How many solutions does a linear equation in two variables have?

A linear equation in two variables has infinitely many solutions. For every value you assign to x, you can calculate a corresponding y that satisfies the equation, giving a new solution. For example, 4x − 3y = 6 has solutions (3,2), (0,−2), (3/2,0) and infinitely more. All these solution points lie on a single straight line when plotted on the Cartesian plane.

How do you find three solutions of a linear equation like 3x + 4y = 7?

Use the substitution method: (1) Set x = 0 and solve for y: 4y = 7, y = 7/4, giving solution (0, 7/4). (2) Set y = 0 and solve for x: 3x = 7, x = 7/3, giving solution (7/3, 0). (3) Set x = 1 and solve for y: 3 + 4y = 7, 4y = 4, y = 1, giving solution (1, 1). The three solutions are (0, 7/4), (7/3, 0), and (1, 1).

How do you find the value of k when a solution of 2x + 3y = k is given?

If (2, 1) is a solution of 2x + 3y = k, substitute x = 2 and y = 1 into the equation: 2(2) + 3(1) = k → 4 + 3 = k → k = 7. Once k is found, the complete equation is 2x + 3y = 7, and you can find additional solutions by setting x = 0 (giving y = 7/3) or y = 0 (giving x = 7/2).

How do you check if a given pair is a solution of 2x − 5y = 10?

Substitute the pair into the equation and compare LHS to RHS = 10. For (0, −2): LHS = 2(0) − 5(−2) = 0 + 10 = 10 = RHS, so it is a solution. For (0, 2): LHS = 2(0) − 5(2) = −10 ≠ 10, so it is not a solution. For (5, 0): LHS = 2(5) − 5(0) = 10 = RHS, so it is a solution.

Exercise 6.2 — Solutions | Class 9 Maths | Linear Equations in Two Variables

Solutions of a Linear Equation in Two Variables

A solution of a linear equation in two variables is an ordered pair (x, y) that makes the equation true when those values are substituted. Unlike a one-variable equation (which has exactly one solution), a linear equation in two variables has infinitely many solutions — because for every value chosen for one variable, you can calculate a corresponding value for the other.

What is a solution?
An ordered pair (x₀, y₀) is a solution of ax + by + c = 0 if and only if substituting x = x₀ and y = y₀ makes LHS = RHS.

Method to find solutions:
◆ Set x = 0 → solve for y → get one solution (0, y).
◆ Set y = 0 → solve for x → get another solution (x, 0).
◆ Set x = any convenient value → solve for y → get a third solution.

Fundamental property: A linear equation in two variables always has infinitely many solutions. The set of all solutions forms a straight line when plotted on the Cartesian plane — this is why it is called a "linear" equation.

Worked Example — Solutions of 4x − 3y = 6

Set x = 3: 4(3) − 3y = 6 → 12 − 3y = 6 → y = 2 → Solution: (3, 2)

Set x = 0: 4(0) − 3y = 6 → −3y = 6 → y = −2 → Solution: (0, −2)

Set y = 0: 4x − 3(0) = 6 → 4x = 6 → x = 3/2 → Solution: (3/2, 0)

Check (2,3): LHS = 4(2)−3(3) = 8−9 = −1 ≠ 6 = RHS → NOT a solution

Check (3,2): LHS = 4(3)−3(2) = 12−6 = 6 = RHS → IS a solution ✓

Solutions include: (3, 2), (0, −2), (3/2, 0) — and infinitely more.

Exercise 6.2 — Question 1: Three Solutions of Each Equation

For each equation, find three solutions by substituting convenient values for x or y. Setting x = 0, y = 0, and x = 1 is the standard strategy used in the textbook.

(i) 3x + 4y = 7

S.No.	Substitution	Calculation	Solution
1	x = 0	3(0) + 4y = 7 → y = 7/4	(0, 7/4)
2	y = 0	3x + 4(0) = 7 → x = 7/3	(7/3, 0)
3	x = 1	3(1) + 4y = 7 → 4y = 4 → y = 1	(1, 1)

(ii) y = 6x

S.No.	Substitution	Calculation	Solution
1	x = 0	y = 6(0) = 0	(0, 0)
2	x = 1	y = 6(1) = 6	(1, 6)
3	x = 2	y = 6(2) = 12	(2, 12)

(iii) 2x − y = 7

S.No.	Substitution	Calculation	Solution
1	x = 0	−y = 7 → y = −7	(0, −7)
2	y = 0	2x = 7 → x = 7/2	(7/2, 0)
3	x = 1	2−y = 7 → y = −5	(1, −5)

(iv) 13x − 12y = 25

S.No.	Substitution	Calculation	Solution
1	x = 0	−12y = 25 → y = −25/12	(0, −25/12)
2	y = 0	13x = 25 → x = 25/13	(25/13, 0)
3	x = 1	13 − 12y = 25 → y = −1	(1, −1)

