Linear Equations in Two Variables — Class 9 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Linear Equations in Two Variables" — 8 important questions with detailed answers for CBSE board exam preparation.

Linear Equations in Two Variables — Class 9 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Linear Equations in Two Variables" — 8 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 9 Mathematics chapter "Linear Equations in Two Variables" — 8 important questions with detailed answers fo…

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NCERT Solutions for Class 9 Mathematics — Linear Equations in Two Variables. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.

Linear Equations in Two Variables — All 8 Questions Solved

Q1. Find four solutions of the linear equation 2x + 3y = 12.

Equation: 2x + 3y = 12

Method: Choose values for x and solve for y (or vice versa).

Solution 1: Let x = 0
2(0) + 3y = 12
3y = 12
y = 4
Solution: (0, 4)

Solution 2: Let x = 3
2(3) + 3y = 12
6 + 3y = 12
3y = 6
y = 2
Solution: (3, 2)

Solution 3: Let x = 6
2(6) + 3y = 12
12 + 3y = 12
3y = 0
y = 0
Solution: (6, 0)

Solution 4: Let x = -3
2(-3) + 3y = 12
-6 + 3y = 12
3y = 18
y = 6
Solution: (-3, 6)

Answer: Four solutions are (0, 4), (3, 2), (6, 0), and (-3, 6)

Q2. Solve the system of equations: 2x + y = 7 and x - y = 2

Equations:
2x + y = 7 ... (1)
x - y = 2 ... (2)

Method: Elimination

Step 1: Add equations (1) and (2) to eliminate y
(2x + y) + (x - y) = 7 + 2
3x = 9

Step 2: Solve for x
x = 9/3 = 3

Step 3: Substitute x = 3 in equation (2)
3 - y = 2
y = 3 - 2 = 1

Step 4: Verify solution in both equations
Equation (1): 2(3) + 1 = 6 + 1 = 7 ✓
Equation (2): 3 - 1 = 2 ✓

Answer: x = 3, y = 1

Q3. Express 2x + 3y - 6 = 0 in the form y = mx + c and identify the slope and y-intercept.

Equation: 2x + 3y - 6 = 0

Step 1: Solve for y
3y = -2x + 6
y = -2x/3 + 6/3
y = (-2/3)x + 2

Step 2: Compare with y = mx + c
This is in the form y = mx + c where:
m = -2/3 (slope)
c = 2 (y-intercept)

Step 3: Interpret the slope
Slope m = -2/3 < 0, so the line is falling (decreasing from left to right).
For every 3 units moved right, the line drops 2 units.

Step 4: Interpret the y-intercept
The line crosses the y-axis at (0, 2).

Answer:
y = (-2/3)x + 2
Slope (m) = -2/3
y-intercept (c) = 2

Q4. Determine if (2, 1) is a solution of the system: x + 2y = 4 and 2x - y = 3

Point: (2, 1) means x = 2, y = 1
Equations:
x + 2y = 4 ... (1)
2x - y = 3 ... (2)

Step 1: Check equation (1)
Substitute x = 2, y = 1:
2 + 2(1) = 2 + 2 = 4 ✓
The point satisfies equation (1).

Step 2: Check equation (2)
Substitute x = 2, y = 1:
2(2) - 1 = 4 - 1 = 3 ✓
The point satisfies equation (2).

Step 3: Conclusion
Since (2, 1) satisfies both equations, it is the solution of the system.

Answer: Yes, (2, 1) is a solution of the given system.

Q5. Solve graphically: x + y = 5 and x - y = 1

Equations:
x + y = 5 ... (1)
x - y = 1 ... (2)

Step 1: Find solutions for equation (1): x + y = 5
Let x = 0: y = 5 → (0, 5)
Let x = 5: y = 0 → (5, 0)
Let x = 2: y = 3 → (2, 3)

Step 2: Find solutions for equation (2): x - y = 1
Let x = 0: y = -1 → (0, -1)
Let x = 1: y = 0 → (1, 0)
Let x = 3: y = 2 → (3, 2)

Step 3: Plot both lines
Line 1 (x + y = 5): passes through (0, 5) and (5, 0)
Line 2 (x - y = 1): passes through (0, -1) and (1, 0)

Step 4: Find intersection point
Both lines intersect at a single point.
From the graphical analysis, the intersection is at (3, 2).

