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Pair of Linear Equations in Two Variables – CBSE NCERT Study Resources

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10th

10th - Mathematics

Pair of Linear Equations in Two Variables

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Overview of the Chapter

This chapter introduces the concept of a pair of linear equations in two variables, their graphical representation, and methods to solve them. Students will learn algebraic techniques like substitution, elimination, and cross-multiplication to find solutions. The chapter also covers real-life applications of these equations.

Key Concepts

Linear Equation in Two Variables: 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.

Solution of a Pair of Linear Equations: A pair of values (x, y) that satisfies both equations simultaneously.

Graphical Representation

A pair of linear equations can be represented graphically as two straight lines. The solution corresponds to the point(s) of intersection of these lines.

  • If the lines intersect at a single point, the system has a unique solution (consistent and independent).
  • If the lines are parallel, the system has no solution (inconsistent).
  • If the lines coincide, the system has infinitely many solutions (consistent and dependent).

Algebraic Methods to Solve Linear Equations

Substitution Method

In this method, one variable is expressed in terms of the other from one equation and substituted into the second equation.

Elimination Method

In this method, one variable is eliminated by adding or subtracting the equations after making their coefficients equal.

Cross-Multiplication Method

This method uses the formula derived from the elimination method to directly find the solution of the pair of equations.

Applications in Real Life

Pair of linear equations are used to solve problems involving two unknown quantities, such as:

  • Finding the cost of two different items.
  • Determining the speed of two vehicles.
  • Calculating ages or other quantities in word problems.

Important Formulas

Cross-Multiplication Formula: For equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the solution is given by:

x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)

Summary

This chapter equips students with the knowledge to solve pairs of linear equations using graphical and algebraic methods. Understanding these concepts is crucial for solving real-world problems and forms the foundation for higher-level mathematics.

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, where a, b ≠ 0

Question 2:
Find the solution of the pair of equations: x + y = 5, x - y = 1.
Answer:

Numeric answer:
(3, 2)

Question 3:
How many solutions does a pair of consistent equations have?
Answer:

Infinitely many solutions

Question 4:
If the lines represented by 2x + 3y = 6 and 4x + 6y = 12 are coincident, what is the relationship between them?
Answer:

They are the same line

Question 5:
What is the graphical representation of a pair of linear equations with no solution?
Answer:

Parallel lines

Question 6:
Solve for x and y: 3x + 2y = 8, 6x + 4y = 16.
Answer:

Infinitely many solutions

Question 7:
What is the condition for a pair of linear equations to be inconsistent?
Answer:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Question 8:
Find the value of k for which the equations 2x + 3y = 7 and kx + 9y = 12 have no solution.
Answer:

Numeric answer:
k = 6

Question 9:
What does the algebraic method of substitution involve?
Answer:

Expressing one variable in terms of another

Question 10:
If x = 2, y = 3 is a solution of 5x + ky = 19, find k.
Answer:

Numeric answer:
k = 3

Question 11:
What is the geometric interpretation of a unique solution in two variables?
Answer:

Intersecting lines

Question 12:
Check if the pair of equations 3x - 5y = 20 and 6x - 10y = 40 are consistent.
Answer:

Yes, consistent

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 pair of consistent and dependent linear equations have?
Answer:

A pair of consistent and dependent linear equations has infinitely many solutions because both equations represent the same line.

Question 15:
If the lines represented by the equations 2x + 3y = 5 and 4x + 6y = 10 are plotted, what can you say about them?
Answer:

The lines are coincident because the second equation is a multiple of the first.
They represent the same line and have infinitely many solutions.

Question 16:
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 17:
Solve the pair of equations x + y = 5 and x - y = 1 using the substitution method.
Answer:

From x + y = 5, we get y = 5 - x.
Substitute in x - y = 1:
x - (5 - x) = 1
2x - 5 = 1
2x = 6
x = 3
Now, y = 5 - 3 = 2.
Solution: (3, 2).

Question 18:
What does it mean if the pair of equations 3x + 2y = 7 and 6x + 4y = 14 are dependent?
Answer:

If the pair is dependent, it means the second equation is a multiple of the first (here, multiplied by 2).
Both equations represent the same line, so they have infinitely many solutions.

Question 19:
Find the value of k for which the system 2x + ky = 8 and 4x + 6y = 12 has no solution.
Answer:

For no solution, the condition is a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
Here, 2/4 = k/6 ≠ 8/12.
Simplifying: 1/2 = k/6 → k = 3.
Check: 8/12 = 2/3 ≠ 1/2.
Thus, k = 3.

Question 20:
What is the graphical representation of a pair of inconsistent linear equations?
Answer:

The graphical representation of a pair of inconsistent linear equations is two parallel lines that never intersect, indicating no solution.

Question 21:
Solve the pair of equations 5x + 2y = 16 and 3x + 4y = 10 using the elimination method.
Answer:

Multiply the first equation by 2: 10x + 4y = 32.
Subtract the second equation:
(10x + 4y) - (3x + 4y) = 32 - 10
7x = 22 → x = 22/7.
Substitute in 5x + 2y = 16:
5(22/7) + 2y = 16
110/7 + 2y = 16
2y = 16 - 110/7
2y = (112 - 110)/7
y = 1/7.
Solution: (22/7, 1/7).

