CH – 3 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

“Master algebraic balance, solve equations with precision, and excel in your board examinations.”

━━━━━━━━━━━━━━━━━━

Chapter Overview

In this chapter, students will explore system of linear equations in two variables and learn how to determine their solutions through graphical and algebraic techniques:

• Definition and general form of a pair of linear equations in two variables.
• Graphical representation and visual interpretation of solutions (intersecting, parallel, and coincident lines).
• Conditions for consistency based on ratios of coefficients.
• Algebraic solution methods: Substitution Method and Elimination Method.
• Formulation and solving of real-world word problems.

━━━━━━━━━━━━━━━━━━

🎯 Board Exam Focus

Students should focus on practicing the following high-priority concepts and skills:

• Understanding linear equations and checking their consistency using coefficient ratios.
• Solving pairs of equations using the Graphical method.
• Mastering algebraic techniques: Substitution method and Elimination method.
• Understanding the Cross-multiplication method (algebraic application and structure).
• Identifying conditions for unique, infinite, or no solutions.
• Translating word problems into a correct pair of linear equations.

━━━━━━━━━━━━━━━━━━

Most Important Concepts

What is a Pair of Linear Equations in Two Variables?

An equation that can be put in the form ax + by + c = 0, where a, b, and c are real numbers, and a and b are not both zero ($a^2 + b^2 \neq 0$), is called a linear equation in two variables x and y.

When two such linear equations are considered together in the same two variables, they form a pair of linear equations in two variables.

The general form for a pair of linear equations in two variables x and y is:

• a₁x + b₁y + c₁ = 0
  
• a₂x + b₂y + c₂ = 0
  

Here, a₁, b₁, c₁, a₂, b₂, and c₂ are real numbers such that $a_1^2 + b_1^2 \neq 0$ and $a_2^2 + b_2^2 \neq 0$.

Example:

• 2x + 3y – 7 = 0
• 9x – 2y – 8 = 0

━━━━━━━━━━━━━━━━━━

Graphical Method

Graphical Method

A pair of linear equations can be solved geometrically by plotting both equations on a graph paper. Each linear equation represents a straight line on the Cartesian plane.

The relative positions of these two lines determine the nature and number of solutions:

• One Solution (Intersecting Lines): The two lines intersect at exactly one point (x, y). This common point is the unique solution to the system. The system is consistent.
• No Solution (Parallel Lines): The two lines are parallel and never intersect. There is no common point, so the system has no solution. The system is inconsistent.
• Infinitely Many Solutions (Coincident Lines): The two lines overlap completely and coincide. Every point on the line is a common solution. The system is dependent and consistent.

Remember: The geometric solution of a pair of linear equations corresponds precisely to the point(s) of intersection of their representative straight lines on a graph sheet.

━━━━━━━━━━━━━━━━━━

Conditions for Consistency

Conditions for Consistency

For a given pair of linear equations:

• a₁x + b₁y + c₁ = 0
  
• a₂x + b₂y + c₂ = 0
  

We can compare the ratio of their coefficients (a₁/a₂, b₁/b₂, and c₁/c₂) without plotting graphs to determine their nature:

• One Unique Solution (Consistent): a₁/a₂ ≠ b₁/b₂ The lines intersect at a single point.
• No Solution (Inconsistent): a₁/a₂ = b₁/b₂ ≠ c₁/c₂ The lines are parallel to each other.
• Infinitely Many Solutions (Dependent / Consistent): a₁/a₂ = b₁/b₂ = c₁/c₂ The lines coincide with each other.

━━━━━━━━━━━━━━━━━━

Substitution Method

Substitution Method

In this method, we express one variable in terms of the other variable from one equation and substitute it into the second equation to reduce it to a single-variable equation.

Steps to Solve:

1. Pick either equation and find the value of one variable (say y) in terms of the other variable (x).
2. Substitute this value of y into the other equation to obtain an equation in a single variable x.
3. Solve this equation to find the value of x.
4. Substitute the value of x back into the equation obtained in Step 1 to calculate the value of y.

Example:

Solve the pair of equations:

• x + y = 14 — (Equation 1)
• x – y = 4 — (Equation 2)
• Step 1: From Equation 2, express x in terms of y: x = 4 + y
• Step 2: Substitute x = 4 + y into Equation 1: (4 + y) + y = 14 ⟹ 4 + 2y = 14
• Step 3: Solve for y: 2y = 10 ⟹ y = 5
• Step 4: Substitute y = 5 into x = 4 + y: x = 4 + 5 = 9

Solution is x = 9 and y = 5.