Step 5: Verify algebraically
Add equations: (x + y) + (x - y) = 5 + 1
2x = 6 → x = 3
Substitute in equation (1): 3 + y = 5 → y = 2

Answer: Solution is (3, 2)

Q6. Find the values of a and b such that (1, 2) is a solution of ax + by = 8 and 2ax - by = 5

Point: (1, 2) means x = 1, y = 2
Equations:
ax + by = 8 ... (1)
2ax - by = 5 ... (2)

Step 1: Substitute (1, 2) into equation (1)
a(1) + b(2) = 8
a + 2b = 8 ... (3)

Step 2: Substitute (1, 2) into equation (2)
2a(1) - b(2) = 5
2a - 2b = 5 ... (4)

Step 3: Solve the system of equations (3) and (4)
From (3): a + 2b = 8
From (4): 2a - 2b = 5

Add equations (3) and (4):
(a + 2b) + (2a - 2b) = 8 + 5
3a = 13
a = 13/3

Step 4: Find b
Substitute a = 13/3 in equation (3):
13/3 + 2b = 8
2b = 8 - 13/3
2b = 24/3 - 13/3
2b = 11/3
b = 11/6

Step 5: Verify
Equation (1): (13/3)(1) + (11/6)(2) = 13/3 + 22/6 = 13/3 + 11/3 = 24/3 = 8 ✓
Equation (2): 2(13/3)(1) - (11/6)(2) = 26/3 - 22/6 = 26/3 - 11/3 = 15/3 = 5 ✓

Answer: a = 13/3, b = 11/6

Q7. A straight line passes through (1, 1) and has slope 2. Write its equation in the form ax + by + c = 0.

Given: Point (1, 1), Slope m = 2

Step 1: Use point-slope form
y - y₁ = m(x - x₁)
y - 1 = 2(x - 1)

Step 2: Expand
y - 1 = 2x - 2
y = 2x - 2 + 1
y = 2x - 1

Step 3: Convert to standard form ax + by + c = 0
2x - y - 1 = 0

Step 4: Verify that (1, 1) lies on the line
2(1) - 1 - 1 = 2 - 1 - 1 = 0 ✓

Step 5: Verify slope
From 2x - y - 1 = 0, solving for y: y = 2x - 1
Slope = coefficient of x = 2 ✓

Answer: 2x - y - 1 = 0

Q8. Solve the system: 3x + 2y = 12 and 2x - 3y = 5 using substitution method.

Equations:
3x + 2y = 12 ... (1)
2x - 3y = 5 ... (2)

Step 1: Express x from equation (2)
2x = 5 + 3y
x = (5 + 3y)/2 ... (3)

Step 2: Substitute (3) into equation (1)
3[(5 + 3y)/2] + 2y = 12

Step 3: Multiply by 2 to clear fraction
3(5 + 3y) + 4y = 24
15 + 9y + 4y = 24
13y = 24 - 15
13y = 9
y = 9/13

Step 4: Find x using equation (3)
x = (5 + 3(9/13))/2
x = (5 + 27/13)/2
x = (65/13 + 27/13)/2
x = (92/13)/2
x = 92/26
x = 46/13

Step 5: Verify in equation (1)
3(46/13) + 2(9/13) = 138/13 + 18/13 = 156/13 = 12 ✓

Step 6: Verify in equation (2)
2(46/13) - 3(9/13) = 92/13 - 27/13 = 65/13 = 5 ✓

Answer: x = 46/13, y = 9/13

Linear Equations in Two Variables — Practice MCQs with Answers

Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.

MCQ 1. Which of the following is a linear equation in two variables?

