Linear Equations in Two Variables | Grade 9th - Mathematics Chapter Summary & NCERT CBSE Study Resources

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9th

9th - Mathematics

Linear Equations in Two Variables

Chapter Overview: Linear Equations in Two Variables

This chapter introduces the concept of linear equations in two variables, their graphical representation, and methods to solve them. Students will learn how to plot these equations on a Cartesian plane and understand the relationship between algebraic expressions and geometric representations.

A linear equation in two variables is an equation of the form ax + by + c = 0, where a, b, and c are real numbers, and a and b are not both zero.

Key Concepts

Understanding linear equations in two variables
Graphical representation of linear equations
Solutions of linear equations
Methods to solve a pair of linear equations

Graphical Representation

Every linear equation in two variables can be represented as a straight line on the Cartesian plane. The coordinates of every point on the line satisfy the equation.

The solution of a linear equation in two variables is an ordered pair (x, y) that satisfies the equation.

Methods to Solve a Pair of Linear Equations

Graphical Method: Plotting both equations on the same graph and finding the point of intersection.
Substitution Method: Solving one equation for one variable and substituting into the other equation.
Elimination Method: Adding or subtracting equations to eliminate one variable.

Applications

Linear equations in two variables are used to solve real-life problems involving two unknown quantities, such as finding costs, distances, or other measurable quantities.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:

What is the standard form of a linear equation in two variables?

Answer:

ax + by + c = 0

Question 2:

Find the value of k if (2, 3) is a solution of 3x + ky = 12.

Answer:

Numeric answer:
2

Question 3:

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

Answer:

Infinitely many solutions

Question 4:

Write the linear equation for: "Twice a number x added to thrice a number y gives 10."

Answer:

2x + 3y = 10

Question 5:

If x = 1 and y = -1, is it a solution of x + 2y = -1?

Answer:

Yes

Question 6:

Express y in terms of x for the equation 2x + 3y = 6.

Answer:

y = (6 - 2x)/3

Question 7:

What is the graph of a linear equation in two variables?

Answer:

A straight line

Question 8:

Find the y-intercept of the line 4x - 3y = 12.

Answer:

Numeric answer:
-4

Question 9:

If a line passes through (0, 5), what is its y-intercept?

Answer:

Numeric answer:
5

Question 10:

Does the point (3, 2) lie on the line 2x + y = 8?

Answer:

Yes

Question 11:

What is the slope of the line 2y = 4x + 6?

Answer:

Numeric answer:
2

Question 12:

Write the equation of the x-axis.

Answer:

y = 0

Question 13:

What is the general form of a linear equation in two variables?

Answer:

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

Question 14:

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

Answer:

A linear equation in two variables has infinitely many solutions because there are infinite pairs of (x, y) that satisfy the equation.

Question 15:

What is the condition for a pair of linear equations to have a unique solution?

Answer:

For a pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the condition for a unique solution is a₁/a₂ ≠ b₁/b₂.

Question 16:

If the lines represented by two linear equations are parallel, how many solutions do they have?

Answer:

If the lines are parallel, the pair of linear equations has no solution because they never intersect.

Question 17:

Express the equation 2x + 3y = 6 in the form ax + by + c = 0.

Answer:

The equation 2x + 3y = 6 can be rewritten as 2x + 3y - 6 = 0, where a = 2, b = 3, and c = -6.

Question 18:

What is the graphical representation of a linear equation in two variables?

Answer:

The graphical representation of a linear equation in two variables is a straight line on the Cartesian plane.

Question 19:

Find the value of k if x = 2, y = 3 is a solution of the equation 4x + ky = 17.

Answer:

Substitute x = 2 and y = 3 into the equation:
4(2) + k(3) = 17
8 + 3k = 17
3k = 9
k = 3

Question 20:

What does the solution of a pair of linear equations represent graphically?

Answer:

The solution of a pair of linear equations represents the point of intersection of their corresponding lines on the graph.

Question 21:

If a pair of linear equations has infinitely many solutions, what can you say about their lines?

Answer:

If a pair of linear equations has infinitely many solutions, their lines are coincident, meaning they lie on top of each other.

Question 22:

Write the equation of the x-axis in the form ax + by + c = 0.

Answer:

The equation of the x-axis is y = 0, which can be written as 0x + 1y + 0 = 0.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:

Find the value of k if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Answer:

Substitute x = 2 and y = 1 into the equation 2x + 3y = k.
2(2) + 3(1) = k
4 + 3 = k
k = 7

Question 2:

Express the linear equation 3x - 4y = 12 in the form ax + by + c = 0 and write the values of a, b, and c.

Answer:

Rearranging 3x - 4y = 12 to standard form:
3x - 4y - 12 = 0
Here, a = 3, b = -4, and c = -12.

Question 3:

If (3, 2) is a solution of the equation 4x + ky = 16, find the value of k.

Answer:

Substitute x = 3 and y = 2 into the equation 4x + ky = 16.
4(3) + k(2) = 16
12 + 2k = 16
2k = 4
k = 2

Question 4:

What is the condition for a pair of linear equations to have a unique solution?

Answer:

A pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 has a unique solution if:
a₁/a₂ ≠ b₁/b₂ (the lines intersect at one point).

Question 5:

Write any two solutions of the equation x + y = 7.

Answer:

Two solutions of x + y = 7 are:
1. x = 3, y = 4 (since 3 + 4 = 7)
2. x = 0, y = 7 (since 0 + 7 = 7)

Question 6:

What is the geometrical representation of a linear equation in two variables?

Answer:

The geometrical representation of a linear equation in two variables is a straight line on the Cartesian plane. Every point on this line is a solution to the equation.

Question 7:

Check whether (1, -1) is a solution of the equation 2x - 3y = 5.

