Exercise 3.6 Class 10 Maths

Exercise 3.6 Class 10 Maths: A full breakdown to Linear Equations in Two Variables

This article provides a practical guide to solving the problems in Exercise 3.Which means 6 of Class 10 mathematics, focusing on linear equations in two variables. We will cover each problem in detail, explaining the concepts involved and providing step-by-step solutions. Understanding linear equations is crucial for further studies in mathematics and related fields, so mastering this exercise is essential. We will explore various methods of solving these equations, including substitution, elimination, and cross-multiplication, ensuring you gain a strong foundational understanding. Let's dive in!

Introduction to Linear Equations in Two Variables

A linear equation in two variables is an equation that can be written in the form ax + by + c = 0, where a, b, and c are constants, and x and y are variables. Solving these equations often involves finding the values of x and y that satisfy the equation. In practice, exercise 3. The graph of a linear equation in two variables is a straight line. 6 typically presents word problems that need to be translated into linear equations before solving.

Understanding the Problems in Exercise 3.6

Exercise 3.6 usually presents a series of word problems that require formulating two linear equations in two variables. These problems often involve real-life scenarios such as:

• Age problems: Determining the current ages of individuals based on their ages in the past or future.
• Speed and distance problems: Calculating speeds and distances based on time taken.
• Number problems: Finding two unknown numbers based on their sum, difference, or other relationships.
• Cost and quantity problems: Determining the cost and quantity of items purchased.

The key to solving these problems is to carefully read the problem statement, identify the unknowns (x and y), and translate the given information into two linear equations.

Methods for Solving Linear Equations in Two Variables

There are several methods to solve a system of two linear equations in two variables:

• Substitution Method: Solve one equation for one variable in terms of the other, and substitute this expression into the second equation. This will give you an equation with only one variable, which you can solve. Then, substitute the value back into either of the original equations to find the value of the other variable.

• Elimination Method: Multiply one or both equations by constants to make the coefficients of one variable opposites. Add the two equations together to eliminate that variable. Solve the resulting equation for the remaining variable. Substitute the value back into either of the original equations to find the value of the other variable.

• Cross-Multiplication Method: This method is particularly useful for solving equations that are already in the standard form (ax + by + c = 0). The solution is given by:

  x = (b1c2 - b2c1) / (a1b2 - a2b1)

  y = (c1a2 - c2a1) / (a1b2 - a2b1)

  where a1, b1, c1 are coefficients of the first equation, and a2, b2, c2 are coefficients of the second equation.

Step-by-Step Solutions to Sample Problems (Illustrative Examples)

Let's walk through a few example problems similar to those found in Exercise 3.6, demonstrating each solution method:

Example 1: Age Problem (Substitution Method)

• Problem: The sum of the ages of a father and his son is 50 years. Five years ago, the father was six times as old as his son. Find their present ages.

• Solution:

  1. Define variables: Let x be the father's current age and y be the son's current age.

  2. Formulate equations:

     • Equation 1: x + y = 50 (Sum of their ages)
     • Equation 2: (x - 5) = 6(y - 5) (Father's age five years ago)

  3. Solve using substitution: Solve Equation 1 for x: x = 50 - y. Substitute this into Equation 2: (50 - y - 5) = 6(y - 5). Simplify and solve for y: 45 - y = 6y - 30 => 7y = 75 => y = 75/7 ≈ 10.71.

  4. Substitute back: Substitute y ≈ 10.71 into x = 50 - y to find x ≈ 39.29.

  5. Answer: The father's age is approximately 39.29 years, and the son's age is approximately 10.71 years. (Note: slight discrepancy due to rounding).

Example 2: Speed and Distance Problem (Elimination Method)

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

  For more on this topic, read our article on words that are plural nouns or check out why do black people smell funny.

• Solution:

  1. Define variables: Let x be the speed of the train (in km/hr) and y be the time taken (in hours).

  2. Formulate equations:

     • Equation 1: xy = 360 (Distance = Speed × Time)
     • Equation 2: (x + 5)(y - 1) = 360 (Speed increased by 5 km/hr, time reduced by 1 hour)

  3. Solve using elimination: From Equation 1, y = 360/x. Substitute this into Equation 2: (x + 5)(360/x - 1) = 360. Simplify and solve for x: 360 - x + 1800/x - 5 = 360 => -x + 1800/x -5 = 0. Multiply by x to get a quadratic equation. Solve the quadratic equation to find the speed.

Example 3: Number Problem (Cross-Multiplication Method)

• Problem: The sum of two numbers is 15 and their difference is 3. Find the numbers. Worth keeping that in mind.

• Solution:

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

  2. Formulate Equations:

     • Equation 1: x + y = 15
     • Equation 2: x - y = 3

  3. Apply Cross-Multiplication: Using the formula mentioned earlier, we can directly calculate x and y:

     x = [(1)(3) - (-1)(15)] / [(1)(-1) - (1)(1)] = 18 / -2 = -9 y = [(15)(1) - (3)(1)] / [(1)(-1) - (1)(1)] = 12 / -2 = -6

  4. Answer: The two numbers are -9 and -6.

Explanation of Scientific Concepts Involved

The problems in Exercise 3.6 fundamentally rely on the concepts of:

• Linear Equations: Understanding the structure and representation of linear equations is essential. The ability to translate word problems into these equations is key.
• Simultaneous Equations: The problems often involve two or more unknown variables that are interconnected. Solving these requires solving simultaneous equations.
• Algebraic Manipulation: Skill in manipulating algebraic expressions is crucial for solving the equations using different methods (substitution, elimination, cross-multiplication).
• Problem-solving strategies: A systematic approach is needed to break down the problem, identify unknowns, and formulate the equations.

Frequently Asked Questions (FAQ)

• Q: What if I get a negative solution for age or speed?

  • A: A negative solution for age or speed usually indicates an error in formulating the equations or solving them. Re-check your equations and calculations.

• Q: Which method is the best for solving these problems?

  • A: There's no single "best" method. The most suitable method depends on the specific equations. Sometimes, substitution is easier, while in other cases, elimination or cross-multiplication may be more efficient. Practice with different methods to develop your skills and choose the most convenient approach for each problem.

• Q: What if I get a solution that doesn't make sense in the context of the problem?

  • A: This often suggests an error in either the formulation of the equations or the solution process. Carefully review your work to identify and correct any mistakes.

• Q: How can I improve my ability to solve these types of problems?

  • A: Practice is key! Work through as many problems as possible. Start with simpler problems and gradually move to more complex ones. Focus on understanding the underlying concepts rather than just memorizing formulas or procedures.

Conclusion

Exercise 3.Now, remember to break down each problem logically, define your variables clearly, and double-check your solutions to ensure they are consistent with the problem's context. With consistent practice and a clear understanding of the concepts involved, you can confidently tackle any problem in Exercise 3.On the flip side, 6 of Class 10 mathematics provides essential practice in solving linear equations in two variables. By mastering the methods discussed – substitution, elimination, and cross-multiplication – and applying them systematically to various word problems, you'll build a strong foundation in this crucial area of mathematics. Think about it: 6 and beyond. Good luck!