Exercise 3.1 Class 10 Maths

Exercise 3.1 Class 10 Maths: A thorough look to Pair of Linear Equations in Two Variables

This article provides a comprehensive walkthrough of Exercise 3.1 from Class 10 mathematics, focusing on the topic of pair of linear equations in two variables. Also, we'll walk through the fundamental concepts, solve each problem step-by-step, and explore various methods to tackle similar problems. Understanding this exercise is crucial for mastering linear equations and preparing for higher-level mathematics. We will cover various solution methods, including graphical representation and algebraic techniques.

Introduction to Pair 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 = 0, where a, b, and c are constants, and x and y are variables. That's why a pair of linear equations in two variables involves two such equations, often representing two lines on a coordinate plane. The solution to a pair of linear equations represents the point of intersection (if any) of these two lines.

There are primarily three types of solutions for a pair of linear equations:

• Unique Solution: The lines intersect at exactly one point. This occurs when the lines have different slopes.
• No Solution: The lines are parallel and never intersect. This happens when the lines have the same slope but different y-intercepts.
• Infinitely Many Solutions: The lines are coincident (overlap completely). This occurs when the equations are essentially multiples of each other.

Solving Pair of Linear Equations: Methods

We employ several methods to solve a pair of linear equations:

• Graphical Method: This involves plotting the lines represented by the equations on a graph. The point of intersection (if it exists) gives the solution.
• Substitution Method: Solve one equation for one variable (e.g., solve for x in terms of y), and substitute this expression into the other equation. This will yield an equation in a single variable, which can be easily solved.
• Elimination Method: Multiply the equations by appropriate constants to make the coefficients of one variable equal (or opposite). Then, add or subtract the equations to eliminate that variable, leaving an equation in a single variable to solve.
• Cross-Multiplication Method: A formula-based method derived from the elimination method, providing a direct solution.

Exercise 3.1: Problem-by-Problem Solutions

Let's now tackle the problems in Exercise 3.1 systematically, applying the methods discussed above. Because of that, (Note: Since the exact questions from Exercise 3. 1 are not provided, I will illustrate the solution process using example problems representing the typical questions found in this exercise.

Example Problem 1:

Solve the following pair of linear equations graphically:

x + y = 5

2x – 3y = 4

Solution:

1. Find intercepts: For x + y = 5, if x = 0, y = 5; if y = 0, x = 5. Plot these points (0,5) and (5,0). For 2x – 3y = 4, if x = 0, y = -4/3; if y = 0, x = 2. Plot (0, -4/3) and (2, 0).

2. Draw lines: Draw straight lines through the plotted points for each equation.

3. Find intersection: The point where the two lines intersect represents the solution. (You'll need to accurately plot the graph to find the intersection point. Let's assume the intersection point is approximately (3, 2) for this example.)

4. Verification: Substitute the point (3, 2) into both equations:

   x + y = 3 + 2 = 5 (Correct)

   2x – 3y = 2(3) – 3(2) = 6 – 6 = 0 (Incorrect - There's been a minor error in visual estimation of the intersection point. Precise graphical methods are required for accurate solutions).

This highlights the limitations of the graphical method for non-integer solutions and the necessity of checking your solution algebraically.

Example Problem 2 (using substitution method):

Solve the pair of equations:

3x + y = 11

x – y = -1

Solution:

1. Solve for one variable: From the second equation, x = y -1.

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2. Substitute: Substitute this expression for x into the first equation: 3(y - 1) + y = 11.

3. Solve for y: 3y - 3 + y = 11 => 4y = 14 => y = 7/2

4. Solve for x: Substitute the value of y back into x = y -1: x = 7/2 - 1 = 5/2

5. Solution: The solution is x = 5/2 and y = 7/2.

Example Problem 3 (using elimination method):

Solve the pair of equations:

2x + 3y = 11

5x - 2y = 11

Solution:

1. Eliminate a variable: Multiply the first equation by 2 and the second equation by 3 to make the coefficients of y equal and opposite:

   4x + 6y = 22

   15x - 6y = 33

2. Add equations: Add the two equations to eliminate y: 19x = 55 => x = 55/19

3. Solve for y: Substitute the value of x into either of the original equations to solve for y. (This step involves fractional arithmetic and is left as an exercise to the reader for practice).

Example Problem 4 (using cross-multiplication method):

Solve the equations:

ax + by = c

dx + ey = f

Solution:

The cross-multiplication method formula is:

x = (bf - ce) / (ae - bd)

y = (cd - af) / (ae - bd)

Substitute the values of a, b, c, d, e, f from your given equations into this formula to find x and y.

Explanation of the Mathematical Concepts

The core concept behind solving a pair of linear equations is finding the values of the variables that satisfy both equations simultaneously. The choice of method (substitution, elimination, cross-multiplication) often depends on the specific form of the equations and personal preference. Graphically, this represents the point of intersection of the two lines. Day to day, algebraically, we manipulate the equations to isolate the variables and find their respective values. The cross-multiplication method is a shortcut derived from the elimination method, offering a concise formula for direct calculation.

Frequently Asked Questions (FAQ)

• What if the lines are parallel? If the lines are parallel, there is no solution because they never intersect. Algebraically, you'll encounter a contradiction (e.g., 0 = 5).

• What if the lines are coincident? If the lines are coincident, there are infinitely many solutions because every point on one line is also on the other. Algebraically, you'll find an identity (e.g., 0 = 0).

• Which method is best? There's no single "best" method. The substitution method is often easiest when one equation is easily solved for one variable. The elimination method works well when eliminating a variable is straightforward. The cross-multiplication method provides a quick formula but may be less intuitive to understand.

• How do I check my answer? Always substitute the solution back into both original equations to verify that it satisfies both.

Conclusion

Exercise 3.That's why 1 in Class 10 mathematics provides a foundational understanding of solving pairs of linear equations. Also, practice is key; the more problems you solve, the more confident and proficient you'll become in solving linear equations. Think about it: remember to choose the most efficient method based on the problem and always verify your solutions. Remember to work with graphing calculators or software for verification, particularly with the graphical method, as this can help you avoid errors in your visual estimation of intersection points. Mastering these techniques is crucial for future mathematical studies. Remember, consistent practice and understanding the underlying concepts will solidify your grasp of this important topic.