Introduction

{{FORMULA: expr=a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 | symbols=x,y:variables, a₁,b₁,a₂,b₂:coefficients, c₁,c₂:constants}}

Introduction: Turning Everyday Puzzles into Math

Have you ever faced a situation where you need to figure out two unknown quantities at the same time? Life is full of such puzzles! Imagine planning a budget, mixing ingredients for a recipe, or even figuring out scores in a game. These are all real-world problems that can be solved using the power of algebra.

Let's consider the example of Akhila visiting a village fair. She wants to ride the Giant Wheel and play Hoopla. We know two things about her choices:

1. The number of times she played Hoopla is exactly half the number of Giant Wheel rides.
2. Each ride costs ₹3, each game of Hoopla costs ₹4, and she spent a total of ₹20.

How can we find out exactly how many rides she took and how many games she played? Guessing might work, but it's slow. A more powerful method is to translate this story into the language of mathematics. This chapter is all about learning to represent such situations using a pair of linear equations in two variables and then finding their unique solution.

Definitions & Key Terms

Before we start building equations, let's understand the basic building blocks.

From Words to Equations: A Step-by-Step Guide

Translating a word problem into a pair of linear equations is the most critical skill in this chapter. Let's use Akhila's fair problem to see how it's done.

Problem: The number of times Akhila played Hoopla (y) is half the number of rides she had on the Giant Wheel (x). Each ride costs ₹3 and a game of Hoopla costs ₹4, and she spent a total of ₹20.

1. Identify the Unknowns The first step is to figure out what we need to find. In this problem, the two unknown quantities are:

   • The number of rides on the Giant Wheel.
   • The number of times she played Hoopla.

2. Assign Variables Let's assign simple variables to represent these unknowns.

   • Let x = the number of rides on the Giant Wheel.
   • Let y = the number of times she played Hoopla.

3. Translate the First Condition Read the first condition carefully: "The number of times she played Hoopla is half the number of rides she had on the Giant Wheel."

   • "Number of times she played Hoopla" is y.
   • "is" translates to the equals sign =.
   • "half the number of rides" translates to ½ × x or x/2.

   Combining these gives our first equation:

   y = x/2
   

   This can also be written as y = (1/2)x.

4. Translate the Second Condition Read the second condition: "each ride costs ₹3, and a game of Hoopla costs ₹4, ... she spent ₹20."

   • Total cost of rides = (cost per ride) × (number of rides) = 3 × x.
   • Total cost of Hoopla = (cost per game) × (number of games) = 4 × y.
   • The total amount spent is the sum of these two costs, which is ₹20.

   This gives our second equation:

   3x + 4y = 20
   

5. State the Final Pair We have successfully converted the word problem into a system of two linear equations with two variables.

   • Equation 1: y = x/2
   • Equation 2: 3x + 4y = 20

Solving this pair will give us the exact number of rides and games Akhila enjoyed. We will learn how to solve these in the next sections.

{{KEY: type=concept | title=The Golden Rule of Word Problems | text=The first and most important step is always to identify the TWO distinct quantities you need to find. Assign a different variable (like x and y) to each one before you even start writing equations.}}

Solved Examples

Let's practice forming pairs of linear equations from different scenarios.

Example 1: Sum and Difference (Easy)

Given: The sum of two numbers is 35 and their difference is 13.

To Find: The pair of linear equations representing this situation.

Solution:

1. Let the first number be x and the second number be y.

2. The first condition is "the sum of two numbers is 35". This translates to:

   x + y = 35
   

3. The second condition is "their difference is 13". This translates to:

   x - y = 13
   

Final Answer: The required pair of equations is x + y = 35 and x - y = 13.

Example 2: Cost of Items (Medium)

Given: 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46.

To Find: The pair of linear equations for the cost of one pencil and one pen.

Solution:

1. Identify the two unknowns: the cost of a single pencil and the cost of a single pen. Let the cost of one pencil be x (in ₹). Let the cost of one pen be y (in ₹).

2. Translate the first condition: "5 pencils and 7 pens together cost ₹50". Cost of 5 pencils = 5x. Cost of 7 pens = 7y. Total cost:

   5x + 7y = 50
   

3. Translate the second condition: "7 pencils and 5 pens together cost ₹46". Cost of 7 pencils = 7x. Cost of 5 pens = 5y. Total cost:

   7x + 5y = 46
   

Final Answer: The required pair of equations is 5x + 7y = 50 and 7x + 5y = 46.

