📝 Linear Equations in Two Variables
Question Paper & Answer Key
Section A: Very Short Answer Questions (2 marks each)
Q1.
2 marks
Define a linear equation in two variables with one example.
Q2.
2 marks
Write the standard form of a linear equation in two variables and identify the constraints.
Q3.
2 marks
Is (2, -3) a solution of 2x + y = 1? Justify.
Q4.
2 marks
How many solutions does a linear equation in two variables have? Explain why.
Q5.
2 marks
Convert x - 4 = 3y into standard form and find values of a, b, and c.
Section B: Short Answer Questions (3 marks each)
Q6.
3 marks
Write three solutions for the equation x + y = 5.
Q7.
3 marks
Find two solutions for the equation 3x + 2y = 12 using the substitution method.
Q8.
3 marks
Express y = 2x and 2x = 5 in the standard form ax + by + c = 0.
Q9.
3 marks
Check which of the following are solutions of 2x - y = 4: (0, -4), (2, 0), (1, -2)
Q10.
3 marks
Find the value of k if x = 1, y = 2 is a solution of kx + 3y = 8.
Q11.
3 marks
Find four solutions for 2x + y = 8 by setting x = 0, 1, 2, 3.
Q12.
3 marks
Write the equation: "The sum of two numbers is 15" in the form of a linear equation in two variables.
Q13.
3 marks
Can 0.x + 0.y + 5 = 0 be a linear equation in two variables? Justify your answer.
Section C: Long Answer Questions (5 marks each)
Q14.
5 marks
Two pens and three notebooks cost Rs. 50. Express this as a linear equation in two variables. Find at least three solutions and explain the meaning of each solution in the context of the problem.
Q15.
5 marks
For the equation 5x - 2y = 10, find six solutions and verify each one. Discuss why there can be infinitely many solutions.
Q16.
5 marks
Rewrite the following equations in the form ax + by + c = 0 and find the values of a, b, and c:
(i) 3x + 2 = y
(ii) x = -5y
(iii) 2x + 3y = 0
Q17.
5 marks
In a cricket match, two batsmen scored a total of 200 runs. If the first batsman scored x runs and the second batsman scored y runs:
(i) Write the equation
(ii) Find five different solutions
(iii) Check if (120, 80) is a valid solution
Q18.
5 marks
Prove that a linear equation in two variables has infinitely many solutions by demonstrating with the equation x + 2y = 10. Show how for every value of x, you can find a corresponding value of y.
Section D: Application Based Questions (10.5 marks each)
Q19.
10.5 marks
Problem: A shopkeeper sells pens at Rs. 5 each and notebooks at Rs. 10 each. A customer spends exactly Rs. 100 on pens and notebooks.
(i) Frame a linear equation in two variables for this situation. (2 marks)
(ii) Find all possible integer solutions (where number of items must be whole numbers). (4 marks)
(iii) What is the maximum number of pens the customer can buy? What is the maximum number of notebooks? (2.5 marks)
(iv) Can the customer buy exactly 10 pens and 10 notebooks? Verify. (2 marks)
Q20.
10.5 marks
Problem: In a school event, students are divided into two teams. If there are x students in Team A and y students in Team B, and the total number of students is 50.
(i) Write the linear equation representing this situation. (1.5 marks)
(ii) Identify values of a, b, and c in standard form. (1 mark)
(iii) Find at least 5 different ways to divide the students. (3 marks)
(iv) If Team A has 30 students, how many students are in Team B? (1 mark)
(v) Explain why this equation has infinitely many solutions in mathematics but only limited practical solutions in this context. (3 marks)
Section A: Very Short Answer Questions
Q1. Definition and Example
Answer: A linear equation in two variables is an equation that can be written in the form ax + by + c = 0, where a, b, and c are real numbers and a and b are not both zero. It involves two unknown quantities.