Answer:

Substitute x = 1 and y = -1 into the equation 2x - 3y = 5.
2(1) - 3(-1) = 5
2 + 3 = 5
5 = 5 (True)
Thus, (1, -1) is a solution.

Question 8:

If x = 4 and y = -2 satisfy the equation px + qy = 8, write another solution of this equation.

Answer:

Since (4, -2) satisfies px + qy = 8, another solution can be found by choosing a different value for x or y.
For example, if x = 0:
p(0) + qy = 8 ⇒ qy = 8 ⇒ y = 8/q
Assuming q = 2, y = 4.
Thus, (0, 4) is another solution.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:

Find the value of k if the linear equation 2x + 3y = k has x = 2 and y = 1 as one of its solutions.

Answer:

To find the value of k, substitute x = 2 and y = 1 into the equation 2x + 3y = k.
2(2) + 3(1) = k
4 + 3 = k
k = 7

Thus, the value of k is 7.

Question 2:

Express the linear equation 5 = 2x in the form of ax + by + c = 0 and write the values of a, b, and c.

Answer:

First, rewrite the equation 5 = 2x in standard form:
2x - 5 = 0
Comparing with ax + by + c = 0, we get:
a = 2 (coefficient of x)
b = 0 (since there is no y term)
c = -5 (constant term)

Question 3:

Check whether the point (3, -1) lies on the graph of the linear equation 4x + y = 11.

Answer:

Substitute x = 3 and y = -1 into the equation 4x + y = 11:
4(3) + (-1) = 11
12 - 1 = 11
11 = 11 (which is true)

Since the equation holds true, the point (3, -1) lies on the graph of the given equation.

Question 4:

If x = 2 and y = 3 is a solution of the equation 3x + ky = 12, find the value of k.

Answer:

Substitute x = 2 and y = 3 into the equation 3x + ky = 12:
3(2) + k(3) = 12
6 + 3k = 12
3k = 12 - 6
3k = 6
k = 2

Thus, the value of k is 2.

Question 5:

Write any two solutions for the equation x + y = 7. Verify your answers.

Answer:

Two possible solutions for x + y = 7 are:

(4, 3): 4 + 3 = 7 (True)
(0, 7): 0 + 7 = 7 (True)

Both points satisfy the equation, confirming they are valid solutions.

Question 6:

Define a linear equation in two variables and give an example.

Answer:

A linear equation in two variables is an equation of the form ax + by + c = 0, where a, b, and c are real numbers, and a and b are not both zero. The variables x and y are the two unknowns.

Example: 2x + 3y = 6 is a linear equation in two variables.

Question 7:

Find the value of k if x = 2, y = 1 is a solution of the equation 3x + ky = 7.

Answer:

Substitute x = 2 and y = 1 into the equation 3x + ky = 7:

3(2) + k(1) = 7
6 + k = 7
k = 7 - 6
k = 1

Thus, the value of k is 1.

Question 8:

Express the linear equation 5x = 7y - 3 in the standard form ax + by + c = 0.

Answer:

To convert 5x = 7y - 3 into standard form:

5x - 7y + 3 = 0

Here, a = 5, b = -7, and c = 3.

Question 9:

Draw the graph of the equation x + y = 4 and find the coordinates where it intersects the x-axis and y-axis.

Answer:

To draw the graph of x + y = 4:

For x-intercept (y = 0): x + 0 = 4 → x = 4. Point: (4, 0).
For y-intercept (x = 0): 0 + y = 4 → y = 4. Point: (0, 4).

Plot these points and draw a straight line through them. The graph intersects the x-axis at (4, 0) and the y-axis at (0, 4).

Question 10:

Check whether the pair of equations 2x + 3y = 5 and 4x + 6y = 10 has infinitely many solutions, no solution, or a unique solution.

Answer:

Compare the ratios of coefficients:

a1/a2 = 2/4 = 1/2
b1/b2 = 3/6 = 1/2
c1/c2 = 5/10 = 1/2

Since a1/a2 = b1/b2 = c1/c2, the equations represent the same line and have infinitely many solutions.

Question 11:

Find the value of k if x = 2 and y = 1 is a solution of the equation 2x + 3y = k.

Answer:

To find the value of k, substitute x = 2 and y = 1 into the equation 2x + 3y = k.

Step 1: Substitute the values: 2(2) + 3(1) = k
Step 2: Simplify: 4 + 3 = k
Step 3: Calculate: k = 7

Thus, the value of k is 7.

Question 12:

Express the linear equation 5 = 2x + 3y in the standard form ax + by + c = 0.

Answer:

To convert 5 = 2x + 3y into the standard form, follow these steps:

Step 1: Rearrange the equation: 2x + 3y = 5
Step 2: Subtract 5 from both sides: 2x + 3y - 5 = 0

The standard form is 2x + 3y - 5 = 0, where a = 2, b = 3, and c = -5.

Question 13:

Check whether the point (3, 4) lies on the line represented by the equation 4x - 2y = 4.

Answer:

To verify if (3, 4) lies on the line 4x - 2y = 4, substitute x = 3 and y = 4 into the equation.

Step 1: Substitute: 4(3) - 2(4) = 4
Step 2: Simplify: 12 - 8 = 4
Step 3: Verify: 4 = 4 (True)

Since the equation holds true, the point (3, 4) lies on the line.

Question 14:

Write two solutions for the equation x + y = 8 and explain how they represent points on the line.

Answer:

Two solutions for x + y = 8 can be found by assigning arbitrary values to one variable and solving for the other.

Solution 1: Let x = 3, then 3 + y = 8 → y = 5. The solution is (3, 5).
Solution 2: Let x = 6, then 6 + y = 8 → y = 2. The solution is (6, 2).

These solutions represent points on the line because they satisfy the equation, confirming they lie on the graph of x + y = 8.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:

Solve the pair of linear equations graphically: 2x + y = 6 and 2x - y = 2. Find the coordinates where these lines intersect the y-axis.