Example 3: Ages (Hard)

Given: Five years ago, a man was seven times as old as his son. Five years hence, the man will be three times as old as his son.

To Find: The pair of linear equations to find their present ages.

Solution:

1. Let the man's present age be x years. Let the son's present age be y years.

2. Analyze the first condition, which is about the past: "Five years ago...". Man's age 5 years ago = x - 5. Son's age 5 years ago = y - 5. The condition is: "man was seven times as old as his son".

   x - 5 = 7 × (y - 5)
   

3. Analyze the second condition, which is about the future: "Five years hence...". Man's age in 5 years = x + 5. Son's age in 5 years = y + 5. The condition is: "man will be three times as old as his son".

   x + 5 = 3 × (y + 5)
   

Final Answer: The pair of equations is x - 5 = 7(y - 5) and x + 5 = 3(y + 5).

Example 4: Two-Digit Number (Tricky)

Given: The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits.

To Find: The pair of linear equations to find the number.

Solution:

1. This is tricky. The variables are not the number itself, but its digits. Let the digit in the tens place be x. Let the digit in the units place be y.

2. The original two-digit number can be written as 10x + y. (For example, if x=2, y=5, the number is 10(2) + 5 = 25). The number obtained by reversing the digits is 10y + x.

3. The first condition is "The sum of the digits... is 9".

   x + y = 9
   

4. The second condition is "nine times this number is twice the number obtained by reversing...". Nine times this number → 9 × (10x + y). Twice the reversed number → 2 × (10y + x). So, the equation is:

   9(10x + y) = 2(10y + x)
   

Final Answer: The pair of equations is x + y = 9 and 9(10x + y) = 2(10y + x).

Tips & Tricks

Use these shortcuts to translate word problems faster and more accurately.

Common Mistakes to Avoid

Many students make small errors when setting up equations. Here's what to watch out for.

Brain-Teaser Questions

(Form the pair of linear equations for each. You don't need to solve them yet!)

1. A fraction becomes 9/11 if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes 5/6. What are the equations?

💡 Answer: Let the numerator be x and the denominator be y. The fraction is x/y. Eq 1: (x + 2) / (y + 2) = 9/11 Eq 2: (x + 3) / (y + 3) = 5/6

2. The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the equations.

💡 Answer: Let the length be l and the breadth be b. The area is A = l × b. Eq 1: (l - 5)(b + 3) = lb - 9 Eq 2: (l + 3)(b + 2) = lb + 67

3. Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. Set up the equations. (Assume Yash attempted all questions).

💡 Answer: Let the number of right answers be x and the number of wrong answers be y. Eq 1: 3x - 1y = 40 Eq 2: 4x - 2y = 50

Mini Cheatsheet

Screenshot this table for a quick revision of today's key concepts!

Graphical Method of Solution of a Pair of Linear Equations — Part 1

Chapter 3: Pair of Linear Equations in Two Variables

Page 2 of 5: Graphical Method of Solution of a Pair of Linear Equations — Part 1

Introduction to Solutions of Linear Equations

When we have a pair of linear equations in two variables, we are interested in finding values of both variables that satisfy both equations simultaneously. The graphical method provides a visual way to understand and solve such systems.

Types of Solutions

A pair of linear equations can have three possible outcomes:

1. Consistent Pair of Equations A pair of linear equations that has at least one solution is called a consistent pair. This means the two lines representing the equations meet at one or more points.

2. Inconsistent Pair of Equations A pair of linear equations that has no solution is called an inconsistent pair. This occurs when the two lines are parallel and never meet.

3. Dependent Pair of Equations A pair of linear equations that are equivalent and have infinitely many solutions is called a dependent pair. Note that a dependent pair is always consistent because it has solutions (infinitely many, in fact).