Question 22:
If the cost of 3 pens and 2 pencils is ₹50, and the cost of 2 pens and 3 pencils is ₹40, represent this situation algebraically.
Answer:

Let the cost of one pen be ₹x and one pencil be ₹y.
The algebraic representation is:
3x + 2y = 50
2x + 3y = 40.

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:
What is the general form of a pair of linear equations in two variables?
Answer:

The general form of a pair of linear equations in two variables x and y is:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
Here, a₁, b₁, c₁ and a₂, b₂, c₂ are real numbers, and at least one of a₁, b₁ or a₂, b₂ is non-zero.

Question 2:
How can you determine if a given pair of linear equations has a unique solution?
Answer:

A pair of linear equations has a unique solution if the ratios of coefficients are unequal:
a₁/a₂ ≠ b₁/b₂
This means the lines represented by the equations intersect at exactly one point.

Question 3:
What does it mean if the pair of linear equations has infinitely many solutions?
Answer:

If a pair of linear equations has infinitely many solutions, it means:
a₁/a₂ = b₁/b₂ = c₁/c₂
This implies both equations represent the same line, so every point on the line is a solution.

Question 4:
Explain the graphical representation of a pair of linear equations with no solution.
Answer:

When a pair of linear equations has no solution, the lines are parallel and never intersect.
This occurs when:
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Graphically, the lines have the same slope but different y-intercepts.

Question 5:
Solve the pair of equations: 2x + 3y = 8 and x - 2y = -3.
Answer:

Using the substitution method:
From the second equation, x = 2y - 3.
Substitute into the first equation:
2(2y - 3) + 3y = 8
4y - 6 + 3y = 8
7y = 14 → y = 2
Then, x = 2(2) - 3 = 1
Solution: x = 1, y = 2

Question 6:
What is the condition for a pair of linear equations to be consistent?
Answer:

A pair of linear equations is consistent if it has at least one solution.
This happens in two cases:
1. Unique solution: a₁/a₂ ≠ b₁/b₂
2. Infinitely many solutions: a₁/a₂ = b₁/b₂ = c₁/c₂

Question 7:
Find the value of k for which the system 3x + y = 1 and (2k - 1)x + (k - 1)y = 2k + 1 has no solution.
Answer:

For no solution, the condition is:
3/(2k - 1) = 1/(k - 1) ≠ 1/(2k + 1)
Solving the first equality:
3(k - 1) = 2k - 1
3k - 3 = 2k - 1 → k = 2
Verify with the inequality:
3/3 = 1/1 ≠ 1/5 → True
Thus, k = 2 satisfies the condition.

Question 8:
How many solutions does the pair of equations y = 0 and y = -5 have? Explain.
Answer:

The pair y = 0 and y = -5 has no solution.
Both equations represent horizontal lines:
y = 0 (x-axis) and y = -5 (parallel line below x-axis).
Since they are parallel and distinct, they never intersect.

Question 9:
If the lines represented by 3x - y = 5 and 6x - 2y = p coincide, find the value of p.
Answer:

For coincident lines, the condition is:
3/6 = -1/-2 = 5/p
Simplifying gives 1/2 = 1/2 = 5/p
Thus, 1/2 = 5/p → p = 10
The value of p is 10.

Question 10:
What is the graphical method of solving a pair of linear equations?
Answer:

The graphical method involves:
1. Plotting both equations on the same coordinate plane.
2. Identifying the point of intersection, which represents the solution.
Possible outcomes:
- One intersection point (unique solution)
- Overlapping lines (infinitely many solutions)
- Parallel lines (no 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:
Solve the pair of linear equations:
2x + 3y = 8
4x + 6y = 16. What can you conclude about the system?
Answer:

The given equations are:
2x + 3y = 8
4x + 6y = 16.
Divide the second equation by 2:
2x + 3y = 8.
Both equations are identical, meaning they represent the same line.
Thus, the system has infinitely many solutions and is dependent.

Question 2:
Find the value of k for which the pair of equations
kx + 2y = 5
3x + y = 1
has a unique solution.
Answer:

For a unique solution, the condition is:
a₁/a₂ ≠ b₁/b₂.
Here, a₁ = k, b₁ = 2, a₂ = 3, b₂ = 1.
So, k/3 ≠ 2/1 ⇒ k ≠ 6.
Thus, the system has a unique solution for all real values of k except k = 6.

Question 3:
Explain the graphical representation of a pair of linear equations in two variables when the lines are parallel.
Answer:

When two linear equations are represented graphically as parallel lines, they have:

  • No common point (no solution).
  • The slopes (a₁/a₂ = b₁/b₂) are equal, but the intercepts (c₁/c₂) are unequal.

For example, the equations
2x + 3y = 5
4x + 6y = 10
are identical (dependent), but if the right-hand side differs (e.g., 4x + 6y = 12), the lines are parallel and inconsistent.

Question 4:
Solve the pair of equations by substitution method:
x + y = 5
2x - 3y = 4.
Answer:

Given equations:
x + y = 5 → (1)
2x - 3y = 4 → (2).
From (1), express x in terms of y:
x = 5 - y.
Substitute into (2):
2(5 - y) - 3y = 4
10 - 2y - 3y = 4
-5y = -6 ⇒ y = 6/5.
Now, x = 5 - (6/5) = 19/5.
The solution is x = 19/5 and y = 6/5.

Question 5:
Solve the pair of linear equations:
2x + 3y = 8
4x + 6y = 16. What can you conclude about their solutions?
Answer:

The given pair of equations is:
2x + 3y = 8
4x + 6y = 16.

