CBSE Class 10 Maths Notes: Pair of Linear Equations in Two Variables

**Form of a Linear Equation in Two Variables and Plotting Lines**

Definitions: 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 real numbers, and $a$ and $b$ are not both zero. Each solution to this equation represents a point that lies on a straight line when plotted on the Cartesian plane.

Core Principles:

• Each linear equation in two variables has infinitely many solutions.
• Each solution is an ordered pair of numbers (x, y) that satisfies the equation.
• The graph of a linear equation is a straight line.

Example: The equation $2x + y – 3 = 0$ is a linear equation in two variables. We can rearrange this to find some solutions. For instance, if $x=1$, then $2(1) + y -3 = 0$ which means $y = 1$. So (1, 1) is a solution.

**Graphical Method to Find Solutions**

Process: To solve a pair of linear equations graphically:

1. Graph each linear equation on the same Cartesian plane.
2. The point of intersection of the two lines represents the solution (x, y) to the system of equations.
3. If the lines are parallel, there is no solution (inconsistent system).
4. If the lines coincide (are the same line), there are infinitely many solutions (dependent system).

Example: Consider the equations $x + y = 5$ and $x – y = 1$. Plotting these lines will reveal they intersect at the point (3, 2). Thus, the solution is $x=3$, $y=2$.

**Algebraic Solution Methods: Substitution and Elimination**

Substitution Method:

1. Solve one equation for one variable in terms of the other.
2. Substitute this expression into the other equation.
3. Solve the resulting equation for the remaining variable.
4. Substitute the value found back into either of the original equations to solve for the other variable.

Example: Solve $y = x + 1$ and $x + 2y = 7$. Substitute $x + 1$ for $y$ in the second equation: $x + 2(x + 1) = 7$. Solve for x: $x = \frac{5}{3}$. Then substitute this back into $y = x + 1$, so $y = \frac{8}{3}$. Solution: $(\frac{5}{3},\frac{8}{3})$.

Elimination Method:

1. Multiply one or both equations by a constant so that the coefficients of one variable are opposites.
2. Add the equations to eliminate one variable.
3. Solve the resulting equation for the remaining variable.
4. Substitute the value found back into either of the original equations to solve for the other variable.

Example: Solve $2x + y = 5$ and $x – y = 1$. Add the equations: $3x = 6$, so $x = 2$. Substituting into the first equation, $2(2) + y = 5$, giving $y=1$. Solution: (2, 1).

**Conditions for Consistency/Inconsistency and Number of Solutions**

Consider a pair of linear equations:

$a_1x + b_1y + c_1 = 0$

$a_2x + b_2y + c_2 = 0$

Conditions:

• Unique Solution (Consistent): $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$. The lines intersect at one point.
• No Solution (Inconsistent): $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. The lines are parallel.
• Infinite Solutions (Consistent & Dependent): $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. The lines coincide.

Terminology: Consistent systems have at least one solution (unique or infinite). Inconsistent systems have no solutions. Dependent systems have infinitely many solutions.

**Simple Word Problems (Contextual Applications)**

Process: To solve word problems using linear equations:

1. Read the problem carefully and identify the unknowns.
2. Assign variables to represent the unknowns (e.g., $x$ and $y$).
3. Translate the given information into two linear equations.
4. Solve the system of equations using any of the methods discussed (graphical, substitution, or elimination).
5. Interpret the solution in the context of the problem.

Example: “The sum of two numbers is 10, and their difference is 2.” Let the numbers be $x$ and $y$. Equations: $x + y = 10$ and $x – y = 2$. Solving (e.g., elimination) gives $x = 6$ and $y = 4$.