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 |
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
- Express one variable (say $y$) in terms of the other variable ($x$) from one equation.
- Substitute this value of $y$ into the second equation to get an equation in one variable ($x$).
- Solve the resulting linear equation to find $x$.
- 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
- Multiply one or both equations by suitable non-zero constants to make the coefficients of one variable numerically equal.
- Add or subtract the equations to eliminate that variable.
- Solve the resulting equation for the remaining variable.
- 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 $y = 4000$.