How to Solve: Linear Equations in Two Variables

How to Solve: Linear Equations in Two Variables

Complete Guide for SSC / Bank / Railway Exams

Introduction

"Master linear equations in two variables, and you unlock 5–10 marks in every SSC, Bank, or Railway exam—questions that test your ability to find hidden values in real-life problems like budgeting, speed-distance, or profit-loss. One method, three steps, and you’ll solve them faster than the clock ticks."

What You Need To Know First

1. What is an equation? A statement with an equals sign (e.g., 2x + 3 = 7).
2. How to solve for one variable (e.g., x in 3x + 2 = 11).
3. Basic algebra rules (add/subtract/multiply/divide both sides equally).

Key Vocabulary

Formulas To Know

1. General Form of a Linear Equation in Two Variables ax + by + c = 0
2. a, b, c = constants (numbers)
3. x, y = variables

4. MEMORISE THIS (used in all problems).

5. Slope-Intercept Form (for graphing) y = mx + c

6. m = slope (steepness)
7. c = y-intercept (where the line crosses the y-axis)
8. Given on exam sheet (but understand it).

Step-by-Step Method

Method 1: Substitution (Best for simple equations)

Step 1: Solve one equation for one variable. - Example: From x + y = 5, get y = 5 – x.

Step 2: Substitute this expression into the second equation. - Example: If second equation is 2x – y = 1, replace y with 5 – x → 2x – (5 – x) = 1.

Step 3: Solve for the remaining variable. - Simplify: 2x – 5 + x = 1 → 3x = 6 → x = 2.

Step 4: Plug the value back to find the other variable. - y = 5 – x → y = 5 – 2 → y = 3.

Step 5: Write the solution as an ordered pair (x, y). - Final answer: (2, 3).

Method 2: Elimination (Best for coefficients that cancel easily)

Step 1: Write both equations in standard form (ax + by = c). - Example: 2x + 3y = 8 and 4x – 3y = 2.

Step 2: Add or subtract equations to eliminate one variable. - Here, +3y and –3y cancel if added: (2x + 3y) + (4x – 3y) = 8 + 2 → 6x = 10.

Step 3: Solve for the remaining variable. - 6x = 10 → x = 10/6 → x = 5/3.

Step 4: Substitute back to find the other variable. - Plug x = 5/3 into 2x + 3y = 8 → 2(5/3) + 3y = 8 → 10/3 + 3y = 8 → 3y = 14/3 → y = 14/9.

Step 5: Write the solution as (x, y). - Final answer: (5/3, 14/9).

Method 3: Cross-Multiplication (For 2 equations in standard form)

Given: a₁x + b₁y + c₁ = 0 a₂x + b₂y + c₂ = 0

Formula: x / (b₁c₂ – b₂c₁) = y / (c₁a₂ – c₂a₁) = 1 / (a₁b₂ – a₂b₁)

Step 1: Identify a₁, b₁, c₁ and a₂, b₂, c₂. - Example: 2x + 3y – 8 = 0 and 4x – 3y – 2 = 0. - a₁ = 2, b₁ = 3, c₁ = –8 - a₂ = 4, b₂ = –3, c₂ = –2

Step 2: Plug into the formula. - x / [(3)(–2) – (–3)(–8)] = y / [(–8)(4) – (–2)(2)] = 1 / [(2)(–3) – (4)(3)] - Simplify denominators: - x / (–6 – 24) = y / (–32 + 4) = 1 / (–6 – 12) - x / (–30) = y / (–28) = 1 / (–18)

Step 3: Solve for x and y. - x = (–30) / (–18) = 5/3 - y = (–28) / (–18) = 14/9

Step 4: Write the solution. - Final answer: (5/3, 14/9).

Worked Examples

Example 1 – Basic (Substitution)

Problem: Solve: x + y = 5 2x – y = 1

Solution:
1. From x + y = 5, get y = 5 – x.
2. Substitute into 2x – y = 1 → 2x – (5 – x) = 1.
3. Simplify: 2x – 5 + x = 1 → 3x = 6 → x = 2.
4. Plug x = 2 into y = 5 – x → y = 3.
5. Solution: (2, 3).

What we did and why: We isolated y first because it was easier (coefficient = 1). Substitution works best when one variable has a coefficient of 1 or –1.

Example 2 – Medium (Elimination)

Problem: Solve: 3x + 2y = 12 5x – 2y = 4

Solution:
1. Add both equations to eliminate y: (3x + 2y) + (5x – 2y) = 12 + 4 → 8x = 16 → x = 2.
2. Plug x = 2 into 3x + 2y = 12 → 6 + 2y = 12 → 2y = 6 → y = 3.
3. Solution: (2, 3).

What we did and why: The y terms had opposite signs (+2y and –2y), so adding them canceled y instantly. Elimination is fastest when coefficients are opposites.

Example 3 – Exam-Style (Disguised Problem)

Problem: The sum of two numbers is 20. Three times the first number minus twice the second number is 10. Find the numbers.

Solution:
1. Let the numbers be x and y.
2. Translate to equations: - x + y = 20 (sum is 20) - 3x – 2y = 10 (three times first minus twice second is 10)
3. Solve x + y = 20 for y → y = 20 – x.
4. Substitute into 3x – 2y = 10 → 3x – 2(20 – x) = 10.
5. Simplify: 3x – 40 + 2x = 10 → 5x = 50 → x = 10.
6. Plug x = 10 into y = 20 – x → y = 10.
7. Solution: The numbers are 10 and 10.

What we did and why: We turned a word problem into equations. Always define variables first, then translate sentences into math.

Common Mistakes

Exam Traps

1-Minute Recap

"Listen up—this is your last-minute checklist for linear equations in two variables:
1. Pick a method: Substitution if one variable is easy to isolate (coefficient = 1). Elimination if coefficients cancel nicely.
2. For substitution: Solve one equation for x or y, plug into the other, solve, then back-substitute.
3. For elimination: Add/subtract equations to kill one variable, solve, then find the other.
4. Cross-multiplication? Use only if equations are in ax + by + c = 0 form—memorize the formula!
5. Word problems? Define variables first, then translate sentences into equations.
6. Double-check: Plug your (x, y) back into both original equations. If they work, you’re golden.
7. Watch signs and fractions—they’re where most mistakes hide. You’ve got this. Now go solve those 5–10 marks like a pro!