━━━━━━━━━━━━━━━━━━

Elimination Method

Elimination Method

In this method, we eliminate one of the variables by making its coefficients numerically equal in both equations.

Steps to Solve:

1. Multiply one or both equations by suitable non-zero numbers so that the coefficients of one variable become numerically equal.
2. Add or subtract the two equations to eliminate that variable, obtaining a linear equation in one variable.
3. Solve this equation for the remaining variable.
4. Substitute this value into either original equation to find the value of the other variable.

Example:

Solve the pair of equations:

• 2x + 3y = 8 — (Equation 1)
• 4x + 5y = 14 — (Equation 2)
• Step 1: Multiply Equation 1 by 2 to make the coefficients of x equal: 4x + 6y = 16 — (Equation 3)
• Step 2: Subtract Equation 2 from Equation 3: (4x + 6y) – (4x + 5y) = 16 – 14 ⟹ y = 2
• Step 3: Substitute y = 2 into Equation 1: 2x + 3(2) = 8 ⟹ 2x + 6 = 8 ⟹ 2x = 2 ⟹ x = 1

Solution is x = 1 and y = 2.

Common Mistake: Forgetting to change the signs of all terms in the subtracted equation during Step 2. Always distribute the negative sign carefully.

━━━━━━━━━━━━━━━━━━

Cross-Multiplication Method

Cross-Multiplication Method

The cross-multiplication method offers a direct algebraic formula derived from solving the standard form equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.

To write the cross-multiplication relation, arrange the coefficients as:

Plaintext

  x         y         1
b₁   c₁   c₁   a₁   a₁   b₁
b₂   c₂   c₂   a₂   a₂   b₂

Cross-multiplying along the diagonal arrows gives:

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

From this relation, provided a₁b₂ – a₂b₁ ≠ 0:

• x = (b₁c₂ – b₂c₁) / (a₁b₂ – a₂b₁)
• y = (c₁a₂ – c₂a₁) / (a₁b₂ – a₂b₁)

Example:

For 2x + y – 5 = 0 and 3x + 2y – 8 = 0:

Here, a₁ = 2, b₁ = 1, c₁ = -5 and a₂ = 3, b₂ = 2, c₂ = -8.

• x / [1(-8) – 2(-5)] = y / [(-5)(3) – (-8)(2)] = 1 / [2(2) – 3(1)]
• x / (-8 + 10) = y / (-15 + 16) = 1 / (4 – 3)
• x / 2 = y / 1 = 1 / 1
• x = 2, y = 1

━━━━━━━━━━━━━━━━━━

How to Choose the Best Method?

How to Choose the Best Method?

Depending on the form of the equations given in the question:

• Graphical Method: Best when explicitly asked in the exam question or when visualizing physical intersections (e.g., finding coordinates forming a triangle with axes).
• Substitution Method: Ideal when one of the variables in either equation has a coefficient of 1 or -1 (e.g., x + 2y = 3).
• Elimination Method: Best and quickest for most standard algebraic problems, especially when coefficients are integers or easily convertible to matching multiples.
• Cross-Multiplication Method: Useful for equations given in standard form with larger coefficient values, provided the formula is carefully applied.

━━━━━━━━━━━━━━━━━━

📌 Important Conditions & Results

• Standard Pair: a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
  
• Unique Solution: a₁/a₂ ≠ b₁/b₂ (Consistent / Intersecting)
• No Solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (Inconsistent / Parallel)
• Infinitely Many Solutions: a₁/a₂ = b₁/b₂ = c₁/c₂ (Dependent & Consistent / Coincident)
• A dependent pair of linear equations is always consistent.

━━━━━━━━━━━━━━━━━━

🧠 Concepts Students Find Difficult

• Understanding Consistency vs. Inconsistency: Students confuse “dependent” and “consistent.” Remember that dependent systems are a sub-category of consistent systems because they have solutions.
• Standardizing Constants before Ratio Checks: If one equation is written as ax + by = c and the second as ax + by + c = 0, sign errors occur. Always convert both equations to the same standard form before calculating ratios.
  
• Choosing the Variable to Eliminate: In elimination, pick the variable that requires multiplying only one equation or has smaller coefficients to keep calculation steps minimal.
• Sign Rules in Elimination: Subtracting an entire equation requires reversing the sign of every single term in that equation.
• Setting up Word Problems: Identifying the unknown quantities and translating word phrases into variables (e.g., speed, age, digits of a number) requires practice.