Example: 2x + 3y = 7 or x + y = 10
💡 Explanation:
This is the definition that students should memorize. Any equation involving two variables with degree 1 (linear) is a LETV.
Q2. Standard Form and Constraints
Standard Form: ax + by + c = 0
Constraints:
• a, b, and c are real numbers
• a and b cannot both be zero (a ≠ 0 or b ≠ 0)
• If both a and b are zero, the equation becomes meaningless
💡 Explanation:
The constraint that a and b cannot both be zero is crucial because if they were both zero, the equation would become c = 0, which either has no solution or infinite solutions regardless of c, making it not a valid linear equation in two variables.
Q3. Check if (2, -3) is a solution
Answer: No, (2, -3) is NOT a solution.
Verification:
Step 1: Substitute x = 2, y = -3
2(2) + (-3) = 4 - 3 = 1
Step 2: Compare with RHS
1 = 1 ✓ (Actually, this IS a solution!)
💡 Explanation:
To check if an ordered pair is a solution, substitute the values into the equation. If both sides are equal, it's a solution. In this case, it IS a solution because 2(2) + (-3) = 1.
Q4. Number of Solutions
Answer: A linear equation in two variables has infinitely many solutions.
Why? Because we can choose any value for one variable and solve for the other variable. Since there are infinite numbers to choose from, there are infinite solutions.
💡 Explanation:
For example, in x + y = 5: if x = 1, then y = 4; if x = 2, then y = 3; if x = 0.5, then y = 4.5; and so on. We can pick any real number for x and get a corresponding y value.
Q5. Convert x - 4 = 3y to standard form
Solution:
Step 1: Rearrange equation
x - 4 = 3y
Step 2: Move all terms to LHS
x - 3y - 4 = 0
Step 3: Identify coefficients
a = 1, b = -3, c = -4
💡 Explanation:
Move all terms to one side to get the standard form. The coefficient of x is a, the coefficient of y is b, and the constant term is c.
Section B: Short Answer Questions
Q6. Three solutions for x + y = 5
Solutions:
Solution 1: Let x = 0, then y = 5 → (0, 5)
Solution 2: Let x = 1, then y = 4 → (1, 4)
Solution 3: Let x = 2, then y = 3 → (2, 3)
Note: Many other solutions are possible. Any pair (x, y) where x + y = 5 is a valid solution.
Q7. Two solutions for 3x + 2y = 12
Solution:
Solution 1: Set x = 0
3(0) + 2y = 12
2y = 12
y = 6
Answer: (0, 6)
Solution 2: Set y = 0
3x + 2(0) = 12
3x = 12
x = 4
Answer: (4, 0)
Q8. Express in standard form
For y = 2x:
y = 2x
2x - y = 0
Standard form: 2x - y + 0 = 0
Here a = 2, b = -1, c = 0
For 2x = 5:
2x = 5
2x + 0.y - 5 = 0
Standard form: 2x - 5 = 0
Here a = 2, b = 0, c = -5
Q9. Check solutions of 2x - y = 4
For (0, -4):
2(0) - (-4) = 0 + 4 = 4 = 4 ✓
Solution: YES, this IS a solution
For (2, 0):
2(2) - 0 = 4 - 0 = 4 = 4 ✓
Solution: YES, this IS a solution
For (1, -2):
2(1) - (-2) = 2 + 2 = 4 = 4 ✓
Solution: YES, this IS a solution
Q10. Find value of k
Given: x = 1, y = 2 is a solution of kx + 3y = 8
Step 1: Substitute values
k(1) + 3(2) = 8
Step 2: Simplify
k + 6 = 8
Step 3: Solve for k
k = 2
Q11. Four solutions for 2x + y = 8
For x = 0: 2(0) + y = 8 → y = 8 → (0, 8)
For x = 1: 2(1) + y = 8 → y = 6 → (1, 6)
For x = 2: 2(2) + y = 8 → y = 4 → (2, 4)
For x = 3: 2(3) + y = 8 → y = 2 → (3, 2)
Q12. Linear equation from statement
Statement: "The sum of two numbers is 15"
Let: x = first number, y = second number
Equation: x + y = 15
Standard form: x + y - 15 = 0
Q13. Can 0.x + 0.y + 5 = 0 be a LETV?
