, c₁=3   |   a₂=4, b₂=6, c₂=−5 Condition: a₁/a₂ = b₁/b₂ 2/4 = (−k)/6 1/2 = −k/6 6 = −2k k = −3 Verify ≠ c₁/c₂: c₁/c₂ = 3/(−5) = −3/5 ≠ 1/2 ✓
Answer: k = −3 makes the lines parallel.
Try This 3
For what value of k do 3x + 4y + 2 = 0 and 9x + 12y + k = 0 represent coincident lines?

Coincident lines → a₁/a₂ = b₁/b₂ = c₁/c₂. All three ratios must be equal.

a₁=3, b₁=4, c₁=2   |   a₂=9, b₂=12, c₂=k a₁/a₂ = 3/9 = 1/3 b₁/b₂ = 4/12 = 1/3 ✓ (matches) Condition: b₁/b₂ = c₁/c₂ 4/12 = 2/k 1/3 = 2/k k = 6 ∴ k = 6
Answer: k = 6 makes the lines coincident.
Try This 4
For what positive value of p do px + 3y − p − 3 = 0 and 12x + py − p = 0 have infinitely many solutions?

Infinitely many solutions → coincident lines → a₁/a₂ = b₁/b₂ = c₁/c₂. Note: this leads to a quadratic in p.

a₁=p, b₁=3, c₁=−(p−3)   |   a₂=12, b₂=p, c₂=−p Use b₁/b₂ = c₁/c₂: 3/p = −(p−3) / (−p) 3/p = (p−3)/p 3p = p(p−3)   (multiply both sides by p) 3p = p² − 3p p² − 6p = 0 p(p − 6) = 0 p = 0 or p = 6 p = 0 rejected (if p=0, equation becomes 0x+3y−3=0, only one equation effectively) ∴ p = 6
Answer: p = 6 (the positive value) gives infinitely many solutions.
📐 Why is p = 0 rejected? The question asks for a positive value. Also, at p = 0 the first equation reduces to 3y − 3 = 0, i.e., y = 1, and the coefficient of x vanishes — it becomes a different type of equation.

Common Mistakes to Avoid

  • Writing equations in ax + by = c form vs ax + by + c = 0: The general form has c on the left side with a positive sign. If your equation is 2x + y = 7, then in standard form it is 2x + y − 7 = 0, making c = −7, not +7. Incorrect sign of c will give wrong ratio comparisons.
  • Confusing c₁/c₂ for the parallel condition: For parallel lines, we need a₁/a₂ = b₁/b₂ BUT c₁/c₂ must be DIFFERENT. Students sometimes check all three ratios and declare coincident when c₁/c₂ also happens to match — always verify carefully.
  • Not verifying the ≠ part for parallel lines (Try This 2): After finding k = −3 in Q2, you must check that c₁/c₂ (= 3/−5 = −3/5) is indeed ≠ a₁/a₂ (= 1/2). If both equalities held, the lines would be coincident, not parallel.
  • Losing the p = 0 solution in Try This 4: The quadratic p(p − 6) = 0 gives two roots. Always mention both, then justify why the unwanted one is rejected.
  • Wrong table values for graphing: Choose x-values that give integer or easy y-values to keep the graph accurate. Messy decimals on a graph make intersection points hard to read precisely.
Board exam alert (CBSE, Telangana & AP): Questions asking for the value of k or p for unique/parallel/coincident conditions are very commonly set as 3–4 mark questions. Always write: (1) identify a₁, b₁, c₁ and a₂, b₂, c₂ explicitly, (2) state the condition used, (3) solve and (4) verify. Each step carries marks.

What This Lesson Prepares You For

The Introduction establishes the graphical foundation for the three algebraic methods of solving a pair of linear equations taught in the subsequent exercises: Exercise 4.1 (Substitution Method), Exercise 4.2 (Elimination Method), and Exercise 4.3 (Cross Multiplication).

The ratio conditions (a₁/a₂, b₁/b₂, c₁/c₂) appear again in Class 10 Chapter 6 (Similar Triangles) when comparing corresponding side ratios, and in Class 11 Mathematics when studying system of linear equations using matrices and determinants.

Students who find this topic challenging should first review Class 9 Linear Equations in Two Variables, which covers how a single equation produces a line, and basic coordinate plotting.

📈 Chapter 4 — Class 10 🔍 Graphical Method ⚖️ Ratio Conditions
📌 Revision checklist (CBSE / Telangana / AP Class 10):
✔ General form: ax + by + c = 0
✔ a₁/a₂ ≠ b₁/b₂ → Unique solution (intersecting lines)
✔ a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → No solution (parallel lines)
✔ a₁/a₂ = b₁/b₂ = c₁/c₂ → Infinitely many solutions (coincident lines)
✔ Always label a, b, c from the STANDARD FORM (rearrange first)
✔ For graphing: use at least 3 points per line
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