Graphical Representation and Solution Behavior

When we represent a pair of linear equations graphically, three situations can arise:

Case (i): Lines Intersect at a Single Point

• The pair of equations has a unique solution (exactly one solution)
• This represents a consistent pair of equations
• The point of intersection gives us the values of x and y

Case (ii): Lines are Parallel

• The equations have no solution
• This represents an inconsistent pair of equations
• The lines never meet, so there are no common points

Case (iii): Lines are Coincident

• The equations have infinitely many solutions
• This represents a dependent (consistent) pair of equations
• The lines lie on top of each other, so every point on one line is also on the other

Algebraic Conditions for Different Cases

Consider two linear equations in the general form:

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

We can determine the nature of solutions by comparing the ratios of coefficients:

Detailed Examples with Ratio Analysis

Let us examine three pairs of equations to understand these conditions:

Example Set 1: Intersecting Lines

• Equation 1: x - 2y = 0
• Equation 2: 3x + 4y - 20 = 0

Here: a₁ = 1, b₁ = -2, c₁ = 0 and a₂ = 3, b₂ = 4, c₂ = -20

Comparing ratios:

• a₁/a₂ = 1/3
• b₁/b₂ = -2/4 = -1/2

Since 1/3 ≠ -1/2, we have a₁/a₂ ≠ b₁/b₂

Conclusion: The lines intersect at exactly one point (unique solution).

Example Set 2: Coincident Lines

• Equation 1: 2x + 3y - 9 = 0
• Equation 2: 4x + 6y - 18 = 0

Here: a₁ = 2, b₁ = 3, c₁ = -9 and a₂ = 4, b₂ = 6, c₂ = -18

Comparing ratios:

• a₁/a₂ = 2/4 = 1/2
• b₁/b₂ = 3/6 = 1/2
• c₁/c₂ = -9/(-18) = 1/2

Since 1/2 = 1/2 = 1/2, we have a₁/a₂ = b₁/b₂ = c₁/c₂

Conclusion: The lines are coincident (infinitely many solutions).

Example Set 3: Parallel Lines

• Equation 1: x + 2y - 4 = 0
• Equation 2: 2x + 4y - 12 = 0

Here: a₁ = 1, b₁ = 2, c₁ = -4 and a₂ = 2, b₂ = 4, c₂ = -12

Comparing ratios:

• a₁/a₂ = 1/2
• b₁/b₂ = 2/4 = 1/2
• c₁/c₂ = -4/(-12) = 1/3

Since 1/2 = 1/2 ≠ 1/3, we have a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Conclusion: The lines are parallel (no solution).

Worked Example 1: Checking Consistency Graphically

Problem: Check graphically whether the pair of equations x + 3y = 6 and 2x - 3y = 12 is consistent. If so, solve them graphically.

Solution:

Step 1: Prepare a table of values for both equations.

For x + 3y = 6, we can write y = (6 - x)/3

For 2x - 3y = 12, we can write y = (2x - 12)/3

Step 2: Plot the points.

• For the first equation: Plot A(0, 2) and B(6, 0)
• For the second equation: Plot P(0, -4) and Q(3, -2)

Step 3: Draw the lines.

• Join points A and B to form line AB
• Join points P and Q to form line PQ

Step 4: Identify the point of intersection. The lines intersect at point B(6, 0).

Step 5: Write the solution. The solution is x = 6 and y = 0.

Conclusion: Since the lines intersect at a point, the given pair of equations is consistent with a unique solution: x = 6, y = 0.

Worked Example 2: Identifying Infinitely Many Solutions

Problem: Graphically, find whether the following pair of equations has no solution, unique solution, or infinitely many solutions:

• 5x - 8y + 1 = 0
• 3x - (24/5)y + 3/5 = 0

Solution:

Step 1: Multiply the second equation by 5/3 to simplify.

(5/3) × [3x - (24/5)y + 3/5 = 0] = 5x - 8y + 1 = 0

Step 2: Compare the equations. After multiplication, the second equation becomes identical to the first equation.

Step 3: Determine the nature of solutions. Since both equations represent the same line, the lines are coincident.

Conclusion: The pair of equations has infinitely many solutions. Every point on the line 5x - 8y + 1 = 0 is a solution to both equations.

Worked Example 3: Real-Life Application

Problem: Champa went to a sale to purchase pants and skirts. The number of skirts is two less than twice the number of pants. Also, the number of skirts is four less than four times the number of pants. How many pants and skirts did she buy?

Solution:

Step 1: Define variables. Let x = number of pants Let y = number of skirts

Step 2: Form equations from the given conditions.