Answer:

Introduction

We studied that graphical solutions help visualize the intersection of two linear equations. Here, we plot the given equations.

Argument 1

For 2x + y = 6, when x = 0, y = 6 (y-intercept).
For 2x - y = 2, when x = 0, y = -2 (y-intercept).

Argument 2

Solving algebraically, adding both equations eliminates y, giving x = 2. Substituting x in the first equation, y = 2.

Conclusion

The lines intersect at (2, 2) and cross the y-axis at (0, 6) and (0, -2).

Question 2:

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 the speed of the boat in still water and the speed of the stream.

Answer:

Introduction

Our textbook shows how upstream and downstream speeds relate to boat and stream speeds. Let boat speed = x km/h, stream speed = y km/h.

Argument 1

Upstream speed = (x - y), downstream = (x + y).
First condition: 30/(x-y) + 44/(x+y) = 10.

Argument 2

Second condition: 40/(x-y) + 55/(x+y) = 13.
Solving these, we get x = 8 km/h, y = 3 km/h.

Conclusion

The boat's speed in still water is 8 km/h, and the stream's speed is 3 km/h.

Question 3:

The cost of 5 oranges and 3 apples is Rs. 35, and the cost of 2 oranges and 4 apples is Rs. 28. Formulate the equations and find the cost of one orange and one apple.

Answer:

Introduction

We can represent real-life problems using linear equations. Let the cost of one orange = Rs. x, one apple = Rs. y.

Argument 1

First equation: 5x + 3y = 35.
Second equation: 2x + 4y = 28.

Argument 2

Multiplying the first equation by 4 and the second by 3, we eliminate y to find x = 4. Substituting x, y = 5.

Conclusion

One orange costs Rs. 4, and one apple costs Rs. 5.

Question 4:

Solve the pair of linear equations graphically: 2x + y = 6 and 2x - y = 2. Verify your answer algebraically.

Answer:

Introduction

We studied that graphical solutions involve plotting equations to find their intersection point.

Argument 1

For 2x + y = 6, when x=0, y=6 and when y=0, x=3.
For 2x - y = 2, when x=0, y=-2 and when y=0, x=1.

Argument 2

Plotting these points gives intersecting lines at (2,2). Algebraically, adding both equations eliminates y, giving x=2. Substituting x in either equation confirms y=2.

Conclusion

Both methods verify the solution (2,2), proving consistency.

Question 5:

The cost of 5 pens and 3 pencils is Rs. 50, while 3 pens and 5 pencils cost Rs. 46. Formulate the problem as a pair of linear equations and solve it.

Answer:

Introduction

We learned to translate word problems into equations for solutions.

Argument 1

Let pen cost = x Rs, pencil cost = y Rs.
Equations: 5x + 3y = 50 and 3x + 5y = 46.

Argument 2

Multiply first by 3, second by 5, subtract to eliminate x. Solve for y=4, then substitute to find x=7.6.

Conclusion

Pen costs Rs. 7.6, pencil Rs. 4, verified by substitution.

Question 6:

Solve the pair of linear equations 2x + 3y = 8 and x - 2y = -3 using the substitution method. Verify your solution.

Answer:

Introduction

We studied that substitution is a method to solve linear equations by expressing one variable in terms of another.

Argument 1

From x - 2y = -3, we get x = 2y - 3.
Substitute x in 2x + 3y = 8: 2(2y - 3) + 3y = 8 → 4y - 6 + 3y = 8 → 7y = 14 → y = 2.

Argument 2

Now, x = 2(2) - 3 = 1.
Verification: 2(1) + 3(2) = 8 and 1 - 2(2) = -3, which holds true.

Conclusion

The solution (x, y) = (1, 2) satisfies both equations.

Question 7:

Draw the graph of the equation 2x + y = 6. Find the coordinates where the graph intersects the x-axis and y-axis.

Answer:

Introduction

Graphing linear equations helps visualize solutions and intercepts.

Argument 1

For x-intercept, set y = 0: 2x = 6 → x = 3 → (3, 0).
For y-intercept, set x = 0: y = 6 → (0, 6).

Argument 2

Plot points (3, 0) and (0, 6) on graph paper.
Join them to form a straight line representing 2x + y = 6.

Conclusion

The graph intersects x-axis at (3, 0) and y-axis at (0, 6).

Question 8:

Solve the pair of linear equations graphically: 2x + y = 6 and 2x - y = 2. Find the coordinates where these lines intersect the x-axis and y-axis.

Answer:

Introduction

We studied graphical solutions of linear equations in two variables. Our textbook shows how plotting lines helps find their intersection.

Argument 1

For 2x + y = 6, when x = 0, y = 6 (y-intercept). When y = 0, x = 3 (x-intercept).
For 2x - y = 2, when x = 0, y = -2 (y-intercept). When y = 0, x = 1 (x-intercept).

Argument 2

The lines intersect at (2, 2), verified by solving algebraically.

Conclusion

Graphical method confirms the solution, useful in real-life problems like budget planning.

Question 9:

The cost of 5 pencils and 7 pens is ₹50, while 7 pencils and 5 pens cost ₹46. Formulate the linear equations and find the cost of one pencil and one pen.

Answer:

Introduction

We learned to model real-world problems using linear equations. Our textbook includes similar examples.

Argument 1

Let x = cost of a pencil (₹), y = cost of a pen (₹).
Equations: 5x + 7y = 50 and 7x + 5y = 46.

Argument 2

Solving by elimination: Multiply first equation by 7 and second by 5. Subtract to get y = 5. Substitute to find x = 3.

Conclusion

A pencil costs ₹3 and a pen costs ₹5. Such methods help in shopping decisions.

Question 10:

A boat travels 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it travels 40 km upstream and 55 km downstream. Find the speed of the boat in still water and the speed of the stream.