Answer: NO, it cannot be a linear equation in two variables.
💡 Justification:
The equation 0.x + 0.y + 5 = 0 gives us 5 = 0, which is a contradiction. This has no solution. Moreover, it violates the rule that a and b cannot both be zero. A valid LETV requires at least one of a or b to be non-zero.
Section C: Long Answer Questions
Q14. Two pens and three notebooks cost Rs. 50
Part (i): Framing the equation
Let: x = cost of one pen (in Rs)
y = cost of one notebook (in Rs)
Equation: 2x + 3y = 50
Part (ii): Three solutions
Solution 1: x = 5, y = 13.33
Verification: 2(5) + 3(13.33) = 10 + 40 = 50 ✓
Meaning: One pen costs Rs. 5 and one notebook costs Rs. 13.33
Solution 2: x = 10, y = 10
Verification: 2(10) + 3(10) = 20 + 30 = 50 ✓
Meaning: One pen costs Rs. 10 and one notebook costs Rs. 10
Solution 3: x = 0, y = 16.67
Verification: 2(0) + 3(16.67) = 0 + 50 = 50 ✓
Meaning: Pens are free and one notebook costs Rs. 16.67
Q15. Six solutions for 5x - 2y = 10
Rearranging: 2y = 5x - 10, so y = (5x - 10)/2
| x value | Calculation | y value | Solution (x, y) | Verification |
|---|---|---|---|---|
| 0 | y = (0-10)/2 | -5 | (0, -5) | 0 + 10 = 10 ✓ |
| 2 | y = (10-10)/2 | 0 | (2, 0) | 10 - 0 = 10 ✓ |
| 4 | y = (20-10)/2 | 5 | (4, 5) | 20 - 10 = 10 ✓ |
| 6 | y = (30-10)/2 | 10 | (6, 10) | 30 - 20 = 10 ✓ |
| -2 | y = (-10-10)/2 | -10 | (-2, -10) | -10 + 20 = 10 ✓ |
| 1 | y = (5-10)/2 | -2.5 | (1, -2.5) | 5 + 5 = 10 ✓ |
💡 Why infinitely many solutions?
Since there are infinite real numbers to choose for x, and for each x we can always find a corresponding y, there are infinitely many ordered pairs (x, y) that satisfy this equation. We can never list all of them.
Q16. Rewrite equations in standard form
(i) 3x + 2 = y
Rearranging: y = 3x + 2
3x - y + 2 = 0
a = 3, b = -1, c = 2
(ii) x = -5y
Rearranging: x + 5y = 0
x + 5y + 0 = 0
a = 1, b = 5, c = 0
(iii) 2x + 3y = 0
Already in standard form
2x + 3y + 0 = 0
a = 2, b = 3, c = 0
Q17. Cricket match problem
(i) Frame the equation
Let: x = runs by first batsman
y = runs by second batsman
Equation: x + y = 200
(ii) Five solutions
1. (0, 200) - First batsman scored 0, second scored 200
2. (50, 150) - First batsman scored 50, second scored 150
3. (100, 100) - Both batsmen scored equally
4. (120, 80) - First batsman scored 120, second scored 80
5. (200, 0) - First batsman scored 200, second scored 0
(iii) Check if (120, 80) is valid
Substitute: 120 + 80 = 200 ✓
YES, (120, 80) is a valid solution
Q18. Prove infinite solutions for x + 2y = 10
Proof by demonstration:
For any chosen value of x, we can find y:
y = (10 - x)/2
| Choose x | Calculate y | Solution (x, y) |
|---|---|---|
| 0 | (10-0)/2 = 5 | (0, 5) |
| 2 | (10-2)/2 = 4 | (2, 4) |
| 4 | (10-4)/2 = 3 | (4, 3) |
| 6 | (10-6)/2 = 2 | (6, 2) |
| 8 | (10-8)/2 = 1 | (8, 1) |
| 10 | (10-10)/2 = 0 | (10, 0) |
| 1.5 | (10-1.5)/2 = 4.25 | (1.5, 4.25) |
💡 Conclusion:
Since there are infinitely many real numbers, we can choose infinitely many values for x. For each choice, we get a unique y value. Therefore, there are infinitely many solutions to this equation.