• First condition: "skirts is two less than twice the pants" y = 2x - 2 ... (1)

• Second condition: "skirts is four less than four times the pants" y = 4x - 4 ... (2)

Step 3: Prepare tables for plotting.

For y = 2x - 2:

For y = 4x - 4:

Step 4: Plot and draw the lines. Plot the points and draw both lines on graph paper.

Step 5: Find the intersection point. The two lines intersect at the point (1, 0).

Step 6: Interpret the solution. x = 1 means she purchased 1 pair of pants y = 0 means she purchased 0 skirts

Verification:

• Check in equation (1): y = 2(1) - 2 = 0 ✓
• Check in equation (2): y = 4(1) - 4 = 0 ✓

Answer: Champa purchased 1 pair of pants and did not buy any skirts.

Key Points to Remember

1. The graphical method involves plotting both equations on the same coordinate plane
2. Always find at least two points for each line (though more points improve accuracy)
3. The intersection point (if it exists) represents the solution
4. Check your solution by substituting back into both original equations
5. Before plotting, compare the ratios a₁/a₂, b₁/b₂, and c₁/c₂ to predict the nature of the solution

Graphical Method of Solution of a Pair of Linear Equations — Part 2

Chapter 3: Pair of Linear Equations in Two Variables

Graphical Method and Conditions for Consistency

In the previous section, we learned how to represent a pair of linear equations graphically by drawing two straight lines. Now, we will delve deeper into what these graphs tell us about the solution of the pair of equations. The way these two lines interact on the graph reveals the nature of the solution.

Behaviour of Lines and Existence of Solutions

When we draw two lines on a Cartesian plane, only three possibilities can occur:

1. The lines intersect at a single point.
2. The lines are parallel and never intersect.
3. The lines are coincident, meaning one line lies exactly on top of the other.

Each of these geometric possibilities corresponds to a specific type of solution for the pair of linear equations. This leads us to two important definitions:

• Consistent Pair: A pair of linear equations that has at least one solution is called a consistent pair. This occurs when the lines are intersecting (one unique solution) or coincident (infinitely many solutions).

  • Dependent Pair: A pair of linear equations which are equivalent and have infinitely many common solutions is called a dependent pair. The lines representing such a pair are coincident. Note that a dependent pair is always consistent.

• Inconsistent Pair: A pair of linear equations that has no solution is called an inconsistent pair. This occurs when the lines representing the equations are parallel.

Condition for Consistency: Comparing Ratios of Coefficients

We can determine the graphical behaviour and the nature of the solutions of a pair of linear equations without even drawing the graph! We can do this by simply comparing the ratios of their coefficients.

Consider the general form of a pair of linear equations: a₁x + b₁y + c₁ = 0 a₂x + b₂y + c₂ = 0

Here, a₁, b₁, c₁ are the coefficients and constant term of the first equation, and a₂, b₂, c₂ are for the second equation.

The relationship between their graphs and solutions can be summarized in the table below.

This table is a powerful tool. By calculating these simple ratios, you can predict whether a pair of equations has one solution, no solution, or infinitely many solutions.

Worked Examples

Let's apply these concepts to solve some problems.

Example 1: Solving Graphically

Check graphically whether the pair of equations is consistent. If so, solve them graphically. Equation (1): x + 3y = 6 Equation (2): 2x – 3y = 12

Solution:

Step 1: Check for consistency using the ratio method. First, let's write the equations in the general form a₁x + b₁y + c₁ = 0. x + 3y – 6 = 0 2x – 3y – 12 = 0

Here, a₁ = 1, b₁ = 3, c₁ = -6 and a₂ = 2, b₂ = -3, c₂ = -12

Now, compare the ratios: a₁/a₂ = 1/2 b₁/b₂ = 3/(-3) = -1

Since a₁/a₂ ≠ b₁/b₂ (because 1/2 ≠ -1), we can predict that the lines will be intersecting. This means the pair of equations has a unique solution and is therefore consistent.

Step 2: Find points to plot the graphs. To draw the graphs, we need at least two points for each line.

For Equation (1): x + 3y = 6 (or y = (6 – x)/3)

For Equation (2): 2x – 3y = 12 (or y = (2x – 12)/3)

Step 3: Plot the points and find the solution. Plot the points A, B, P, and Q on a graph paper and draw the lines passing through them.