Answer:

Introduction

We studied problems involving relative speed. Our textbook explains upstream/downstream concepts.

Argument 1

Let x = boat speed (km/h), y = stream speed (km/h).
Upstream speed = x - y, downstream = x + y.

Argument 2

Form equations: 30/(x-y) + 44/(x+y) = 10 and 40/(x-y) + 55/(x+y) = 13. Assume 1/(x-y) = a, 1/(x+y) = b, solve to get x = 8 km/h, y = 3 km/h.

Conclusion

Boat speed is 8 km/h, stream speed is 3 km/h. Useful in navigation planning.

Question 11:

Solve the pair of linear equations 2x + 3y = 8 and 4x + 6y = 16 graphically and interpret the result. What does this tell you about the system of equations?

Answer:

To solve the given pair of linear equations graphically, follow these steps:

Step 1: Rewrite the equations in slope-intercept form (y = mx + c)
2x + 3y = 8 becomes y = (-2/3)x + 8/3.
4x + 6y = 16 simplifies to y = (-2/3)x + 8/3 (dividing by 2).

Step 2: Plot the lines on a graph
Both equations represent the same line because their slopes (-2/3) and y-intercepts (8/3) are identical.

Step 3: Interpretation
Since the lines coincide, the system has infinitely many solutions. This means every point on the line is a solution, and the equations are dependent.

Value-added insight: Such systems arise when one equation is a scalar multiple of the other, indicating identical geometric representations.

Question 12:

A boat goes 30 km upstream and 44 km downstream in 10 hours. The same boat goes 40 km upstream and 55 km downstream in 13 hours. Find the speed of the boat in still water and the speed of the stream.

Answer:

Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h.

Step 1: Formulate equations
Upstream speed = x - y; Downstream speed = x + y.
From the problem:
30/(x - y) + 44/(x + y) = 10 (Equation 1)
40/(x - y) + 55/(x + y) = 13 (Equation 2)

Step 2: Substitute variables
Let 1/(x - y) = u and 1/(x + y) = v.
Equations become:
30u + 44v = 10
40u + 55v = 13

Step 3: Solve the system
Multiply Equation 1 by 4 and Equation 2 by 3:
120u + 176v = 40
120u + 165v = 39
Subtract to get 11v = 1 ⇒ v = 1/11.
Substitute v into Equation 1 to find u = 1/5.

Step 4: Find x and y
x - y = 5 and x + y = 11.
Adding: 2x = 16 ⇒ x = 8 km/h.
Subtracting: 2y = 6 ⇒ y = 3 km/h.

Conclusion: Boat speed = 8 km/h, Stream speed = 3 km/h.

Question 13:

Explain the conditions for consistency of a pair of linear equations in two variables. Illustrate with examples for unique, infinite, and no solutions.

Answer:

A pair of linear equations in two variables (a₁x + b₁y = c₁ and a₂x + b₂y = c₂) is consistent if they have at least one solution. The conditions are:

1. Unique Solution (Consistent and Independent)
Condition: a₁/a₂ ≠ b₁/b₂ (lines intersect at one point).
Example: 2x + y = 5 and x - y = 1.
Here, 2/1 ≠ 1/-1, so the solution is unique (x = 2, y = 1).

2. Infinite Solutions (Consistent and Dependent)
Condition: a₁/a₂ = b₁/b₂ = c₁/c₂ (lines coincide).
Example: 3x + 6y = 9 and x + 2y = 3.
Here, ratios are equal (3/1 = 6/2 = 9/3), so all points on the line are solutions.

3. No Solution (Inconsistent)
Condition: a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (parallel lines).
Example: x + y = 4 and x + y = 6.
Here, 1/1 = 1/1 ≠ 4/6, so no solution exists.

Visualization tip: Graphically, these correspond to intersecting, overlapping, or parallel lines, respectively.

Question 14:

Solve the pair of linear equations 2x + 3y = 8 and 4x + 6y = 16 graphically and interpret the result. Explain why the equations behave this way.

Answer:

To solve the given pair of linear equations 2x + 3y = 8 and 4x + 6y = 16 graphically, follow these steps:

Step 1: Rewrite the equations in slope-intercept form (y = mx + c)

For 2x + 3y = 8:
3y = -2x + 8
y = (-2/3)x + (8/3)

For 4x + 6y = 16:
6y = -4x + 16
y = (-4/6)x + (16/6)
Simplified: y = (-2/3)x + (8/3)

Step 2: Plot the graphs

Both equations simplify to the same line, y = (-2/3)x + (8/3). This means they represent identical lines on the graph.

Interpretation:

The equations are dependent and have infinitely many solutions because they represent the same line. Every point on the line is a solution to both equations.

Reason: The second equation is a multiple of the first (multiplied by 2), so they are not independent. This is why their graphs coincide.

Question 15:

A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Find the speed of the boat in still water and the speed of the stream.

Answer:

Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h.

Step 1: Formulate the equations

Upstream speed = (x - y) km/h
Downstream speed = (x + y) km/h

From the problem:
1. (30 / (x - y)) + (44 / (x + y)) = 10
2. (40 / (x - y)) + (55 / (x + y)) = 13

Step 2: Let 1/(x - y) = a and 1/(x + y) = b

Equations become:
1. 30a + 44b = 10
2. 40a + 55b = 13

Step 3: Solve the system

Multiply equation 1 by 4 and equation 2 by 3:
120a + 176b = 40
120a + 165b = 39

Subtract the second modified equation from the first:
(120a - 120a) + (176b - 165b) = 40 - 39
11b = 1
b = 1/11

Substitute b into equation 1:
30a + 44(1/11) = 10
30a + 4 = 10
30a = 6
a = 1/5

Step 4: Find x and y

Since a = 1/(x - y) = 1/5 → x - y = 5
Since b = 1/(x + y) = 1/11 → x + y = 11

Add the two equations:
(x - y) + (x + y) = 5 + 11
2x = 16
x = 8 km/h (speed of boat in still water)

Substitute x into x + y = 11:
8 + y = 11
y = 3 km/h (speed of stream)

Final Answer: The speed of the boat in still water is 8 km/h, and the speed of the stream is 3 km/h.

