Class 10 Mathematics - Chapter 3: Pair of Linear Equations in Two Variables
Chapter 3: Pair of Linear Equations in Two Variables
1. Algebraic Form & Graphical Interpretation
General Form: A pair of linear equations in two variables $x$ and $y$ can be represented as:
$$a_1 x + b_1 y + c_1 = 0$$
$$a_2 x + b_2 y + c_2 = 0$$
where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers such that $a_1^2 + b_1^2 \neq 0$ and $a_2^2 + b_2^2 \neq 0$.
Consistency & Ratio Comparison

The behavior of lines representing a pair of linear equations depends on the ratios of their coefficients:

Ratio Comparison Graphical Representation Algebraic Interpretation Consistency
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ Intersecting lines Exactly one solution (Unique) Consistent
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ Coincident lines Infinitely many solutions Dependent (Consistent)
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ Parallel lines No solution Inconsistent
Intersecting (1 Solution) Parallel (No Solution) Coincident (Infinite Sol.)
Example 1.1: Testing Consistency using Ratios

Check whether the pair of equations $2x - 3y = 7$ and $4x - 6y = 9$ is consistent or inconsistent.

Solution:
Rewrite in standard form $ax + by + c = 0$:
$2x - 3y - 7 = 0 \implies a_1 = 2, b_1 = -3, c_1 = -7$
$4x - 6y - 9 = 0 \implies a_2 = 4, b_2 = -6, c_2 = -9$

Comparing ratios:

$$\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$$
$$\frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}$$
$$\frac{c_1}{c_2} = \frac{-7}{-9} = \frac{7}{9}$$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel. Therefore, the pair of linear equations has no solution and is inconsistent.

2. Algebraic Methods of Solution
Method 1: Substitution Method
  1. Express one variable (say $y$) in terms of the other variable ($x$) from one equation.
  2. Substitute this value of $y$ into the second equation to get an equation in one variable ($x$).
  3. Solve the resulting linear equation to find $x$.
  4. Substitute $x$ back into the first expression to find $y$.
Example 2.1: Solving by Substitution

Solve the following system by substitution:

$$7x - 15y = 2 \quad \text{--- (1)}$$
$$x + 2y = 3 \quad \text{--- (2)}$$

Solution:
From equation (2), isolate $x$:

$$x = 3 - 2y \quad \text{--- (3)}$$
Substitute expression (3) into equation (1):
$$7(3 - 2y) - 15y = 2$$
$$21 - 14y - 15y = 2 \implies 21 - 29y = 2$$
$$-29y = 2 - 21 = -19 \implies y = \frac{19}{29}$$
Substitute $y = \frac{19}{29}$ back into equation (3):
$$x = 3 - 2\left(\frac{19}{29}\right) = 3 - \frac{38}{29} = \frac{87 - 38}{29} = \frac{49}{29}$$
Thus, the solution is $x = \frac{49}{29}$ and $y = \frac{19}{29}$.

Method 2: Elimination Method
  1. Multiply one or both equations by suitable non-zero constants to make the coefficients of one variable numerically equal.
  2. Add or subtract the equations to eliminate that variable.
  3. Solve the resulting equation for the remaining variable.
  4. Substitute the value obtained into either original equation to find the second variable.
Example 2.2: Solving by Elimination

Solve the system of equations using elimination:

$$9x - 4y = 2000 \quad \text{--- (1)}$$
$$7x - 3y = 2000 \quad \text{--- (2)}$$

Solution:
Multiply Equation (1) by 3 and Equation (2) by 4 to align the coefficients of $y$:

$$3 \times (9x - 4y = 2000) \implies 27x - 12y = 6000 \quad \text{--- (3)}$$
$$4 \times (7x - 3y = 2000) \implies 28x - 12y = 8000 \quad \text{--- (4)}$$
Subtract Equation (3) from Equation (4):
$$(28x - 12y) - (27x - 12y) = 8000 - 6000$$
$$x = 2000$$
Substitute $x = 2000$ into Equation (1):
$$9(2000) - 4y = 2000 \implies 18000 - 4y = 2000$$
$$4y = 16000 \implies y = 4000$$
Thus, $x = 2000$ and $y = 4000$.