We observe that the two lines intersect at point B(6, 0). This point of intersection is the solution to the pair of equations.

Answer: The solution is x = 6, y = 0. The pair of equations is consistent.

Example 2: Identifying Coincident Lines

Graphically, find whether the following pair of equations has no solution, a unique solution, or infinitely many solutions: Equation (1): 5x – 8y + 1 = 0 Equation (2): 3x – (24/5)y + 3/5 = 0

Solution:

Step 1: Compare the ratios of the coefficients. Here, a₁ = 5, b₁ = -8, c₁ = 1 and a₂ = 3, b₂ = -24/5, c₂ = 3/5

Let's compute the ratios: a₁/a₂ = 5 / 3 b₁/b₂ = (-8) / (-24/5) = (-8) × (5/-24) = 40/24 = 5/3 c₁/c₂ = 1 / (3/5) = 1 × (5/3) = 5/3

Step 2: Analyse the ratios. We see that a₁/a₂ = b₁/b₂ = c₁/c₂ (since all ratios are equal to 5/3).

According to our table, this condition means the lines are coincident. Coincident lines overlap completely, so every point on the line is a solution.

Answer: The pair of equations has infinitely many solutions. The pair is dependent and consistent.

Note: If you multiply Equation (2) by 5/3, you get 5x – 8y + 1 = 0, which is identical to Equation (1). This confirms they are the same line.

Example 3: Word Problem

Champa went to a sale to purchase some pants and skirts. The number of skirts she bought is two less than twice the number of pants. Also, the number of skirts is four less than four times the number of pants. Find how many pants and skirts Champa bought.

Solution:

Step 1: Formulate the pair of linear equations. Let the number of pants purchased be x. Let the number of skirts purchased be y.

From the first condition: "The number of skirts (y) is two less than twice the number of pants (2x)". Equation (1): y = 2x – 2

From the second condition: "The number of skirts (y) is four less than four times the number of pants (4x)". Equation (2): y = 4x – 4

Step 2: Find points to plot the graphs.

For Equation (1): y = 2x – 2

For Equation (2): y = 4x – 4

Step 3: Plot the lines and find the intersection. Draw the lines using the points above. You will find that the two lines intersect at a specific point. Let's find that point by solving. The lines appear to intersect at the point (1, 0).

Step 4: Verify the solution. Let's check if (x=1, y=0) satisfies both equations. In Eq (1): 0 = 2(1) – 2 → 0 = 2 – 2 → 0 = 0. (True) In Eq (2): 0 = 4(1) – 4 → 0 = 4 – 4 → 0 = 0. (True) The solution is correct.

Answer: Champa bought 1 pant (x=1) and 0 skirts (y=0).

Substitution Method

Substitution Method

Concept Introduction

Have you ever noticed how supermarkets price their combo offers? Imagine a shop where 2 pencils and 3 erasers cost ₹9, while 4 pencils and 6 erasers cost ₹18. If you tried solving this problem graphically by plotting two lines, you'd find both equations represent the same line — meaning infinitely many solutions exist!

The substitution method is an algebraic technique that helps us solve such systems without drawing graphs. Instead of plotting points and hoping they intersect clearly, we express one variable in terms of the other and substitute it into the second equation. This transforms a two-variable problem into a single-variable equation that we can solve directly.

This method is particularly powerful when one equation is already isolated (or easily isolatable) for a variable. Real-world applications include calculating ages from relationship clues, determining ticket prices from sales data, and mixing solutions in chemistry labs. Let's master this fundamental algebraic tool step by step.

{{FORMULA: expr=x = (expression in y) | symbols=x:first variable, y:second variable}}

Definitions & Formulas

Derivation / Logic

The substitution method works through a systematic transformation process:

Step 1: Choose the simpler equation (or the one where a variable has coefficient 1 or -1) from the given pair.

Step 2: Isolate one variable (usually the one with the simplest coefficient) in terms of the other. For example, from x + 2y = 3, express x as x = 3 - 2y.

Step 3: Substitute this expression into the other equation. This eliminates one variable, reducing the system to a single-variable linear equation.

Step 4: Solve the resulting single-variable equation using basic algebraic operations (transpose, simplify, divide).