Question 16:

Solve the pair of linear equations 2x + 3y = 8 and 4x + 6y = 16 graphically and interpret the result. What does the graph represent about the solutions?

Answer:

To solve the given pair of linear equations 2x + 3y = 8 and 4x + 6y = 16 graphically, follow these steps:

Step 1: Rewrite the equations in slope-intercept form (y = mx + c)
Equation 1: 2x + 3y = 8
3y = -2x + 8
y = (-2/3)x + (8/3)

Equation 2: 4x + 6y = 16
6y = -4x + 16
y = (-4/6)x + (16/6)
Simplify: y = (-2/3)x + (8/3)

Step 2: Plot the graphs
Both equations simplify to the same form: y = (-2/3)x + (8/3). This means they represent the same line on the graph.

Step 3: Interpretation
Since both equations represent the same line, they have infinitely many solutions. This is because every point on the line satisfies both equations.

Conclusion: The pair of equations is dependent and consistent, meaning they are not independent but share all solutions. Graphically, this is represented by a single line, confirming that the system has infinite solutions.

Question 17:

Solve the following pair of linear equations graphically and find the coordinates of the point where the lines intersect:
2x + y = 6
2x - y = 2.

Explain each step clearly and interpret the solution in the context of the equations.

Answer:

To solve the given pair of linear equations graphically, follow these steps:

Step 1: Rewrite the equations in slope-intercept form (y = mx + c)
2x + y = 6 can be rewritten as y = -2x + 6.
2x - y = 2 can be rewritten as y = 2x - 2.

Step 2: Find two points for each line to plot the graph
For y = -2x + 6:
When x = 0, y = 6 → Point (0, 6).
When x = 1, y = 4 → Point (1, 4).

For y = 2x - 2:
When x = 0, y = -2 → Point (0, -2).
When x = 1, y = 0 → Point (1, 0).

Step 3: Plot the points and draw the lines
Draw the lines using the points obtained. The lines will intersect at a unique point.

Step 4: Find the intersection point
The intersection point is the solution to the pair of equations. From the graph, the lines intersect at (2, 2).

Verification:
Substitute x = 2 and y = 2 in both equations:
2(2) + 2 = 6 → 6 = 6 (True).
2(2) - 2 = 2 → 2 = 2 (True).

The solution (2, 2) satisfies both equations, confirming it is correct. Graphically, this represents the point where the two lines cross, indicating the common solution to both equations.

Question 18:

Solve the following pair of linear equations graphically and find the coordinates of the point where the lines intersect:
2x + y = 6
2x - y = 2. Also, verify your answer algebraically.

Answer:

To solve the given pair of linear equations graphically and algebraically, follow these steps:

Graphical Method:

Step 1: Find at least two solutions for each equation.
For 2x + y = 6:
When x = 0, y = 6 → (0, 6)
When y = 0, x = 3 → (3, 0)
For 2x - y = 2:
When x = 0, y = -2 → (0, -2)
When y = 0, x = 1 → (1, 0)
Step 2: Plot these points on a graph and draw the lines representing each equation.
Step 3: The point of intersection is the solution. Here, the lines intersect at (2, 2).

Algebraic Verification:

Step 1: Add the two equations to eliminate y:
2x + y = 6
+ 2x - y = 2
----------------
4x = 8 → x = 2
Step 2: Substitute x = 2 into 2x + y = 6:
2(2) + y = 6 → y = 2
Conclusion: The solution (2, 2) satisfies both equations, confirming the graphical result.

Key Concept: The graphical method visually represents the solution, while algebraic methods provide precise verification. Both methods confirm the solution as (2, 2).

Question 19:

Solve the following pair of linear equations graphically and find the coordinates of the point where the two lines intersect:
2x + y = 6
2x - y = 2

Answer:

To solve the given pair of linear equations graphically, we follow these steps:

Step 1: Rewrite the equations in slope-intercept form (y = mx + c)
For the first equation 2x + y = 6:
y = -2x + 6

For the second equation 2x - y = 2:
y = 2x - 2

Step 2: Find two points for each line to plot them on the graph

For y = -2x + 6:
When x = 0, y = 6 → (0, 6)
When x = 1, y = 4 → (1, 4)
For y = 2x - 2:
When x = 0, y = -2 → (0, -2)
When x = 1, y = 0 → (1, 0)

Step 3: Plot the points and draw the lines
Draw the lines passing through the respective points for each equation.

Step 4: Identify the intersection point
The two lines intersect at the point where both equations are satisfied simultaneously.
Solving algebraically:
2x + y = 6
2x - y = 2
Adding both equations:
4x = 8 → x = 2
Substituting x = 2 into the first equation:
2(2) + y = 6 → y = 2

Final Answer: The lines intersect at the point (2, 2).

Value-Added Information: Graphical solutions help visualize the relationship between two variables. The intersection point represents the unique solution to the system of equations, indicating the values of x and y that satisfy both equations simultaneously.

Question 20:

Solve the pair of linear equations 2x + 3y = 8 and 4x + 6y = 16 graphically and interpret the result. What does the solution represent?

Answer:

To solve the given pair of linear equations 2x + 3y = 8 and 4x + 6y = 16 graphically, follow these steps:

Step 1: Rewrite the equations in slope-intercept form (y = mx + c)

For the first equation 2x + 3y = 8:

3y = -2x + 8

y = (-2/3)x + (8/3)

For the second equation 4x + 6y = 16:

6y = -4x + 16

y = (-4/6)x + (16/6)

Simplifying, y = (-2/3)x + (8/3)

Step 2: Compare the equations

Both equations simplify to the same form: y = (-2/3)x + (8/3). This means they represent the same line on the graph.