  • A. 2x + 3 = 0
  • B. x^2 + y = 5
  • C. 3x - 4y = 7
  • D. xy = 1

Answer: C. 3x - 4y = 7 — A linear equation in two variables is of the form ax + by + c = 0, where a and b are not both zero. Only 3x - 4y = 7 fits this definition as the powers of x and y are both 1.

MCQ 2. Which of the following is a solution to the equation x + 2y = 7?

  • A. (1, 2)
  • B. (2, 1)
  • C. (3, 2)
  • D. (1, 3)

Answer: D. (1, 3) — Substitute x=1 and y=3 into the equation: 1 + 2(3) = 1 + 6 = 7. Since 7 = 7, (1, 3) is a solution.

MCQ 3. How many solutions does a linear equation in two variables have?

  • A. One
  • B. Two
  • C. Infinite
  • D. None

Answer: C. Infinite — A linear equation in two variables has infinitely many solutions because for every chosen value of one variable, there is a corresponding unique value for the other variable.

MCQ 4. The equation of the x-axis is:

  • A. x = 0
  • B. y = 0
  • C. x + y = 0
  • D. x = y

Answer: B. y = 0 — Every point on the x-axis has its y-coordinate as 0. Therefore, the equation of the x-axis is y = 0.

MCQ 5. The equation of the y-axis is:

  • A. x = 0
  • B. y = 0
  • C. x + y = 0
  • D. x = y

Answer: A. x = 0 — Every point on the y-axis has its x-coordinate as 0. Therefore, the equation of the y-axis is x = 0.

MCQ 6. Which point satisfies the equation x - y = 4?

  • A. (1, 3)
  • B. (3, 1)
  • C. (5, 1)
  • D. (1, 5)

Answer: C. (5, 1) — Substitute x=5 and y=1 into the equation: 5 - 1 = 4. This is true, so (5, 1) is a solution.

MCQ 7. The sum of two numbers is 15. If the first number is x and the second number is y, write the linear equation.

  • A. x - y = 15
  • B. x + y = 15
  • C. xy = 15
  • D. x = 15y

Answer: B. x + y = 15 — The statement 'The sum of two numbers is 15' directly translates to x + y = 15 when x and y are the two numbers.

MCQ 8. In a cricket match, two batsmen M and N together scored 176 runs. Represent this situation as a linear equation in two variables.

  • A. M - N = 176
  • B. M + N = 176
  • C. MN = 176
  • D. M / N = 176

Answer: B. M + N = 176 — The phrase 'together scored 176 runs' implies that the sum of their individual scores is 176. So, M + N = 176.

MCQ 9. For the equation x + 2y = 5, if y = 0, what is the value of x?

  • A. 5
  • B. 0
  • C. 2.5
  • D. 1

Answer: A. 5 — Substitute y=0 into the equation x + 2y = 5: x + 2(0) = 5. This simplifies to x = 5.

MCQ 10. The general form of a linear equation in two variables x and y is:

  • A. ax + by + c = 0
  • B. ax^2 + by + c = 0
  • C. ax + by^2 + c = 0
  • D. ax + c = 0

Answer: A. ax + by + c = 0 — The standard or general form of a linear equation in two variables is ax + by + c = 0, where a, b, c are real numbers and a, b are not both zero.

MCQ 11. If (2, 3) is a solution to the equation ax - 3y = 5, then what is the value of a?

  • A. 5
  • B. 7
  • C. 3
  • D. 14

Answer: B. 7 — Substitute x=2 and y=3 into the equation: a(2) - 3(3) = 5. This simplifies to 2a - 9 = 5, so 2a = 14, which gives a = 7.

MCQ 12. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.

  • A. x = 2y
  • B. y = 2x
  • C. x + y = 2
  • D. x - y = 2

Answer: A. x = 2y — Let x be the cost of the notebook and y be the cost of the pen. According to the statement, the cost of the notebook (x) is twice the cost of the pen (2y), so x = 2y.

MCQ 13. Which of the following equations is equivalent to 2x + 3y - 6 = 0?