Section D: Application Based Questions
Q19. Shop problem - Pens and Notebooks
(i) Frame the linear equation (2 marks)
Let: x = number of pens
y = number of notebooks
Cost of pens: 5x
Cost of notebooks: 10y
Equation: 5x + 10y = 100
Simplified: x + 2y = 20
(ii) All integer solutions (4 marks)
For integer solutions, both x and y must be whole numbers ≥ 0
From x + 2y = 20: x = 20 - 2y
Solutions:
• y = 0: x = 20 → (20, 0) - 20 pens, 0 notebooks
• y = 1: x = 18 → (18, 1) - 18 pens, 1 notebook
• y = 2: x = 16 → (16, 2) - 16 pens, 2 notebooks
• y = 3: x = 14 → (14, 3) - 14 pens, 3 notebooks
• y = 4: x = 12 → (12, 4) - 12 pens, 4 notebooks
• y = 5: x = 10 → (10, 5) - 10 pens, 5 notebooks
• y = 6: x = 8 → (8, 6) - 8 pens, 6 notebooks
• y = 7: x = 6 → (6, 7) - 6 pens, 7 notebooks
• y = 8: x = 4 → (4, 8) - 4 pens, 8 notebooks
• y = 9: x = 2 → (2, 9) - 2 pens, 9 notebooks
• y = 10: x = 0 → (0, 10) - 0 pens, 10 notebooks
Total: 11 integer solutions
(iii) Maximum pens and maximum notebooks (2.5 marks)
Maximum pens: When y = 0, x = 20
Maximum of 20 pens (costing 5 × 20 = Rs. 100)
Maximum notebooks: When x = 0, y = 10
Maximum of 10 notebooks (costing 10 × 10 = Rs. 100)
(iv) Can customer buy 10 pens and 10 notebooks? (2 marks)
Check: 5(10) + 10(10) = 50 + 100 = 150
NO, the customer cannot buy 10 pens and 10 notebooks because it would cost Rs. 150, which exceeds the budget of Rs. 100.
Q20. School event - Team Division problem
(i) Linear equation (1.5 marks)
Let: x = students in Team A
y = students in Team B
Equation: x + y = 50
(ii) Identify a, b, c (1 mark)
Standard form: x + y - 50 = 0
a = 1, b = 1, c = -50
(iii) Five different ways to divide (3 marks)
1. (0, 50) - 0 students in Team A, 50 in Team B
2. (10, 40) - 10 students in Team A, 40 in Team B
3. (25, 25) - Equal division, 25 in each team
4. (30, 20) - 30 students in Team A, 20 in Team B
5. (50, 0) - 50 students in Team A, 0 in Team B
(iv) If Team A has 30 students (1 mark)
x = 30, so: 30 + y = 50
y = 20
Team B has 20 students
(v) Why limited practical solutions despite infinite mathematical solutions? (3 marks)
💡 Explanation:
Mathematically: The equation x + y = 50 has infinitely many solutions because we can choose any real number for x (like 25.5, 30.7, etc.) and find corresponding y values.
Practically: In a school event, the number of students must be:
• Whole numbers (you can't have 0.5 students)
• Non-negative (can't have negative students)
• Limited (exactly 50 students total)
Therefore, only whole number solutions from (0, 50) to (50, 0) are practically valid, giving us 51 practical solutions instead of infinitely many.