Step 3: Graphical interpretation

When plotted, both equations will overlap completely, forming a single straight line. This indicates that the system has infinitely many solutions because every point on the line satisfies both equations.

Conclusion:

The pair of equations is dependent and consistent. The solution represents all the points (x, y) lying on the line y = (-2/3)x + (8/3). This shows that the two equations are not independent but are essentially the same equation scaled by a factor.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:

A shopkeeper sells pencils and erasers. The cost of 2 pencils and 3 erasers is ₹20, while 4 pencils and 1 eraser cost ₹25. Problem Interpretation: Identify the variables and equations. Mathematical Modeling: Formulate the equations. Solution: Find the cost of one pencil and one eraser.

Answer:

Problem Interpretation:

Let the cost of one pencil be ₹x and one eraser be ₹y.

Mathematical Modeling:

2x + 3y = 20
4x + y = 25

Solution:

Solving the equations, we get x = 5 and y = 5. Thus, one pencil costs ₹5 and one eraser costs ₹5.

Question 2:

A car travels 300 km in 5 hours with a mix of speeds: 60 km/h on highways and 40 km/h in cities. Problem Interpretation: Define variables for time. Mathematical Modeling: Frame equations for distance and time. Solution: Find time taken on highways and in cities.

Answer:

Problem Interpretation:

Let time on highways be t₁ hours and in cities be t₂ hours.

Mathematical Modeling:

t₁ + t₂ = 5
60t₁ + 40t₂ = 300

Solution:

Solving, we get t₁ = 2.5 hours and t₂ = 2.5 hours. Thus, the car took 2.5 hours each on highways and in cities.

Question 3:

A shopkeeper sells notebooks and pens. The cost of 3 notebooks and 2 pens is ₹110, while 5 notebooks and 4 pens cost ₹200.
Problem Interpretation: Represent this situation algebraically.
Mathematical Modeling: Formulate the equations.
Solution: Find the cost of one notebook and one pen.

Answer:

Problem Interpretation: We studied how to represent real-life problems using linear equations. Here, we need to find the cost of a notebook (x) and a pen (y).
Mathematical Modeling: The equations are:

3x + 2y = 110
5x + 4y = 200

Solution: Solving by elimination, we multiply the first equation by 2: 6x + 4y = 220. Subtracting the second equation, we get x = 20. Substituting x in the first equation, y = 25. Thus, a notebook costs ₹20 and a pen ₹25.

Question 4:

A car travels 60 km in 1 hour and 100 km in 2 hours at constant speed.
Problem Interpretation: Express distance (d) as a linear equation in terms of time (t).
Mathematical Modeling: Derive the equation.
Solution: Predict the distance covered in 4 hours.

Answer:

Problem Interpretation: Our textbook shows that distance-time relationships can be linear. Here, speed is constant.
Mathematical Modeling: The equation is d = mt + c. Using given data:

60 = m(1) + c
100 = m(2) + c

Solving, m = 40 km/h and c = 20 km. Thus, d = 40t + 20.
Solution: For t = 4 hours, d = 40(4) + 20 = 180 km. The car covers 180 km in 4 hours.

Question 5:

A shopkeeper sells pens and pencils. The cost of 2 pens and 3 pencils is ₹50, while 4 pens and 2 pencils cost ₹70.
Problem Interpretation: Represent this situation algebraically and find the cost of one pen and one pencil.

Answer:

Problem Interpretation: We need to form two linear equations representing the costs.
Mathematical Modeling: Let cost of 1 pen = ₹x, 1 pencil = ₹y. Equations: 2x + 3y = 50 and 4x + 2y = 70.
Solution: Solving by elimination:

Multiply first equation by 2: 4x + 6y = 100
Subtract second equation: 4y = 30 → y = 7.5
Substitute y in first equation: x = ₹17.5

Cost of 1 pen = ₹17.5, 1 pencil = ₹7.5.

Question 6:

A car travels 300 km in 5 hours. If it covers equal distances in two parts of the journey, first at speed x km/h and then at y km/h, write the equations and find possible speeds if x + y = 120.

Answer:

Problem Interpretation: The journey is split into two equal parts (150 km each).
Mathematical Modeling: Time = Distance/Speed. Equations: (150/x) + (150/y) = 5 and x + y = 120.
Solution:

From second equation: y = 120 - x
Substitute in first equation: (150/x) + 150/(120-x) = 5
Simplify to quadratic: x² - 120x + 3600 = 0 → (x-60)² = 0

Thus, x = 60 km/h, y = 60 km/h.

Question 7:

A shopkeeper sells notebooks and pens. The cost of 3 notebooks and 2 pens is ₹110, while 2 notebooks and 3 pens cost ₹90. Formulate linear equations and find the cost of one notebook and one pen.

Answer:

Problem Interpretation

We need to find the cost of one notebook (x) and one pen (y) using given conditions.

Mathematical Modeling

3x + 2y = 110
2x + 3y = 90

Solution

Solving by elimination, we multiply the first equation by 2 and the second by 3. Subtracting them gives y = ₹10. Substituting y in the first equation gives x = ₹30. Thus, one notebook costs ₹30 and one pen costs ₹10.

Question 8:

A train covers a distance of 300 km at a uniform speed. If the speed were 5 km/h more, it would take 1 hour less. Represent this situation algebraically and find the original speed.

Answer:

Problem Interpretation

We need to find the original speed (x) of the train using time-distance relations.

Mathematical Modeling

Time taken at speed x: 300/x hours
Time taken at speed (x+5): 300/(x+5) hours
Difference: (300/x) - (300/(x+5)) = 1

Solution

Simplifying, we get x² + 5x - 1500 = 0. Solving this quadratic equation gives x = 40 km/h (discarding negative value). Thus, the original speed is 40 km/h.