━━━━━━━━━━━━━━━━━━

⚠️ Common Mistakes Students Make

1. Incorrect Sign Application in Ratios
   • Mistake: Comparing ratios without bringing terms to one side. (e.g., x + y = 5 and x + y + 5 = 0).
   • How to avoid: Ensure both equations follow ax + by + c = 0 before identifying a, b, and c.
2. Incorrect Distribution During Elimination
   • Mistake: Multiplying only the left side of the equation when matching coefficients (e.g., 2 × (x + 3y = 5) written as 2x + 6y = 5).
   • How to avoid: Multiply every term on both sides of the equal sign.
3. Incomplete Substitution Steps
   • Mistake: Substituting the expression for a variable back into the same equation it was derived from, resulting in trivial identity statements like 0 = 0.
   • How to avoid: Always substitute the isolated expression into the other un-used equation.
4. Assuming Parallel Lines Mean Unique Intersects
   • Mistake: Misinterpreting a ratio like a₁/a₂ = b₁/b₂ ≠ c₁/c₂ as having a unique solution.
   • How to avoid: Memorize that equality of variable ratios with unequal constant ratios signifies zero intersections (parallel).
5. Reversing Digits in Number Word Problems
   • Mistake: Writing a two-digit number as x + y instead of 10x + y.
   • How to avoid: Remember that place value dictates two-digit numbers as 10 × (tens digit) + (units digit).

━━━━━━━━━━━━━━━━━━

🔥 Priority Revision

⭐⭐⭐ HIGH-PRIORITY PRACTICE

• Finding values of unknown constants (e.g., k or p) given conditions for unique, no, or infinite solutions.
• Solving equations using the Elimination Method.
• Word problems based on two-digit numbers, ages, and simple linear relations.

⭐⭐ IMPORTANT PRACTICE

• Solving equations using the Substitution Method.
• Plotting graphical solutions and determining the vertices/area of triangles formed with the axes.
• Application of Cross-Multiplication Method.

⭐ ADDITIONAL REVISION

• Verifying algebraic consistency before choosing solution methods.
• Word problems related to fractions, fixed/variable charges, and geometry.

━━━━━━━━━━━━━━━━━━

⚡ Quick Revision Sheet

• Linear Pair: a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
• Intersecting Lines: a₁/a₂ ≠ b₁/b₂ ⟹ 1 Unique Solution ⟹ Consistent
• Parallel Lines: a₁/a₂ = b₁/b₂ ≠ c₁/c₂ ⟹ No Solution ⟹ Inconsistent
• Coincident Lines: a₁/a₂ = b₁/b₂ = c₁/c₂ ⟹ Infinitely Many Solutions ⟹ Dependent/Consistent
• Substitution: Express one variable from an equation and substitute in the other.
• Elimination: Multiply equations to match coefficients of one variable, then add or subtract.
• Board Tip: Always double-check your x and y values by substituting them back into both original equations to verify correctness!

━━━━━━━━━━━━━━━━━━

📝 Test Yourself

1. For what value of k will the equations 3x + y = 1 and (2k – 1)x + (k – 1)y = 2k + 1 have no solution?
2. Check whether the pair of equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0 is consistent or inconsistent.
3. Solve the pair of linear equations graphically: 2x + y = 6 and 2x – y = 2.
4. Solve using the Substitution Method: 3x – y = 3 and 9x – 3y = 9.
5. Solve using the Elimination Method: 3x + 4y = 10 and 2x – 2y = 2.
6. Find the solution of the equations 2x + 3y = 11 and 2x – 4y = -24. Hence, find the value of ‘m’ for which y = mx + 3.
7. Solve using the Cross-Multiplication Method: 2x + 3y = 17 and 3x – 2y = 6.
8. The sum of a two-digit number and the number obtained by reversing its digits is 99. If the digits differ by 3, find the number.
9. Five years ago, a father was thrice as old as his son. Ten years later, the father will be twice as old as his son. Find their present ages.
10. The perimeter of a rectangular field is 32 meters. If its length is increased by 2 meters and breadth is reduced by 1 meter, the area remains unchanged. Find the dimensions of the field.

━━━━━━━━━━━━━━━━━━

🎯 Before Your Exam

☐ I understand what a pair of linear equations is. ☐ I can solve equations graphically. ☐ I understand the conditions for one, no, and infinitely many solutions. ☐ I can use the substitution method. ☐ I can use the elimination method. ☐ I understand the cross-multiplication method. ☐ I can form equations from word problems.

Stay confident, double-check your algebraic steps, and approach your examination with total clarity!