Step 5: Substitute the obtained value back into the expression from Step 2 to find the value of the second variable.

Step 6: Verify the solution by substituting both values into both original equations. If both equations are satisfied, the solution is correct.

{{KEY: type=concept | title=Why Substitution Works | text=Substitution works because we're maintaining equality throughout. If x = 3 - 2y is true, then replacing every x with (3 - 2y) preserves the truth of the equation. This transforms two equations in two unknowns into one equation in one unknown.}}

Solved Examples

Example 1: Basic Substitution (Easy)

Given: 7x - 15y = 2 and x + 2y = 3

To Find: Values of x and y

Solution:

1. From the second equation, isolate x:

x = 3 - 2y

2. Substitute this expression for x into the first equation:

7(3 - 2y) - 15y = 2

3. Expand and simplify:

21 - 14y - 15y = 2

21 - 29y = 2

4. Solve for y:

-29y = 2 - 21

-29y = -19

y = 19/29

5. Substitute y = 19/29 back into x = 3 - 2y:

x = 3 - 2(19/29)

x = 3 - 38/29

x = 87/29 - 38/29

x = 49/29

6. Verify in both original equations (verification confirms both are satisfied).

Final Answer: x = 49/29, y = 19/29

Example 2: Age Problem (Medium)

Given: Seven years ago, Aftab was seven times as old as his daughter. Three years from now, he will be three times as old as she will be.

To Find: Present ages of Aftab and his daughter

Solution:

1. Let s = Aftab's present age and t = daughter's present age. Seven years ago: s - 7 = 7(t - 7):

s - 7 = 7t - 49

s - 7t = -42

s = 7t - 42 ... (1)

2. Three years from now: s + 3 = 3(t + 3):

s + 3 = 3t + 9

s = 3t + 6 ... (2)

3. From equation (2), we already have s isolated. Substitute into equation (1):

3t + 6 = 7t - 42

4. Solve for t:

6 + 42 = 7t - 3t

48 = 4t

t = 12

5. Substitute t = 12 into equation (2):

s = 3(12) + 6

s = 36 + 6

s = 42

6. Verification: Seven years ago: 42 - 7 = 35, 12 - 7 = 5, and 35 = 7 × 5 ✓. Three years later: 42 + 3 = 45, 12 + 3 = 15, and 45 = 3 × 15 ✓.

Final Answer: Aftab's age = 42 years, Daughter's age = 12 years

Example 3: Dependent System (Hard)

Given: 2x + 3y = 9 and 4x + 6y = 18

To Find: Cost of pencils and erasers (if unique solution exists)

Solution:

1. From the first equation, isolate x:

2x = 9 - 3y

x = (9 - 3y)/2

2. Substitute into the second equation:

4[(9 - 3y)/2] + 6y = 18

3. Simplify:

2(9 - 3y) + 6y = 18

18 - 6y + 6y = 18

18 = 18

4. This is a true statement for all values of y. The variable y has been eliminated, resulting in an identity.

5. This means the two equations represent the same line. Every point on the line 2x + 3y = 9 is a solution.

6. Conclusion: The system has infinitely many solutions. We cannot determine unique costs.

Final Answer: Infinitely many solutions (dependent system)

{{KEY: type=warning | title=Recognizing Dependent Systems | text=When substitution leads to a true statement like 18 = 18 or 0 = 0 with no variables remaining, the system has infinitely many solutions. The equations are proportional and represent the same geometric line.}}

Example 4: Inconsistent System (Tricky)

Given: Two railway tracks represented by x + 2y - 4 = 0 and 2x + 4y - 12 = 0

To Find: Will the rails cross each other?

Solution:

1. Rewrite in standard form. First equation: x + 2y = 4. Second equation: 2x + 4y = 12.

2. From the first equation, isolate x:

x = 4 - 2y

3. Substitute into the second equation:

2(4 - 2y) + 4y = 12

4. Expand and simplify:

8 - 4y + 4y = 12

8 = 12

5. This is a false statement. No value of y can make this true.

6. Conclusion: The system has no solution (inconsistent). The rails are parallel and will never cross.

Final Answer: No solution (rails are parallel and never cross)

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Question 1: A pair of equations 3x - y = 3 and 9x - 3y = 9 is given. Without fully solving, determine the nature of the solution set and explain why.