First, simplify the second equation by dividing all terms by 2:
2x + 3y = 8.

Now, both equations are identical:
2x + 3y = 8.

This means the equations represent the same line and have infinitely many solutions. The system is dependent.

Question 6:
Find the value of k for which the pair of equations
3x + y = 1
(2k - 1)x + (k - 1)y = 2k + 1 has no solution.
Answer:

For the system to have no solution, the lines must be parallel (i.e., their coefficients must satisfy a₁/a₂ = b₁/b₂ ≠ c₁/c₂).

Given equations:
3x + y = 1
(2k - 1)x + (k - 1)y = 2k + 1.

Condition 1: 3/(2k - 1) = 1/(k - 1)
Cross-multiplying:
3(k - 1) = 2k - 1
3k - 3 = 2k - 1
k = 2.

Condition 2: 1/(2k + 1) ≠ 3/(2k - 1)
Substitute k = 2:
1/5 ≠ 3/3 → 1/5 ≠ 1 (True).

Thus, k = 2 makes the system inconsistent.

Question 7:
Graphically, check whether the pair of equations
x + 3y = 6
2x - 3y = 12 is consistent. If so, find the solution.
Answer:

To check consistency, plot the lines:

Line 1: x + 3y = 6
When x = 0, y = 2 → (0, 2)
When y = 0, x = 6 → (6, 0).

Line 2: 2x - 3y = 12
When x = 0, y = -4 → (0, -4)
When y = 0, x = 6 → (6, 0).

The lines intersect at (6, 0), which is the unique solution. Thus, the system is consistent.

Solution: x = 6, y = 0.

Question 8:
The sum of the digits of a two-digit number is 9. If 27 is added to the number, the digits interchange their places. Find the number.
Answer:

Let the tens digit = x and units digit = y.

Given:
x + y = 9 (Sum of digits)
Original number = 10x + y
Reversed number = 10y + x.

According to the problem:
10x + y + 27 = 10y + x
Simplify: 9x - 9y = -27
Divide by 9: x - y = -3.

Now, solve the system:
x + y = 9
x - y = -3
Add: 2x = 6 → x = 3
Substitute: 3 + y = 9 → y = 6.

Thus, the number is 36.

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: 2x + 3y = 8 and 4x + 6y = 16 graphically. What does this represent?
Answer:
Introduction

We studied that a pair of linear equations can have unique, infinite, or no solutions. Here, we analyze the given equations.


Argument 1
  • Divide the second equation by 2: 2x + 3y = 8, which matches the first equation.
  • Both equations represent the same line, so they have infinitely many solutions.

Argument 2

Graphically, the lines coincide, showing all points on the line satisfy both equations.


Conclusion

This is a case of dependent equations, as our textbook shows in Example 3.4.

Question 2:
A boat covers 32 km upstream and 36 km downstream in 7 hours. If the speed of the stream is 2 km/h, find the boat's speed in still water.
Answer:
Introduction

We use linear equations to model real-life problems like this, where upstream/downstream speeds depend on stream speed.


Argument 1
  • Let boat speed in still water = x km/h.
  • Upstream speed = (x - 2) km/h, downstream = (x + 2) km/h.

Argument 2
  • Time equation: 32/(x - 2) + 36/(x + 2) = 7.
  • Solving (like NCERT Example 3.6), we get x = 10 km/h.

Conclusion

The boat's speed in still water is 10 km/h, verified by substituting back into the equation.

Question 3:
Solve the pair of linear equations 2x + 3y = 8 and 4x + 6y = 16 graphically. What does this represent?
Answer:
Introduction

We studied that a pair of linear equations represents two lines on a graph. Here, the equations are 2x + 3y = 8 and 4x + 6y = 16.


Argument 1
  • Divide the second equation by 2: 2x + 3y = 8, which matches the first equation.
  • Both equations represent the same line, meaning infinite solutions.

Argument 2

Graphically, the lines coincide, showing all points on the line satisfy both equations. Our textbook shows such cases as dependent equations.


Conclusion

This represents infinitely many solutions, as both equations describe the same line.

Question 4:
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 and the stream.
Answer:
Introduction

We studied upstream/downstream problems where boat speed (x) and stream speed (y) form linear equations.


Argument 1
  • Let boat speed = x km/h, stream speed = y km/h.
  • Upstream speed = x - y, downstream = x + y.
  • From given data: 30/(x-y) + 44/(x+y) = 10 and 40/(x-y) + 55/(x+y) = 13.

Argument 2

Assume 1/(x-y) = u and 1/(x+y) = v. Solve the simplified equations to get x = 8 km/h, y = 3 km/h.


Conclusion

The boat speed is 8 km/h, and the stream speed is 3 km/h, as verified by substitution.

Question 5:
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 and stream using substitution method.
Answer:
Introduction

We studied upstream/downstream problems where boat speed (x) and stream speed (y) form linear equations.


Argument 1
  • Let upstream speed = x - y, downstream = x + y.
  • From given data: 30/(x-y) + 44/(x+y) = 10 and 40/(x-y) + 55/(x+y) = 13.

Argument 2

Assume 1/(x-y) = u, 1/(x+y) = v. Solve the simplified equations 30u + 44v = 10 and 40u + 55v = 13 to find x = 8 km/h, y = 3 km/h.


Conclusion

The boat speed is 8 km/h, and stream speed is 3 km/h.

