10th Class Maths Exercise 3.3

Conquering Class 10 Maths: A Deep Dive into Exercise 3.3 (Linear Equations in Two Variables)

Exercise 3.3 in most Class 10 mathematics textbooks typically focuses on the application of linear equations in two variables to solve real-world problems. This exercise often involves word problems that require translating everyday situations into mathematical equations and then solving them to find the unknown quantities. Mastering this exercise is crucial for building a solid foundation in algebra and preparing for more advanced mathematical concepts. This comprehensive guide will break down the key concepts, provide step-by-step solutions to common problem types, and offer strategies to tackle even the most challenging questions.

Introduction: Linear Equations in Two Variables

Before we dive into Exercise 3.3, let's refresh our understanding of linear equations in two variables. A linear equation in two variables is an equation that can be written in the form ax + by = c, where a, b, and c are constants, and x and y are the variables. The graph of a linear equation is always a straight line. Solving these equations often involves finding the values of x and y that satisfy the equation. Exercise 3.3 typically presents problems where we need to formulate these equations from word problems and then solve them using methods like substitution, elimination, or graphical methods.

Types of Problems in Exercise 3.3

Exercise 3.3 usually encompasses a variety of word problems, each requiring a slightly different approach to formulating the equations. Common themes include:

• Age Problems: These problems involve finding the current ages of individuals based on relationships between their ages at different points in time.
• Number Problems: These problems involve finding two or more unknown numbers based on relationships between them, such as their sum, difference, or product.
• Speed, Distance, and Time Problems: These problems involve using the formula distance = speed × time to solve for unknown variables like speed, distance, or time.
• Geometry Problems: These problems involve applying linear equations to solve for unknown lengths, angles, or other geometric properties.
• Cost and Quantity Problems: These problems involve finding the cost or quantity of items based on given relationships between prices and quantities.

Step-by-Step Approach to Solving Word Problems

Solving word problems involving linear equations follows a systematic approach:

1. Understand the Problem: Carefully read the problem multiple times to understand what information is given and what is being asked for. Identify the unknown variables and assign them appropriate variables (e.g., x, y).

2. Formulate the Equations: Translate the information given in the problem into mathematical equations. This is often the most challenging step. Look for keywords that indicate mathematical operations (e.g., "sum," "difference," "product," "equals").

3. Solve the Equations: Use an appropriate method to solve the system of equations. Common methods include:

   • Substitution: Solve one equation for one variable and substitute it into the other equation.
   • Elimination: Add or subtract the equations to eliminate one variable.
   • Graphical Method: Graph the equations and find the point of intersection.

4. Check Your Solution: Substitute the values you found for the variables back into the original equations to make sure they satisfy the conditions of the problem. Also, ensure your solution makes logical sense in the context of the word problem.

Detailed Examples and Explanations

Let's work through a few examples representing different problem types typically found in Exercise 3.3:

Example 1: Age Problem

Problem: The sum of the ages of a father and his son is 50 years. After 5 years, the father's age will be thrice the son's age. Find their present ages.

Solution:

1. Variables: Let the father's present age be x and the son's present age be y.

2. Equations:

   • Equation 1: x + y = 50 (Sum of present ages)
   • Equation 2: x + 5 = 3(y + 5) (Father's age after 5 years is thrice the son's age after 5 years)

3. Solving: We can use the substitution method. From Equation 1, we get x = 50 - y. Substituting this into Equation 2: (50 - y) + 5 = 3(y + 5) 55 - y = 3y + 15 4y = 40 y = 10

   Substituting y = 10 back into x = 50 - y, we get x = 40.

4. Check: The father's present age is 40 and the son's present age is 10. Their sum is 50. After 5 years, the father will be 45 and the son will be 15 (45 = 3 × 15).

Example 2: Number Problem

Problem: The difference between two numbers is 26, and one number is three times the other. Find the numbers.

Solution:

1. Variables: Let the two numbers be x and y.

2. Equations:

   • Equation 1: x - y = 26 (Difference between the numbers)
   • Equation 2: x = 3y (One number is three times the other)

3. Solving: We can use substitution. Substituting Equation 2 into Equation 1: 3y - y = 26 2y = 26 y = 13

   Substituting y = 13 into Equation 2: x = 3(13) = 39

4. Check: The two numbers are 39 and 13. Their difference is 26, and 39 is three times 13.

Example 3: Speed, Distance, Time Problem

Problem: A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution:

1. Variables: Let the speed of the train be x km/h.

2. Equations:

   • Time taken at speed x: Time = Distance/Speed = 360/x

   • Time taken at speed (x+5): Time = 360/(x+5)

   • Equation: 360/x - 360/(x+5) = 1 (The difference in time is 1 hour)

3. Solving: 360(x+5) - 360x = x(x+5) 360x + 1800 - 360x = x² + 5x x² + 5x - 1800 = 0 This is a quadratic equation. We can solve it by factoring or using the quadratic formula. Factoring gives (x-40)(x+45) = 0. Since speed cannot be negative, x = 40 km/h.

4. Check: At 40 km/h, the time taken is 360/40 = 9 hours. At 45 km/h, the time taken is 360/45 = 8 hours. The difference is 1 hour.

Frequently Asked Questions (FAQs)

• What if I get a negative solution for a variable representing a real-world quantity like age or speed? A negative solution indicates an error in either formulating the equations or solving them. Review your work carefully. Negative solutions are often meaningless in the context of these problems.

• Which method is best for solving systems of linear equations – substitution or elimination? Both methods are equally valid. The best method depends on the specific equations. If one equation is easily solved for one variable, substitution is often easier. If the coefficients of one variable are the same or opposites in the two equations, elimination is often easier.

• What if I can't solve the equations? Double-check your equations to ensure they accurately represent the problem. Look for algebraic errors in your solution steps. If you're still stuck, try a different method (e.g., if you used substitution, try elimination).

• How important is it to check my solution? Checking your solution is crucial. It helps identify any errors made in the problem-solving process and ensures that your answer is logically consistent with the context of the word problem.

Conclusion

Mastering Exercise 3.3 requires a systematic approach, careful attention to detail, and practice. By understanding the different problem types, following a step-by-step solution process, and regularly checking your work, you can build confidence and proficiency in solving word problems involving linear equations. Remember, the key is to translate the words into mathematical equations and then apply your algebraic skills to find the solution. Consistent practice will help you develop the skills necessary not only to solve these problems but also to apply these techniques to more advanced mathematical concepts in the future. Don't be afraid to seek help if needed – understanding the concepts is key to long-term success in mathematics.