(v) 10x + 11y = 21

S.No.	Substitution	Calculation	Solution
1	x = 0	11y = 21 → y = 21/11	(0, 21/11)
2	y = 0	10x = 21 → x = 21/10	(21/10, 0)
3	x = 1	10 + 11y = 21 → y = 1	(1, 1)

(vi) x + y = 0

S.No.	Substitution	Calculation	Solution
1	x = 0	y = 0	(0, 0)
2	x = 1	y = −1	(1, −1)
3	x = 2	y = −2	(2, −2)

Exercise 6.2 — Question 2: Find 'a' and 'b' from Given Solution Points

If (0, a) is a solution, substitute x = 0, y = a to find a. If (b, 0) is a solution, substitute x = b, y = 0 to find b. These are the y-intercept and x-intercept of the line respectively.

(i) 8x − y = 34

For (0, a): 8(0) − a = 34 → a = −34 For (b, 0): 8b − 0 = 34 → b = 34/8 = 17/4

a = −34, b = 17/4

(ii) 3x = 7y − 21

For (0, a): 0 = 7a − 21 → a = 3 For (b, 0): 3b = −21 → b = −7

a = 3, b = −7

(iii) 5x − 2y + 3 = 0

For (0, a): −2a + 3 = 0 → a = 3/2 For (b, 0): 5b + 3 = 0 → b = −3/5

a = 3/2, b = −3/5

Exercise 6.2 — Question 3: Check Solutions of 2x − 5y = 10

Substitute each given pair into 2x − 5y = 10. If LHS = RHS = 10, it is a solution. If LHS ≠ 10, it is not.

(i) (0, 2)

LHS = 2(0) − 5(2) = 0 − 10 = −10 RHS = 10 LHS ≠ RHS ❌ Not a solution

(ii) (0, −2)

LHS = 2(0) − 5(−2) = 0 + 10 = 10 RHS = 10 LHS = RHS ✓ ✅ Is a solution

(iii) (5, 0)

LHS = 2(5) − 5(0) = 10 − 0 = 10 RHS = 10 LHS = RHS ✓ ✅ Is a solution

(iv) (2√3, −√3)

LHS = 2(2√3) − 5(−√3) = 4√3 + 5√3 = 9√3 9√3 ≈ 15.6 ≠ 10 ❌ Not a solution

(v) (1/2, 2)

LHS = 2(1/2) − 5(2) = 1 − 10 = −9 RHS = 10 LHS ≠ RHS ❌ Not a solution

Exercise 6.2 — Question 4: Find k, Then Two More Solutions

Question 4 — Equation: 2x + 3y = k, Solution: (2, 1)

Since (2,1) is a solution: substitute x = 2, y = 1

2(2) + 3(1) = k → 4 + 3 = k → k = 7

Resultant equation: 2x + 3y = 7

Set x = 0: 3y = 7 → y = 7/3 → Solution: (0, 7/3)

Set y = 0: 2x = 7 → x = 7/2 → Solution: (7/2, 0)

k = 7. Two more solutions: (0, 7/3) and (7/2, 0)

Exercise 6.2 — Question 5: Find α from Parametric Solution

Question 5 — Equation: 3x − 2y + 6 = 0, Solution: x = 2−α, y = 2+α

Substitute into equation: 3(2−α) − 2(2+α) + 6 = 0

6 − 3α − 4 − 2α + 6 = 0

8 − 5α = 0 → α = 8/5

Resultant equation: 3x − 2y + 6 = 0

Set x = 0: −2y + 6 = 0 → y = 3 → Solution: (0, 3)

Set y = 0: 3x + 6 = 0 → x = −2 → Solution: (−2, 0)

Set y = 6: 3x − 12 + 6 = 0 → x = 2 → Solution: (2, 6)

α = 8/5. Three more solutions: (0, 3), (−2, 0), (2, 6)

Exercise 6.2 — Question 6: Find 'a' from a Given Solution

Question 6 — Equation: 3x + ay = 6, Solution: (1, 1)

Since (1,1) satisfies 3x + ay = 6:

3(1) + a(1) = 6

3 + a = 6

a = 3

a = 3. The equation becomes 3x + 3y = 6, or simplified: x + y = 2.

Summary — Key Rules for Solutions

A linear equation in two variables always has infinitely many solutions — one for each value of x or y you choose.
To verify a solution (a, b): substitute x = a and y = b into the equation and check if LHS equals RHS.
The point (0, a) gives the y-intercept of the line. The point (b, 0) gives the x-intercept.
To find k or a in an equation: substitute the given solution values and solve the resulting one-variable equation.
Every solution (x, y) corresponds to a point on the line that the equation represents on the Cartesian plane.

Quick verification rule: Always check your solutions by substituting back. A solution (x₀, y₀) must satisfy LHS = RHS exactly. If even one side differs by a single unit, the pair is not a solution of that equation. Arithmetic errors in substitution are the most common mistake — be especially careful with negative numbers and fractions.