Question 6:
Explain how the graphical method helps solve a pair of linear equations. Use the example x + y = 5 and 2x - y = 4.
Answer:
Introduction

Graphical method visually represents solutions by plotting lines. We studied this in NCERT for pairs like x + y = 5 and 2x - y = 4.


Argument 1
  • Find intercepts: For x + y = 5, points (5,0) and (0,5). For 2x - y = 4, points (2,0) and (0,-4).
  • Plot both lines on graph paper.

Argument 2

The intersection point (3,2) is the solution, as it satisfies both equations. Our textbook confirms this as the unique solution.


Conclusion

Graphical method helps identify solutions as intersection points, making it easy to understand.

Question 7:
Solve the pair of linear equations 2x + 3y = 8 and 4x + 6y = 7 graphically. What does this represent?
Answer:
Introduction

We studied that a pair of linear equations represents two straight lines. Here, we analyze their graphical solution.


Argument 1
  • Equation 1: 2x + 3y = 8 → Slope = -2/3, y-intercept = 8/3.
  • Equation 2: 4x + 6y = 7 → Simplified to 2x + 3y = 3.5, same slope but different intercept.

Argument 2

Both lines are parallel (same slope) but never meet. Thus, no solution exists, making them inconsistent.


Conclusion

This represents an inconsistent pair with no common solution, as shown in NCERT Example 4.

Question 8:
Solve the following pair of linear equations graphically and find the point of intersection:
2x + y = 6
2x - y = 2. Also, verify the solution algebraically.
Answer:
Introduction

We studied that a pair of linear equations can be solved graphically by plotting the lines and finding their intersection.


Argument 1
  • Graphical Solution: Plotting 2x + y = 6 (y = -2x + 6) and 2x - y = 2 (y = 2x - 2), the lines intersect at (2, 2).

Argument 2
  • Algebraic Verification: Adding the equations eliminates y, giving 4x = 8 → x = 2. Substituting x = 2 into the first equation gives y = 2.

Conclusion

The solution (2, 2) satisfies both equations, confirming consistency.

Question 9:
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

Our textbook shows how to model real-life problems using linear equations. Here, we relate time, distance, and speed.


Argument 1
  • Let boat speed = x km/h, stream speed = y km/h. Upstream speed = (x - y), downstream = (x + y).
  • Equations: 30/(x - y) + 44/(x + y) = 10 and 40/(x - y) + 55/(x + y) = 13.

Argument 2
  • Assume 1/(x - y) = a, 1/(x + y) = b. Solve 30a + 44b = 10 and 40a + 55b = 13 to get a = 1/5, b = 1/11.
  • Thus, x - y = 5, x + y = 11 → x = 8 km/h, y = 3 km/h.

Conclusion

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

Question 10:
Solve the pair of linear equations 2x + 3y = 8 and 4x + 6y = 16 graphically. What does this represent? Explain with a real-life example.
Answer:
Introduction

We studied graphical solutions for pairs of linear equations. Here, both equations represent the same line.


Argument 1
  • Divide the second equation by 2: 2x + 3y = 8, identical to the first.
  • Graphically, they overlap completely, showing infinitely many solutions.

Argument 2

Our textbook shows this represents dependent equations. A real-life example is buying pens (x) and pencils (y) where total cost remains ₹8 for any combination satisfying 2x + 3y = 8.


Conclusion

This system is consistent and dependent, applicable in budget constraints.

Question 11:
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 boat's speed in still water and the stream's speed using substitution method.
Answer:
Introduction

We use the substitution method to solve speed-based problems. Let boat speed = x km/h and stream speed = y km/h.


Argument 1
  • Upstream speed = x - y, downstream = x + y.
  • Form equations: 30/(x-y) + 44/(x+y) = 10 and 40/(x-y) + 55/(x+y) = 13.

Argument 2

Let 1/(x-y) = a and 1/(x+y) = b. Solve to get x = 8 km/h, y = 3 km/h (NCERT-style).


Conclusion

Boat speed = 8 km/h, stream speed = 3 km/h, verified by substitution.

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 and the stream.
Answer:
Introduction

We use Pair of Linear Equations to solve real-life problems like this, as in NCERT Example 3.6.


Argument 1
  • Let boat speed = x km/h, stream speed = y 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.
  • Solving gives x = 8 km/h, y = 3 km/h.

Conclusion

The boat speed is 8 km/h, and the stream speed is 3 km/h.

Question 13:
Solve the following 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 graphical methods to solve pairs of linear equations. Here, we plot two equations to find their solution.


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

Argument 2

Plotting these points, the lines intersect at (2,2). The y-intercepts are (0,6) and (0,-2).


Conclusion

The solution is (2,2), and the y-intercepts are confirmed.

Question 14:
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 to model real-life problems using linear equations. Here, we relate boat speeds to upstream/downstream distances.


Argument 1
  • Let boat speed = x km/h, stream speed = y km/h.
  • Upstream speed = x - y, downstream = x + y.

Argument 2
  • Equations: 30/(x-y) + 44/(x+y) = 10 and 40/(x-y) + 55/(x+y) = 13.
  • Solving, we get x = 8 km/h, y = 3 km/h.

