Chapter 3: Pair of Linear Equations in Two Variables
Your complete guide to understanding and solving linear equations. Explore graphical methods, algebraic techniques, and practice questions.
Updated till October 20251. Graphical Method of Solution
A pair of linear equations can be represented on a graph. The way the two lines interact tells us the nature of the solution.
Case 1: Intersecting Lines
The lines cross at a single point. This means there is exactly one unique solution.
Condition:
a₁/a₂ ≠ b₁/b₂
Result: Unique Solution (Consistent)
Case 2: Parallel Lines
The lines never cross. This means there is no possible solution to the system.
Condition:
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Result: No Solution (Inconsistent)
Case 3: Coincident Lines
The lines are identical and overlap completely. This means there are infinitely many solutions.
Condition:
a₁/a₂ = b₁/b₂ = c₁/c₂
Result: Infinite Solutions (Consistent)
2. Consistency and Ratio Comparison
The relationship between the coefficients of the equations `a₁x + b₁y + c₁ = 0` and `a₂x + b₂y + c₂ = 0` determines the type of solution.
| Ratio Comparison | Graphical Representation | Algebraic Solution | Consistency |
|---|---|---|---|
| a₁/a₂ ≠ b₁/b₂ | Intersecting Lines | Exactly one solution (Unique) | Consistent |
| a₁/a₂ = b₁/b₂ = c₁/c₂ | Coincident Lines | Infinitely many solutions | Consistent (Dependent) |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | Parallel Lines | No solution | Inconsistent |
3. Algebraic Methods of Solution
Substitution Method
This method involves solving one equation for one variable and then substituting that expression into the other equation.
Solve one equation for one variable (e.g., solve for `y` in terms of `x`).
Substitute this expression for `y` into the *other* equation. This creates an equation with only `x`.
Solve the new equation for `x`.
Substitute the `x` value back into your expression from Step 1 to find `y`.
4. Equations Reducible to Linear Form
Sometimes, equations are not linear at first glance but can be transformed into a linear system. This often involves variables in the denominator.
How to Solve Reducible Equations
The strategy is to make a substitution to create a new, linear system.
Identify the non-linear parts (e.g., `1/x` or `1/(y-2)`) and substitute them with new variables, like `p` and `q`.
Rewrite the original equations as a new linear system in terms of `p` and `q`.
Solve this new system for `p` and `q` using elimination or substitution.
Substitute the values of `p` and `q` back into your original substitutions (from Step 1) to solve for `x` and `y`.
Example:
Solve: `2/x + 3/y = 13` and `5/x - 4/y = -2`
1. Let `p = 1/x` and `q = 1/y`.
2. The system becomes:
2p + 3q = 13
5p - 4q = -2
3. Solve for p and q. Multiply first eq. by 4, second by 3:
8p + 12q = 52
15p - 12q = -6
Adding them: `23p = 46` -> `p = 2`.
Put `p=2` in `2p + 3q = 13`: `2(2) + 3q = 13` -> `4 + 3q = 13` -> `3q = 9` -> `q = 3`.
4. Back-substitute:
`p = 1/x` -> `2 = 1/x` -> `x = 1/2`
`q = 1/y` -> `3 = 1/y` -> `y = 1/3`
Solution: x = 1/2, y = 1/3
5. Interactive Graph Visualizer
Use the sliders to change the coefficients of the two equations. Watch the graph and the solution type update in real time.
Line 1:
Line 2:
6. Strategies for Word Problems
The key to word problems is to translate the text into mathematical equations. Always start by identifying the two unknown quantities and assigning them variables (like `x` and `y`).
Solving Age Problems
These problems involve the relationship between the ages of two people at different points in time.
- Let the present age of the first person be `x` and the second be `y`.
- "Five years ago": Their ages were `x - 5` and `y - 5`.
- "Ten years from now": Their ages will be `x + 10` and `y + 10`.
- Form two equations based on the statements. For example, "A father is three times as old as his son" translates to `x = 3y`.
7. Practice Q&A
Q1: For what value of 'k' will the equations `x + 2y = 3` and `5x + ky = 15` have infinitely many solutions?
1. The condition for infinite solutions is `a₁/a₂ = b₁/b₂ = c₁/c₂`.
2. Here, a₁=1, b₁=2, c₁=3 and a₂=5, b₂=k, c₂=15.
3. Set up the ratios: `1/5 = 2/k = 3/15`.
4. From `1/5 = 2/k`, we get `k = 5 * 2`, so `k = 10`.
5. We check this with the third ratio: `2/10 = 1/5` and `3/15 = 1/5`. The condition holds.
Thus, `k = 10`.
Q2: Solve the system: `2x + 3y = 11` and `2x - 4y = -24`
1. Use elimination. Subtract the second equation from the first:
`(2x + 3y) - (2x - 4y) = 11 - (-24)`
`2x + 3y - 2x + 4y = 11 + 24`
`7y = 35`
2. Solve for `y`: `y = 35 / 7`, so `y = 5`.
3. Substitute `y = 5` into the first equation: `2x + 3(5) = 11`
`2x + 15 = 11` -> `2x = -4`
4. Solve for `x`: `x = -2`.
The solution is `x = -2, y = 5`.
Q3: The sum of two numbers is 25 and their difference is 5. Find the numbers.
1. Let the numbers be `x` and `y`.
2. `x + y = 25` (Equation 1)
3. `x - y = 5` (Equation 2)
4. Add both equations: `(x + y) + (x - y) = 25 + 5` -> `2x = 30` -> `x = 15`.
5. Substitute `x = 15` into Equation 1: `15 + y = 25` -> `y = 10`.
The two numbers are 15 and 10.
8. Frequently Asked Questions (FAQs)
Q: What is a linear equation in two variables?
A: An equation in the form `ax + by + c = 0`, where a, b, and c are real numbers and `a` and `b` are not both zero, is a linear equation in two variables, `x` and `y`.
Q: What does "consistent" mean for a system of equations?
A: A system is "consistent" if it has at least one solution. Intersecting lines (one solution) and coincident lines (infinite solutions) are both consistent systems.
Q: What does "inconsistent" mean?
A: A system is "inconsistent" if it has no solution. This happens when the lines are parallel.
Q: Can I use any method to solve a system of equations?
A: Yes. All three algebraic methods (substitution, elimination, cross-multiplication) will give the same answer. The graphical method also gives the answer, but it can be less precise if the solution involves fractions or decimals.