Question 9:

A shopkeeper sells pencils and erasers. The cost of 2 pencils and 3 erasers is ₹18, while 4 pencils and 6 erasers cost ₹36. Represent this situation algebraically and check if the equations are consistent.

Answer:

Problem Interpretation

We need to form equations for the given situation and check consistency.

Mathematical Modeling

Let cost of 1 pencil = ₹x
Let cost of 1 eraser = ₹y

Solution

First equation: 2x + 3y = 18
Second equation: 4x + 6y = 36
Dividing second equation by 2 gives 2x + 3y = 18, which matches the first equation. Hence, the equations are dependent and consistent.

Question 10:

A boat covers 32 km upstream and 36 km downstream in 7 hours. It also covers 40 km upstream and 48 km downstream in 9 hours. Find the speed of the boat in still water and the speed of the stream.

Answer:

Problem Interpretation

We studied upstream/downstream problems in our textbook. Here, we need to find two speeds.

Mathematical Modeling

Let boat speed in still water = x km/h
Let stream speed = y km/h

Solution

Upstream speed = (x - y) km/h
Downstream speed = (x + y) km/h
Form equations: 32/(x-y) + 36/(x+y) = 7 and 40/(x-y) + 48/(x+y) = 9. Solving gives x = 10 km/h, y = 2 km/h.

Question 11:

A shopkeeper sells apples and oranges. The cost of 3 apples and 2 oranges is ₹120, while the cost of 2 apples and 3 oranges is ₹130. Represent this situation algebraically and find the cost of one apple and one orange.

Answer:

Let the cost of one apple be ₹x and one orange be ₹y.

According to the problem:
1. 3x + 2y = 120 (Equation 1)
2. 2x + 3y = 130 (Equation 2)

To solve the equations:
Step 1: Multiply Equation 1 by 2 → 6x + 4y = 240
Step 2: Multiply Equation 2 by 3 → 6x + 9y = 390
Step 3: Subtract the modified Equation 1 from modified Equation 2 → (6x + 9y) - (6x + 4y) = 390 - 240
→ 5y = 150
→ y = 30
Step 4: Substitute y = 30 in Equation 1 → 3x + 2(30) = 120
→ 3x + 60 = 120
→ 3x = 60
→ x = 20

Thus, the cost of one apple is ₹20 and one orange is ₹30.

Question 12:

A train covers a certain distance at a uniform speed. If the train had been 10 km/h faster, it would have taken 2 hours less, and if it had been 10 km/h slower, it would have taken 3 hours more. Find the original speed and distance of the train.

Answer:

Let the original speed of the train be x km/h and the distance be y km.

Time taken at original speed: y/x hours.

Case 1: If speed increases by 10 km/h → New speed = (x + 10) km/h
Time taken = y/(x + 10) hours
According to the problem: y/x - y/(x + 10) = 2 (Equation 1)

Case 2: If speed decreases by 10 km/h → New speed = (x - 10) km/h
Time taken = y/(x - 10) hours
According to the problem: y/(x - 10) - y/x = 3 (Equation 2)

Simplifying Equation 1:
(y(x + 10) - yx) / (x(x + 10)) = 2
→ 10y = 2x(x + 10)
→ 5y = x² + 10x (Equation A)

Simplifying Equation 2:
(yx - y(x - 10)) / (x(x - 10)) = 3
→ 10y = 3x(x - 10)
→ 10y = 3x² - 30x (Equation B)

From Equation A and B:
5y = x² + 10x → Multiply by 2 → 10y = 2x² + 20x
But 10y = 3x² - 30x (from Equation B)
Thus, 2x² + 20x = 3x² - 30x
→ x² - 50x = 0
→ x(x - 50) = 0
→ x = 50 (since speed cannot be 0)

Substitute x = 50 in Equation A:
5y = (50)² + 10(50)
→ 5y = 2500 + 500
→ y = 600

Thus, the original speed is 50 km/h and the distance is 600 km.

Question 13:

A shopkeeper sells apples and oranges. The cost of 2 apples and 3 oranges is ₹80, while the cost of 4 apples and 1 orange is ₹90. Represent this situation algebraically and find the cost of one apple and one orange.

Answer:

Let the cost of one apple be ₹x and one orange be ₹y.

According to the problem:
1. Equation 1: 2x + 3y = 80
2. Equation 2: 4x + y = 90

Step 1: Solve Equation 2 for y.
y = 90 - 4x

Step 2: Substitute y in Equation 1.
2x + 3(90 - 4x) = 80
2x + 270 - 12x = 80
-10x = -190
x = 19

Step 3: Substitute x back into the expression for y.
y = 90 - 4(19) = 90 - 76 = 14

Final Answer:
Cost of one apple = ₹19
Cost of one orange = ₹14

Question 14:

A train covers a certain distance at a uniform speed. If the train had been 10 km/h faster, it would have taken 2 hours less for the same journey. If it had been 10 km/h slower, it would have taken 3 hours more. Find the original speed and distance of the train.

Answer:

Let the original speed of the train be x km/h and the distance be d km.

According to the problem:
1. Case 1 (Faster): d/(x + 10) = (d/x) - 2
2. Case 2 (Slower): d/(x - 10) = (d/x) + 3

Step 1: Simplify Case 1.
d/(x + 10) = (d - 2x)/x
Cross-multiply: dx = (d - 2x)(x + 10)
dx = dx + 10d - 2x² - 20x
0 = 10d - 2x² - 20x
5d = x² + 10x → Equation A

Step 2: Simplify Case 2.
d/(x - 10) = (d + 3x)/x
Cross-multiply: dx = (d + 3x)(x - 10)
dx = dx - 10d + 3x² - 30x
0 = -10d + 3x² - 30x
10d = 3x² - 30x → Equation B

Step 3: Solve Equations A and B.
From Equation A: d = (x² + 10x)/5
Substitute in Equation B:
10[(x² + 10x)/5] = 3x² - 30x
2x² + 20x = 3x² - 30x
0 = x² - 50x
x(x - 50) = 0
x = 0 or x = 50

Since speed cannot be 0, x = 50 km/h.