💡 Answer: The second equation is exactly 3 times the first equation (multiply 3x - y = 3 by 3 to get 9x - 3y = 9). Since both equations represent the same line, the system has infinitely many solutions. Every point satisfying 3x - y = 3 automatically satisfies the second equation.

Question 2: Solve the system: s/3 - t/2 = 6 and s - t = 3. Find s + t.

💡 Answer: From the second equation: s = 3 + t. Substitute into the first equation: (3 + t)/3 - t/2 = 6 Multiply through by 6: 2(3 + t) - 3t = 36 6 + 2t - 3t = 36 -t = 30 → t = -30 Then s = 3 + (-30) = -27 Therefore, s + t = -27 + (-30) = -57

Question 3: A system has equations 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3. If you multiply the first equation by 5 and the second by 10 to clear decimals before substitution, what values do you get?

💡 Answer: First equation × 5: x + 1.5y = 6.5 → multiply by 2 → 2x + 3y = 13 Second equation × 10: 4x + 5y = 23 From 2x + 3y = 13: x = (13 - 3y)/2 Substitute into 4x + 5y = 23: 4[(13 - 3y)/2] + 5y = 23 2(13 - 3y) + 5y = 23 26 - 6y + 5y = 23 -y = -3 → y = 3 Then x = (13 - 9)/2 = 2 Answer: x = 2, y = 3

Mini Cheatsheet

Pro Tip for Exams: When dealing with word problems, always define your variables clearly at the start (Let x = ...). This prevents confusion during substitution and makes verification straightforward. The substitution method is especially powerful when coefficients are simple (±1, ±2) or when one equation is already partially solved.

Summary & Quick Revision

Chapter 3: Pair of Linear Equations in Two Variables — Summary & Quick Revision

Welcome to the final page of this chapter! We have explored various ways to represent and solve pairs of linear equations. This page serves as a comprehensive summary and a quick revision guide to consolidate your understanding and prepare you for examinations.

1. What is a Pair of Linear Equations in Two Variables?

A pair of linear equations in two variables, say x and y, is a set of two equations that can be written in the general form:

Equation 1: a₁x + b₁y + c₁ = 0 Equation 2: a₂x + b₂y + c₂ = 0

Here, a₁, b₁, c₁, a₂, b₂, and c₂ are real numbers, such that a₁² + b₁² ≠ 0 and a₂² + b₂² ≠ 0. The solution to this pair of equations is a pair of values (x, y) that satisfies both equations simultaneously.

2. Methods of Solving

We learned two primary approaches to find the solution: the Graphical Method and the Algebraic Method.

A. Graphical Method

In this method, we represent each linear equation as a straight line on a Cartesian plane. The relationship between these two lines tells us about the solution of the pair of equations.

Think of the solution as the common point(s) between the two lines. If they meet at one point, there's one solution. If they are the same line, all their points are common. If they never meet, there are no common points.

B. Algebraic Methods

These methods provide a precise way to find the solution without drawing a graph.

I. Substitution Method

This method involves expressing one variable in terms of the other from one equation and substituting this value into the second equation.

Steps:

1. Take one of the equations and express one variable (e.g., y) in terms of the other (x).
2. Substitute this expression for y into the other equation. This will create a linear equation in just one variable (x).
3. Solve this equation to find the value of x.
4. Substitute the value of x back into the expression from Step 1 to find the value of y.

Worked Example: Solve the pair of equations: x + 2y = 3 ---(1) 7x - 15y = 2 ---(2)

• Step 1: From equation (1), express x in terms of y: x = 3 - 2y ---(3)

• Step 2: Substitute this expression for x into equation (2): 7(3 - 2y) - 15y = 2

• Step 3: Solve for y: 21 - 14y - 15y = 2 21 - 29y = 2 -29y = 2 - 21 -29y = -19 y = 19/29

• Step 4: Substitute y = 19/29 into equation (3) to find x: x = 3 - 2(19/29) x = 3 - 38/29 x = (87 - 38) / 29 x = 49/29

Solution: The solution is (x = 49/29, y = 19/29).

II. Elimination Method

This method involves eliminating one of the variables by making its coefficients equal in both equations and then adding or subtracting the equations.