Conclusion

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

Question 15:
Solve the following pair of linear equations graphically and find the coordinates of the points where the lines meet the y-axis:
2x + 3y = 12 and x - y = 1.
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 2x + 3y = 12:
3y = -2x + 12
y = (-2/3)x + 4

For x - y = 1:
-y = -x + 1
y = x - 1

Step 2: Plot the lines on a graph
For y = (-2/3)x + 4:
When x = 0, y = 4 → Point (0, 4)
When y = 0, x = 6 → Point (6, 0)

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

Step 3: Identify the intersection point
The lines intersect at (3, 2), which is the solution.

Step 4: Find where the lines meet the y-axis
The first line meets the y-axis at (0, 4).
The second line meets the y-axis at (0, -1).

Thus, the solution is (3, 2), and the y-intercepts are (0, 4) and (0, -1).

Question 16:
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 first condition:
(30 / (x - y)) + (44 / (x + y)) = 10

From the second condition:
(40 / (x - y)) + (55 / (x + y)) = 13

Step 2: Simplify the equations.
Let 1/(x - y) = u and 1/(x + y) = v.
The equations become:
30u + 44v = 10
40u + 55v = 13

Step 3: Solve the simplified equations.
Multiply the first equation by 4 and the second by 3:
120u + 176v = 40
120u + 165v = 39

Subtract the second from the first:
11v = 1 → v = 1/11
Substitute v into the first equation:
30u + 44(1/11) = 10 → 30u + 4 = 10 → u = 1/5

Step 4: Find x and y.
u = 1/(x - y) = 1/5 → x - y = 5
v = 1/(x + y) = 1/11 → x + y = 11

Add the two equations:
2x = 16 → x = 8 km/h
Subtract the second from the first:
-2y = -6 → y = 3 km/h

Final Answer:
Speed of the boat in still water = 8 km/h
Speed of the stream = 3 km/h
Question 17:
The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the digits. Find the number.
Answer:

Let the two-digit number be represented as 10a + b, where a is the tens digit and b is the units digit.


Step 1: Formulate the equations
From the first condition (sum of digits):
a + b = 9

From the second condition (nine times the number equals twice the reversed number):
9(10a + b) = 2(10b + a)
90a + 9b = 20b + 2a
88a - 11b = 0
8a - b = 0 → b = 8a

Step 2: Solve the system
Substitute b = 8a into the first equation:
a + 8a = 9
9a = 9 → a = 1

Then, b = 8(1) = 8

Step 3: Construct the number
The number is 10a + b = 10(1) + 8 = 18.

Thus, the required two-digit number is 18.

Question 18:
Solve the following pair of linear equations graphically and find the coordinates of the points where the lines meet the y-axis:
2x + 3y = 12
x - y = 1
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)
For the equation 2x + 3y = 12:
3y = -2x + 12
y = (-2/3)x + 4

For the equation x - y = 1:
-y = -x + 1
y = x - 1

Step 2: Plot the lines on a graph
For y = (-2/3)x + 4:
- When x = 0, y = 4 → (0, 4)
- When y = 0, x = 6 → (6, 0)

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

Step 3: Identify the point of intersection
The two lines intersect at (3, 2), which is the solution to the pair of equations.

Step 4: Find where the lines meet the y-axis
- The line 2x + 3y = 12 meets the y-axis at (0, 4).
- The line x - y = 1 meets the y-axis at (0, -1).

Thus, the solution is (3, 2), and the y-intercepts are (0, 4) and (0, -1).

Question 19:
Solve the following pair of linear equations graphically and interpret the result:
2x + y = 6
2x - y = 2. Also, find the area of the triangle formed by these lines and the x-axis.
Answer:

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


Step 1: Find the intercepts for both equations.
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 the lines on a graph.
Draw the lines using the intercepts. The lines intersect at (2, 2), which is the solution.

Step 3: Interpret the result.
The lines intersect at a unique point, so the equations are consistent with a unique solution.

Step 4: Find the area of the triangle.
The triangle is formed by the points (3, 0), (1, 0), and (2, 2).
Base = 3 - 1 = 2 units
Height = 2 units
Area = (1/2) × base × height = (1/2) × 2 × 2 = 2 square units.
Question 20:
Solve the following pair of linear equations graphically and find the coordinates of the points where the lines meet the y-axis:
2x + y = 6
2x - y + 2 = 0.
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 = 0:
    y = 2x + 2

Step 2: Find two points for each line to plot 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 = 4 → (1, 4)

Step 3: Plot the points and draw the lines

On plotting the points and drawing the lines, we observe that the two lines intersect at the point (1, 4), which is the solution to the pair of equations.


Step 4: Find where the lines meet the y-axis
  • The line y = -2x + 6 meets the y-axis at (0, 6) (since x = 0).
  • The line y = 2x + 2 meets the y-axis at (0, 2) (since x = 0).

Conclusion: The solution to the pair of equations is (1, 4). The lines meet the y-axis at (0, 6) and (0, 2), respectively.

Question 21:
Solve the following pair of linear equations graphically and find the coordinates of the points where the lines meet the y-axis:
2x + 3y = 12
x - y = 1
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)
For 2x + 3y = 12:
3y = -2x + 12
y = (-2/3)x + 4

For x - y = 1:
-y = -x + 1
y = x - 1

Step 2: Plot the lines on a graph
For y = (-2/3)x + 4:
- When x = 0, y = 4 → (0, 4)
- When y = 0, x = 6 → (6, 0)

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

Step 3: Identify the point of intersection
The two lines intersect at (3, 2), which is the solution to the pair of equations.

Step 4: Find where the lines meet the y-axis
- The first line (2x + 3y = 12) meets the y-axis at (0, 4).
- The second line (x - y = 1) meets the y-axis at (0, -1).

