Graphical Method of Solution of a Pair of Linear Equations for CBSE Class 10

In elementary algebra, a linear equation in two variables represents a relationship between two unknown quantities. Geometrically, every such equation corresponds to a straight line on the Cartesian coordinate plane. When two linear equations are considered together, they form a pair of linear equations in two variables (also called a system of simultaneous linear equations).

In CBSE Class 10 Mathematics, the graphical method provides an intuitive, visual approach to solving these systems. By graphing both lines simultaneously, students can determine whether a unique solution exists, whether infinitely many solutions are possible, or whether the system is inconsistent.

What You Will Learn

• Standard algebraic form of a pair of linear equations in two variables
• Geometric representation of linear systems on the coordinate plane
• The three graphical possibilities: Intersecting, Coincident, and Parallel lines
• Algebraic conditions for consistency and inconsistency based on coefficient ratios ($\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}$)
• Step-by-step protocol for plotting lines and finding solutions graphically
• Finding the vertices and area of geometric shapes formed with coordinate axes
• Common graph-plotting mistakes and board exam tips

1. Standard Form of a Pair of Linear Equations

A linear equation in two variables $x$ and $y$ has the standard form: $ax + by + c = 0, \quad \text{where } a, b, c \in \mathbb{R} \text{ and } a^2 + b^2 \ne 0$

A pair of linear equations in two variables is represented as:

$\begin{cases} a_1 x + b_1 y + c_1 = 0 \\ a_2 x + b_2 y + c_2 = 0 \end{cases}$

where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers such that $a_1^2 + b_1^2 \ne 0$ and $a_2^2 + b_2^2 \ne 0$.

2. Geometric Interpretation and Three Cases

Since each linear equation corresponds to a straight line in a two-dimensional plane, only three mutually exclusive geometric relationships can occur between the two lines:

         Case 1: Intersecting               Case 2: Coincident                 Case 3: Parallel
                 /                                 //                                /      /
                /   Point of                      //                                /      /
               /    Intersection                 //                                /      /
              X                                 //                                /      /
             / \                               //                                /      /
            /   \                             //                                /      /
      Unique Solution (Consistent)     Infinitely Many (Dependent)         No Solution (Inconsistent)

Case 1: Intersecting Lines (Unique Solution)

• The two lines intersect at exactly one point $(x_1, y_1)$.
• The coordinates of this intersection point represent the unique common solution of the system.
• The system of equations is called consistent.
• Algebraic Condition: $\frac{a_1}{a_2} \ne \frac{b_1}{b_2}$

Case 2: Coincident Lines (Infinitely Many Solutions)

• The two lines lie directly on top of each other, sharing every single point.
• Every point on the line satisfies both equations simultaneously $\implies$ infinitely many solutions.
• The system of equations is called consistent and dependent.
• Algebraic Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Case 3: Parallel Lines (No Solution)

• The two lines run in the same direction and never intersect, no matter how far they are extended.
• There is no common point $\implies$ no solution.
• The system of equations is called inconsistent.
• Algebraic Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$

Important: <u>A system of linear equations is consistent if it has at least one solution (either unique or infinitely many). It is inconsistent if it has zero solutions.</u>

3. Step-by-Step Procedure for Graphical Solution

To solve a pair of linear equations graphically in board examinations:

1. Express $y$ in terms of $x$: For each equation, solve explicitly for $y$: $y = \frac{-c - ax}{b}$.
2. Generate Table of Values: Choose at least three convenient integer values for $x$ and calculate the corresponding $y$ values.
3. Plot Points: Plot the coordinates $(x, y)$ carefully on graph paper using an appropriate scale (e.g., $1\text{ cm} = 1\text{ unit}$).
4. Draw Straight Lines: Join the points with a ruler and extend the lines in both directions, labeling each line with its equation.
5. Identify Intersection: Locate the point where the two lines intersect. Read its coordinates $(x, y)$ to state the final solution.

Remember: Always calculate three points for each line rather than two. While two points define a straight line, the third point serves as a verification check against arithmetic errors!

4. Solved CBSE Board Examination Problems

Solved Example 1: Solving Graphically and Finding Shaded Triangle

Problem: Solve the following pair of linear equations graphically: $x - y + 1 = 0 \quad \text{and} \quad 3x + 2y - 12 = 0$ Determine the coordinates of the vertices of the triangle formed by these lines and the $x$-axis, and shade the triangular region.

Solution:

Step 1: Table of Values for Line 1 ($x - y + 1 = 0 \implies y = x + 1$)

• When $x = 0 \implies y = 1 \implies (0, 1)$
• When $x = -1 \implies y = 0 \implies (-1, 0)$
• When $x = 2 \implies y = 3 \implies (2, 3)$

Step 2: Table of Values for Line 2 ($3x + 2y - 12 = 0 \implies y = \frac{12 - 3x}{2}$)

• When $x = 0 \implies y = \frac{12}{2} = 6 \implies (0, 6)$
• When $x = 2 \implies y = \frac{12 - 6}{2} = 3 \implies (2, 3)$
• When $x = 4 \implies y = \frac{12 - 12}{2} = 0 \implies (4, 0)$

Step 3: Graphical Intersection

Plotting both lines reveals that they intersect at point $A(2, 3)$. Therefore, <u>the unique solution is $x = 2$ and $y = 3$</u>.

Step 4: Vertices of the Triangle with the $x$-axis

• Line 1 intersects the $x$-axis ($y=0$) at $B(-1, 0)$.
• Line 2 intersects the $x$-axis ($y=0$) at $C(4, 0)$.
• The intersection point of the two lines is $A(2, 3)$.
• Thus, the vertices of the triangle are $A(2, 3)$, $B(-1, 0)$, and $C(4, 0)$.

Area Calculation: $\text{Base } BC = 4 - (-1) = 5 \text{ units}$ $\text{Height } h = y\text{-coordinate of } A = 3 \text{ units}$ $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 5 \times 3 = 7.5 \text{ sq. units}$

Solved Example 2: Determining Consistency without Graphing

Problem: On comparing the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2},$ and $\frac{c_1}{c_2}$, find out whether the lines representing $2x - 3y = 8$ and $4x - 6y = 9$ intersect at a point, are parallel, or are coincident.

Solution:

1. Write in standard form:
   • $2x - 3y - 8 = 0 \implies a_1 = 2, b_1 = -3, c_1 = -8$
   • $4x - 6y - 9 = 0 \implies a_2 = 4, b_2 = -6, c_2 = -9$
2. Compute the ratios: $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-8}{-9} = \frac{8}{9}$
3. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}$, the lines are parallel.
4. Therefore, the system has no solution and is inconsistent.

5. Master Comparison Table

Exam Tip: In board exam graph questions, always mention the chosen scale at the top right of the graph paper (e.g., Scale: 1 cm = 1 unit on both axes). Missing the scale loses 1 mark!

Common Mistake: Forgetting to convert equations to standard form with $c_1, c_2$ on the same side. If one equation has the constant on the LHS and the other on the RHS, sign errors will corrupt the ratio $\frac{c_1}{c_2}$.