Steps:

1. Multiply one or both equations by suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal.
2. Add or subtract the equations to eliminate that variable. This results in an equation with a single variable.
3. Solve the resulting equation to find the value of that variable.
4. Substitute this value into either of the original equations to find the value of the other variable.

Worked Example: Solve the same pair of equations: x + 2y = 3 ---(1) 7x - 15y = 2 ---(2)

• Step 1: Let's eliminate x. To make the coefficients of x equal, multiply equation (1) by 7. 7(x + 2y) = 7(3) 7x + 14y = 21 ---(3)

• Step 2: Now, subtract equation (2) from the new equation (3). (7x + 14y) - (7x - 15y) = 21 - 2 7x + 14y - 7x + 15y = 19 29y = 19

• Step 3: Solve for y: y = 19/29

• Step 4: Substitute y = 19/29 into the original equation (1): x + 2(19/29) = 3 x + 38/29 = 3 x = 3 - 38/29 x = (87 - 38) / 29 x = 49/29

Solution: The solution is (x = 49/29, y = 19/29), which matches the result from the substitution method.

3. Conditions for Consistency of a Pair of Linear Equations

Without solving the equations, we can predict the nature of their solution by comparing the ratios of the coefficients of the variables and the constant terms.

For the general pair of equations: a₁x + b₁y + c₁ = 0 a₂x + b₂y + c₂ = 0

We compare the ratios a₁/a₂, b₁/b₂, and c₁/c₂.

Worked Example: Check the consistency of the following pairs of equations.

a) 2x + 3y - 7 = 0 and 6x + 5y - 11 = 0

• Here, a₁=2, b₁=3, a₂=6, b₂=5.
• a₁/a₂ = 2/6 = 1/3
• b₁/b₂ = 3/5
• Since a₁/a₂ ≠ b₁/b₂ (1/3 ≠ 3/5), the pair of equations is consistent with a unique solution.

b) x + 2y - 4 = 0 and 2x + 4y - 12 = 0

• Here, a₁=1, b₁=2, c₁=-4, a₂=2, b₂=4, c₂=-12.
• a₁/a₂ = 1/2
• b₁/b₂ = 2/4 = 1/2
• c₁/c₂ = -4 / -12 = 1/3
• Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (1/2 = 1/2 ≠ 1/3), the pair of equations is inconsistent with no solution.

c) 3x - y - 5 = 0 and 9x - 3y - 15 = 0

• Here, a₁=3, b₁=-1, c₁=-5, a₂=9, b₂=-3, c₂=-15.
• a₁/a₂ = 3/9 = 1/3
• b₁/b₂ = -1/-3 = 1/3
• c₁/c₂ = -5/-15 = 1/3
• Since a₁/a₂ = b₁/b₂ = c₁/c₂ (1/3 = 1/3 = 1/3), the pair of equations is dependent and consistent with infinitely many solutions.

4. Equations Reducible to a Linear Pair

Some equations are not linear in their original form but can be transformed into a linear pair by making a suitable substitution. This often occurs when variables are in the denominator.

Example: Consider the equations: 2/x + 3/y = 13 5/x - 4/y = -2

These are not linear. However, if we make the substitution: Let p = 1/x and q = 1/y.

The equations transform into: 2p + 3q = 13 5p - 4q = -2

This is now a standard pair of linear equations in variables p and q. We can solve for p and q using any algebraic method. Once we find the values of p and q, we must remember to substitute back to find x and y: x = 1/p and y = 1/q.

This technique is powerful for solving a wider variety of problems that can be modeled by reducible equations.

Quick Revision Checklist

• Can I identify the coefficients a₁, b₁, c₁, a₂, b₂, c₂ from any given pair of linear equations?
• Can I solve a pair of linear equations using the Substitution Method?
• Can I solve a pair of linear equations using the Elimination Method?
• Can I use the ratio of coefficients (a₁/a₂, b₁/b₂, c₁/c₂) to determine if a pair of equations will have a unique solution, no solution, or infinitely many solutions?
• Do I know the difference between consistent, inconsistent, and dependent pairs of equations?
• Can I identify and solve equations that are reducible to a linear pair?

This chapter provides foundational skills for solving systems of equations, a concept that is widely used in science, engineering, and economics. Mastering these methods will serve you well in higher mathematics. Happy revising