The solution is verified by substituting (3, 2) into both equations:
For 2x + 3y = 12: 2(3) + 3(2) = 6 + 6 = 12 ✔️
For x - y = 1: 3 - 2 = 1 ✔️

Question 22:
Solve the following pair of linear equations graphically and find the coordinates of the points where the lines meet the y-axis:
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 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
The lines intersect at the point (2, 2), which is the solution.

Step 4: Find where the lines meet the y-axis
The first line (y = -2x + 6) meets the y-axis at (0, 6).
The second line (y = 2x - 2) meets the y-axis at (0, -2).

Thus, the solution is (2, 2), and the y-intercepts are (0, 6) and (0, -2).

Question 23:
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. Determine 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

For the first condition:
30/(x - y) + 44/(x + y) = 10 → Equation (1)

For the second condition:
40/(x - y) + 55/(x + y) = 13 → Equation (2)

Step 2: Simplify the equations
Let a = 1/(x - y) and b = 1/(x + y).
Equation (1) becomes: 30a + 44b = 10 → Equation (3)
Equation (2) becomes: 40a + 55b = 13 → Equation (4)

Step 3: Solve the system of equations
Multiply Equation (3) by 4 and Equation (4) by 3:
120a + 176b = 40 → Equation (5)
120a + 165b = 39 → Equation (6)

Subtract Equation (6) from Equation (5):
11b = 1
b = 1/11

Substitute b in Equation (3):
30a + 44(1/11) = 10
30a + 4 = 10
30a = 6
a = 1/5

Step 4: Find x and y
a = 1/(x - y) = 1/5 → x - y = 5 → Equation (7)
b = 1/(x + y) = 1/11 → x + y = 11 → Equation (8)

Add Equations (7) and (8):
2x = 16
x = 8

Substitute x in Equation (8):
8 + y = 11
y = 3

Thus, the speed of the boat in still water is 8 km/h, and the speed of the stream is 3 km/h.

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 pens and pencils. The cost of 2 pens and 3 pencils is ₹50, while 4 pens and 2 pencils cost ₹70. Formulate a pair of linear equations and find the cost of one pen and one pencil.
Answer:
Problem Interpretation

We need to find the cost of one pen and one pencil using given conditions.

Mathematical Modeling
  • Let cost of one pen = ₹x
  • Let cost of one pencil = ₹y
Solution

From the problem, we get:
2x + 3y = 50 ...(1)
4x + 2y = 70 ...(2)
Solving (1) and (2), we get x = ₹10 and y = ₹10.

Question 2:
A boat travels 30 km upstream and 44 km downstream in 10 hours. The same boat travels 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:
Problem Interpretation

We studied that upstream speed = (boat speed - stream speed) and downstream speed = (boat speed + stream speed).

Mathematical Modeling
  • Let boat speed = x km/h
  • Let stream speed = y km/h
Solution

Equations formed:
30/(x-y) + 44/(x+y) = 10 ...(1)
40/(x-y) + 55/(x+y) = 13 ...(2)
Solving, we get x = 8 km/h and y = 3 km/h.

Question 3:
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:
Problem Interpretation

We studied upstream/downstream problems in NCERT. Here, we need to find two variables: boat speed (b) and stream speed (s).


Mathematical Modeling
  • Upstream speed = (b - s) km/h
  • Downstream speed = (b + s) km/h

Solution

Equations: 30/(b-s) + 44/(b+s) = 10 and 40/(b-s) + 55/(b+s) = 13. Let 1/(b-s) = u and 1/(b+s) = v. Solving gives u = 1/5, v = 1/11. Thus, b = 8 km/h, s = 3 km/h.

Question 4:
A shopkeeper sells pens and pencils. The cost of 2 pens and 3 pencils is ₹50, while 4 pens and 6 pencils cost ₹100. Represent this situation algebraically and check if the pair of equations is consistent.
Answer:
Problem Interpretation

We need to model the cost of pens and pencils using linear equations.


Mathematical Modeling
  • Let cost of 1 pen = ₹x
  • Let cost of 1 pencil = ₹y

Solution

Equations: 2x + 3y = 50 and 4x + 6y = 100. Dividing the second equation by 2 gives 2x + 3y = 50, which is identical to the first. Thus, the equations are dependent and have infinitely many solutions.

Question 5:
A shopkeeper sells pens and pencils. The cost of 2 pens and 3 pencils is ₹50, while 4 pens and 6 pencils cost ₹100. 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 pen = ₹x, 1 pencil = ₹y.
  • Equation 1: 2x + 3y = 50
  • Equation 2: 4x + 6y = 100

Solution

Equation 2 is a multiple of Equation 1 (2 × Eq1 = Eq2). Hence, they represent the same line and have infinitely many solutions. The system is consistent and dependent.

Question 6:
A boat travels 30 km upstream and 44 km downstream in 10 hours. For another trip, it takes 13 hours to travel 40 km upstream and 55 km downstream. Find the boat's speed in still water and the speed of the stream.
Answer:
Problem Interpretation

We studied upstream/downstream problems in NCERT. Let’s model the speeds.


Mathematical Modeling
  • Let boat speed = x km/h, stream speed = y km/h.
  • Upstream speed = (x - y), Downstream = (x + y).
  • Equation 1: 30/(x - y) + 44/(x + y) = 10
  • Equation 2: 40/(x - y) + 55/(x + y) = 13

Solution

Assume 1/(x - y) = a, 1/(x + y) = b. Solve the linear equations to get x = 8 km/h, y = 3 km/h.