  • A. y = (6 - 2x) / 3
  • B. x = (3y - 6) / 2
  • C. 2x = 6 + 3y
  • D. 3y = 2x + 6

Answer: A. y = (6 - 2x) / 3 — From 2x + 3y - 6 = 0, move 2x and -6 to the right side: 3y = 6 - 2x. Then divide by 3: y = (6 - 2x) / 3.

MCQ 14. Which of the following pairs are solutions to the equation 2x - y = 5?

  • A. (3, 1) and (2, -1)
  • B. (1, -3) and (3, 1)
  • C. (2, 1) and (3, 1)
  • D. (0, -5) and (1, 3)

Answer: B. (1, -3) and (3, 1) — For (1, -3): 2(1) - (-3) = 2 + 3 = 5. For (3, 1): 2(3) - 1 = 6 - 1 = 5. Both pairs satisfy the equation.

MCQ 15. Which equation represents a line passing through the origin?

  • A. x + y = 1
  • B. x - y = 2
  • C. 2x + 3y = 0
  • D. x = 5

Answer: C. 2x + 3y = 0 — A line passes through the origin if (0, 0) is a solution to its equation. For 2x + 3y = 0, substituting x=0, y=0 gives 2(0) + 3(0) = 0, which is true.

MCQ 16. What is the y-intercept of the line 3x + 4y = 12?

  • A. (4, 0)
  • B. (0, 3)
  • C. (3, 0)
  • D. (0, 4)

Answer: B. (0, 3) — The y-intercept is the point where the line crosses the y-axis, meaning x=0. Substitute x=0 into the equation: 3(0) + 4y = 12, which gives 4y = 12, so y = 3. The y-intercept is (0, 3).

MCQ 17. A taxi fare in a city is Rs. 8 for the first kilometre and Rs. 5 for each subsequent kilometre. If x is the distance covered in kilometres and y is the total fare, write a linear equation for this.

  • A. y = 8 + 5x
  • B. y = 8x + 5
  • C. y = 5x + 3
  • D. y = 8 + 5(x - 1)

Answer: D. y = 8 + 5(x - 1) — For the first kilometre, the fare is Rs. 8. For the remaining (x - 1) kilometres, the fare is 5(x - 1). So, the total fare y = 8 + 5(x - 1).

MCQ 18. If the line y = mx + 3 passes through the point (-1, 0), find the value of m.

  • A. 3
  • B. -3
  • C. 0
  • D. 1

Answer: A. 3 — Since the point (-1, 0) lies on the line, substitute x=-1 and y=0 into the equation: 0 = m(-1) + 3. This simplifies to 0 = -m + 3, so m = 3.

MCQ 19. A part of the monthly expenses of a family is fixed and the other part varies with the number of members in the family. When there are 4 members, the monthly expense is Rs. 1800. When there are 7 members, the monthly expense is Rs. 2400. Write linear equations to represent this scenario.

  • A. f + 4v = 1800, f + 7v = 2400
  • B. 4f + v = 1800, 7f + v = 2400
  • C. f - 4v = 1800, f - 7v = 2400
  • D. f + v = 1800, f + v = 2400

Answer: A. f + 4v = 1800, f + 7v = 2400 — Let f be the fixed expense and v be the variable expense per member. For 4 members, f + 4v = 1800. For 7 members, f + 7v = 2400.

MCQ 20. Write the equation x/2 + y/3 = 1 in the form ax + by + c = 0.

  • A. 3x + 2y - 6 = 0
  • B. 2x + 3y - 1 = 0
  • C. 3x + 2y + 6 = 0
  • D. x + y - 1 = 0

Answer: A. 3x + 2y - 6 = 0 — To eliminate denominators, multiply the entire equation by the LCM of 2 and 3 (which is 6): 6(x/2) + 6(y/3) = 6(1). This simplifies to 3x + 2y = 6, which rearranges to 3x + 2y - 6 = 0.

How to Use These Solutions

Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.

Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.

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