Step 4: Find d using Equation A.
d = (50² + 10×50)/5 = (2500 + 500)/5 = 600 km

Final Answer:
Original speed = 50 km/h
Distance = 600 km

Question 15:

A shopkeeper sells apples and oranges. The cost of 2 apples and 3 oranges is ₹100, while the cost of 4 apples and 1 orange is ₹120. Represent this situation algebraically and find the cost of one apple and one orange.

Answer:

Let the cost of one apple be ₹x and one orange be ₹y.
The given situation can be represented by the following linear equations in two variables:

1. 2x + 3y = 100 (Equation 1)
2. 4x + y = 120 (Equation 2)

To solve these equations, we can use the substitution method:

From Equation 2: y = 120 - 4x
Substitute y in Equation 1:
2x + 3(120 - 4x) = 100
2x + 360 - 12x = 100
-10x = -260
x = 26

Now, substitute x = 26 in Equation 2:
4(26) + y = 120
104 + y = 120
y = 16

Thus, the cost of one apple is ₹26 and one orange is ₹16.

Question 16:

Rahul and Riya are saving money. Rahul has ₹50 more than twice the amount Riya has. Together, they have ₹500. Formulate the equations and find how much each has saved.

Answer:

Let the amount saved by Riya be ₹x and by Rahul be ₹y.
The given conditions can be expressed as:

1. y = 2x + 50 (Rahul has ₹50 more than twice Riya's amount)
2. x + y = 500 (Total savings together)

Substitute y from Equation 1 into Equation 2:
x + (2x + 50) = 500
3x + 50 = 500
3x = 450
x = 150

Now, substitute x = 150 in Equation 1:
y = 2(150) + 50
y = 300 + 50
y = 350

Thus, Riya has saved ₹150 and Rahul has saved ₹350.

Question 17:

Answer:

Let the cost of one apple be ₹x and one orange be ₹y.

According to the problem:
1. 3x + 2y = 120 (Equation 1)
2. 2x + 3y = 130 (Equation 2)

To solve these linear equations in two variables, we can use the elimination method.

Step 1: Multiply Equation 1 by 2 and Equation 2 by 3 to make coefficients of x equal.
→ 6x + 4y = 240 (Equation 3)
→ 6x + 9y = 390 (Equation 4)

Step 2: Subtract Equation 3 from Equation 4 to eliminate x.
→ (6x + 9y) - (6x + 4y) = 390 - 240
→ 5y = 150
→ y = 30

Step 3: Substitute y = 30 in Equation 1 to find x.
→ 3x + 2(30) = 120
→ 3x + 60 = 120
→ 3x = 60
→ x = 20

Thus, the cost of one apple is ₹20 and one orange is ₹30.

Question 18:

A train travels 300 km at a uniform speed. If the speed had been 10 km/h more, it would have taken 1 hour less for the same journey. Represent this situation algebraically and find the original speed of the train.

Answer:

Let the original speed of the train be x km/h.

According to the problem:
1. Time taken at original speed = 300 / x hours.
2. Time taken at increased speed = 300 / (x + 10) hours.

The difference in time is 1 hour, so:
→ (300 / x) - (300 / (x + 10)) = 1

Simplify the equation:
→ 300(x + 10) - 300x = x(x + 10)
→ 300x + 3000 - 300x = x² + 10x
→ 3000 = x² + 10x
→ x² + 10x - 3000 = 0

Solve the quadratic equation using the quadratic formula:
→ x = [-10 ± √(100 + 12000)] / 2
→ x = [-10 ± √12100] / 2
→ x = [-10 ± 110] / 2

Ignore the negative value since speed cannot be negative:
→ x = (-10 + 110) / 2 = 100 / 2 = 50

Thus, the original speed of the train is 50 km/h.

Question 19:

Answer:

Let the original speed of the train be x km/h and the distance be d km.

According to the problem:
1. Time taken at original speed: t = d/x
2. If speed increases by 10 km/h: t - 2 = d/(x + 10)
3. If speed decreases by 10 km/h: t + 3 = d/(x - 10)

Step 1: Rewrite the equations using t = d/x.
d/x - 2 = d/(x + 10) (Equation 1)
d/x + 3 = d/(x - 10) (Equation 2)

Step 2: Simplify Equation 1.
d(x + 10) - 2x(x + 10) = dx
dx + 10d - 2x² - 20x = dx
10d - 2x² - 20x = 0
5d - x² - 10x = 0 (Equation 3)

Step 3: Simplify Equation 2 similarly.
d(x - 10) + 3x(x - 10) = dx
dx - 10d + 3x² - 30x = dx
-10d + 3x² - 30x = 0
10d = 3x² - 30x (Equation 4)

Step 4: Substitute 10d from Equation 4 into Equation 3.
(3x² - 30x)/2 - x² - 10x = 0
1.5x² - 15x - x² - 10x = 0
0.5x² - 25x = 0
x(0.5x - 25) = 0
x = 0 (invalid) or x = 50

Step 5: Substitute x = 50 into Equation 4 to find d.
10d = 3(50)² - 30(50)
10d = 7500 - 1500
d = 600

Thus, the original speed is 50 km/h and the distance is 600 km.

Linear Equations in Two Variables – CBSE NCERT Study Resources

Chapter Overview: Linear Equations in Two Variables

Key Concepts

Graphical Representation

Methods to Solve a Pair of Linear Equations

Applications

All Question Types with Solutions – CBSE Exam Pattern

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)