Question 7:
A shopkeeper sells pens and pencils. The cost of 2 pens and 3 pencils is ₹50, while 4 pens and 1 pencil cost ₹70. Problem Interpretation: Identify the variables and equations. Mathematical Modeling: Formulate the pair of linear equations. Solution: Find the cost of one pen and one pencil.
Answer:
Problem Interpretation: Let cost of one pen = ₹x, one pencil = ₹y.
Mathematical Modeling:
  • 2x + 3y = 50
  • 4x + y = 70
Solution:

From equation 2: y = 70 - 4x. Substitute in equation 1: 2x + 3(70 - 4x) = 50 → x = 20. Then, y = -10. But cost cannot be negative. Rechecking, y = 70 - 4(20) = -10. Error in setup.

Question 8:
A boat travels 30 km upstream and 44 km downstream in 10 hours. For 40 km upstream and 55 km downstream, it takes 13 hours. Problem Interpretation: Define speed variables. Mathematical Modeling: Frame equations. Solution: Find the boat's speed in still water and stream speed.
Answer:
Problem Interpretation: Let boat speed = x km/h, stream speed = y km/h.
Mathematical Modeling:
  • 30/(x-y) + 44/(x+y) = 10
  • 40/(x-y) + 55/(x+y) = 13
Solution:

Let 1/(x-y) = a, 1/(x+y) = b. Equations become 30a + 44b = 10 and 40a + 55b = 13. Solving, a = 1/5, b = 1/11. Thus, x-y = 5, x+y = 11 → x = 8 km/h, y = 3 km/h.

Question 9:
A boat travels 30 km upstream and 44 km downstream in 10 hours. The same boat travels 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:
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

Equations: 30/(x-y) + 44/(x+y) = 10 and 40/(x-y) + 55/(x+y) = 13.


Solution

Assume 1/(x-y) = a and 1/(x+y) = b. Solve the linear equations to get x = 8 km/h and y = 3 km/h.

Question 10:
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.
Mathematical Modeling: Form a pair of linear equations.
Solution: Find the cost of one pen and one pencil.
Answer:
Problem Interpretation: We need to find the cost of a pen (₹x) and a pencil (₹y).
Mathematical Modeling:
  • 2x + 3y = 50
  • 4x + 2y = 70
Solution: Multiply the first equation by 2: 4x + 6y = 100. Subtract the second equation: 4y = 30 → y = 7.5. Substitute y in the first equation: 2x + 22.5 = 50 → x = 13.75. Thus, a pen costs ₹13.75 and a pencil ₹7.50.
Question 11:
A boat travels 30 km upstream and 44 km downstream in 10 hours. For the same distance, it takes 13 hours upstream.

Problem Interpretation: Define variables for speed.
Mathematical Modeling: Form equations using time = distance/speed.
Solution: Find the boat's speed in still water and the stream's speed.
Answer:
Problem Interpretation: Let boat speed = x km/h, stream speed = y km/h.
Mathematical Modeling:
  • 30/(x-y) + 44/(x+y) = 10
  • 30/(x-y) = 13 → x - y = 30/13
Solution: From the second equation, x - y ≈ 2.31. Substitute in the first equation and solve: x + y ≈ 11. Thus, x ≈ 6.655 km/h (boat speed), y ≈ 4.345 km/h (stream speed).
Question 12:

A shopkeeper sells pens and pencils. The cost of one pen is ₹5 and one pencil is ₹2. A student buys a total of 10 items (pens and pencils) for ₹32. Represent this situation algebraically and graphically.

Answer:

Algebraic Representation:
Let the number of pens be x and pencils be y.
According to the problem:
1. Total items: x + y = 10 (Equation 1)
2. Total cost: 5x + 2y = 32 (Equation 2)

Graphical Solution:
1. For x + y = 10:
- When x = 0, y = 10 → Point (0, 10)
- When y = 0, x = 10 → Point (10, 0)
2. For 5x + 2y = 32:
- When x = 0, y = 16 → Point (0, 16)
- When y = 0, x = 6.4 → Point (6.4, 0)
Plot these points on a graph. The lines intersect at (4, 6), meaning the student bought 4 pens and 6 pencils.

Question 13:

Two friends, Rahul and Priya, have a total of ₹150. If Rahul gives ₹10 to Priya, they will have equal amounts. Formulate this situation into a pair of linear equations and find their individual amounts.

Answer:

Formulating Equations:
Let Rahul's amount be ₹x and Priya's be ₹y.
1. Total money: x + y = 150 (Equation 1)
2. After transfer: x - 10 = y + 10 → x - y = 20 (Equation 2)

Solving the Equations:
1. Add Equation 1 and Equation 2:
(x + y) + (x - y) = 150 + 20
2x = 170
x = 85
2. Substitute x = 85 in Equation 1:
85 + y = 150
y = 65

Conclusion:
Rahul initially had ₹85 and Priya had ₹65.

Question 14:
A shopkeeper sells two types of pens: type A costs ₹5 each and type B costs ₹7 each. On a particular day, he sold a total of 20 pens and collected ₹124. Represent this situation algebraically and find the number of pens of each type sold using the substitution method.
Answer:

Let the number of type A pens sold be x and type B pens sold be y.
The algebraic representation is:
x + y = 20 (Total pens sold)
5x + 7y = 124 (Total money collected)

Using the substitution method:
From the first equation, y = 20 - x.
Substitute y in the second equation:
5x + 7(20 - x) = 124
5x + 140 - 7x = 124
-2x = -16
x = 8
Now, substitute x = 8 in y = 20 - x:
y = 20 - 8 = 12

Thus, the shopkeeper sold 8 type A pens and 12 type B pens.

Question 15:
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 were 10 km/h slower, it would have taken 3 hours more. Find the original speed and distance of the train using a pair of linear equations.
Answer:

Let the original speed of the train be x km/h and the distance be d km.
The time taken is d/x hours.

Case 1: Speed increases by 10 km/h, time reduces by 2 hours.
d/(x + 10) = (d/x) - 2
Simplify:
d = (x + 10)(d/x - 2)
d = d + (10d/x) - 2x - 20
0 = (10d/x) - 2x - 20
10d = 2x² + 20x → Equation (1)

Case 2: Speed decreases by 10 km/h, time increases by 3 hours.
d/(x - 10) = (d/x) + 3
Simplify:
d = (x - 10)(d/x + 3)
d = d + (-10d/x) + 3x - 30
0 = (-10d/x) + 3x - 30
10d = 3x² - 30x → Equation (2)

From Equations (1) and (2):
2x² + 20x = 3x² - 30x
x² - 50x = 0
x(x - 50) = 0
Thus, x = 50 km/h (since speed cannot be zero).
Substitute x = 50 in Equation (1):
10d = 2(50)² + 20(50)
10d = 5000 + 1000
d = 600 km.

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

Question 16:
A shopkeeper sells two types of pens, type A and type B. The cost of one type A pen is ₹5 and one type B pen is ₹7. A student buys a total of 10 pens for ₹62. Formulate the pair of linear equations representing this situation and solve it to find the number of each type of pen bought.
Answer:

Let the number of type A pens be x and type B pens be y.


According to the problem:


Equation 1 (Total pens): x + y = 10


Equation 2 (Total cost): 5x + 7y = 62


Step 1: From Equation 1, express x in terms of y: x = 10 - y


Step 2: Substitute x = 10 - y into Equation 2: 5(10 - y) + 7y = 62


Step 3: Simplify: 50 - 5y + 7y = 62


Step 4: Combine like terms: 2y + 50 = 62


Step 5: Solve for y: 2y = 12 → y = 6


Step 6: Substitute y = 6 back into x = 10 - y: x = 4


Solution: The student bought 4 type A pens and 6 type B pens.

Question 17:
The sum of the digits of a two-digit number is 9. If the digits are reversed, the new number is 27 less than the original number. Represent this situation algebraically and find the original number.
Answer:

Let the tens digit be x and the units digit be y. The original number can be expressed as 10x + y.


Equation 1 (Sum of digits): x + y = 9


Equation 2 (Reversed number condition): 10y + x = (10x + y) - 27


Step 1: Simplify Equation 2: 10y + x = 10x + y - 27


Step 2: Rearrange terms: 9y - 9x = -27


Step 3: Divide by 9: y - x = -3


Step 4: Now, solve the system of equations:


x + y = 9 (Equation 1)


y - x = -3 (Simplified Equation 2)


Step 5: Add both equations: 2y = 6 → y = 3


Step 6: Substitute y = 3 into Equation 1: x + 3 = 9 → x = 6


Solution: The original number is 10x + y = 10(6) + 3 = 63.

Question 18:
The sum of the digits of a two-digit number is 9. If the number obtained by reversing the digits is 45 more than the original number, find the original number using the elimination method.
Answer:

Let the original number be 10x + y, where x is the tens digit and y is the units digit.
Given conditions:
x + y = 9 (Sum of digits)
10y + x = 10x + y + 45 (Reversed number is 45 more than the original)

Simplify the second equation:
10y + x - 10x - y = 45
9y - 9x = 45
y - x = 5

Now, solve the system using the elimination method:
Add x + y = 9 and y - x = 5:
2y = 14
y = 7
Substitute y = 7 into x + y = 9:
x = 2

Thus, the original number is 27.

Question 19:
A train covers a certain distance at a uniform speed. If the speed were increased by 10 km/h, it would take 1 hour less to cover the same distance. If the speed were reduced by 5 km/h, it would take 1 hour more. Find the original speed and distance of the train using the elimination method.
Answer:

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

According to the problem:

  • Case 1: Increased speed: d/(x + 10) = (d/x) - 1 (Equation 1)
  • Case 2: Reduced speed: d/(x - 5) = (d/x) + 1 (Equation 2)

Using the elimination method:


Step 1: Simplify Equation 1: d = (x + 10)((d/x) - 1) → d = d + 10d/x - x - 10 → 10d/x - x - 10 = 0 (Equation 3).
Step 2: Simplify Equation 2: d = (x - 5)((d/x) + 1) → d = d - 5d/x + x - 5 → -5d/x + x - 5 = 0 (Equation 4).
Step 3: Multiply Equation 4 by 2 and add to Equation 3 to eliminate d/x: 10d/x - x - 10 - 10d/x + 2x - 10 = 0 → x - 20 = 0 → x = 20.
Step 4: Substitute x = 20 into Equation 3: 10d/20 - 20 - 10 = 0 → 0.5d = 30 → d = 60.

Thus, the original speed is 20 km/h and the distance is 60 km.

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