Mathematics : Chapter 1: Pairs of Equations

 
🧩 Topic 1: Form
ing and Solving Linear Equation Pairs

Concept:
In many problems, we are asked to find two unknown quantities based on two conditions. These can be solved by forming and solving a pair of linear equations.
🔍 Example 1: Beads Problem

Problem: A box contains 100 beads. There are 10 more black beads than white beads. Find the number of each.

Step-by-step:

    Let number of black beads = x

    Then white beads = x − 10

    Total beads = x + (x − 10) = 100
    → 2x − 10 = 100

    Solve:
    2x = 110
    x = 55 (black),
    white = 55 − 10 = 45

✅ Answer: Black = 55, White = 45
🔍 Example 2: Table & Chair

Problem: A table and a chair cost ₹11000 together. The table and four chairs cost ₹14000. Find the cost of each.

Step-by-step:

    Let chair = x → table = 11000 − x

    Then: (11000 − x) + 4x = 14000
    → 11000 + 3x = 14000
    → 3x = 3000 → x = 1000

    Table = 11000 − 1000 = 10000

✅ Answer: Chair = ₹1000, Table = ₹10000
🔍 Example 3: Number Relation

Problem: A number is 5 times another number. Their difference is 32. Find both numbers.

Step-by-step:

    Let smaller number = x → larger = 5x

    5x − x = 32
    → 4x = 32 → x = 8

    Larger = 5 × 8 = 40

✅ Answer: 8 and 40
🧩 Topic 2: Solving Using Substitution Method

Concept: Solve one equation for one variable, substitute into the other.
🔍 Example 1: Chairs and Tables (Revisit)

    x + y = 11000 → y = 11000 − x

    Substitute into:
    4x + y = 14000
    → 4x + (11000 − x) = 14000
    → 3x + 11000 = 14000
    → x = 1000

✅ Same Result
🔍 Example 2: Fractions Problem

Problem: A fraction becomes ½ when 1 is added to numerator. It becomes ⅓ when 1 is added to denominator.

Step-by-step:

    Let fraction = x/y

    Given: (x+1)/y = ½ → y = 2(x+1)

    Also: x/(y+1) = ⅓ → y + 1 = 3x

    Now substitute:
    2(x + 1) + 1 = 3x
    → 2x + 2 + 1 = 3x
    → 3 = x → x = 3

    y = 2(3 + 1) = 8

✅ Answer: Fraction = 3/8
🔍 Example 3: Investment

Problem: Total ₹1,00,000 invested in two schemes at 6% and 7%. Interest received = ₹6750.

Step-by-step:

    Let amount in 6% scheme = x
    → in 7% scheme = 100000 − x

    Interest: 0.06x + 0.07(100000 − x) = 6750

    Simplify:
    0.06x + 7000 − 0.07x = 6750
    → −0.01x = −250 → x = 25000

    So, ₹25000 at 6%, ₹75000 at 7%

✅ Answer: ₹25000 (6%), ₹75000 (7%)
🧩 Topic 3: Solving by Elimination Method
🔍 Example 1: Pen and Notebook

Problem:
2 pens + 3 notebooks = ₹110
2 pens + 5 notebooks = ₹170

Step-by-step:

    Let pen = x, notebook = y
    → 2x + 3y = 110 (i)
    → 2x + 5y = 170 (ii)

    Subtract:
    (ii) − (i): 2y = 60 → y = 30

    Substitute in (i):
    2x + 3(30) = 110 → x = 10

✅ Answer: Pen = ₹10, Notebook = ₹30
🔍 Example 2: Pencils and Pens

Problem:
3 pencils + 4 pens = ₹66
6 pencils + 3 pens = ₹72

Step-by-step:

    Let pencil = x, pen = y
    → 3x + 4y = 66 (i)
    → 6x + 3y = 72 (ii)

    Multiply (i) × 2: 6x + 8y = 132
    Subtract from (ii):
    (6x + 8y) − (6x + 3y) → 5y = 60 → y = 12

    Substitute:
    3x + 4(12) = 66 → x = 6

✅ Answer: Pencil = ₹6, Pen = ₹12
🔍 Example 3: Vessel Capacity

Problem:
5 small + 2 large vessels = 20L
2 small + 3 large = 19L

Step-by-step:

    5x + 2y = 20 (i)
    2x + 3y = 19 (ii)

    Multiply: (i) × 2 → 10x + 4y = 40
    (ii) × 5 → 10x + 15y = 95

    Subtract: 11y = 55 → y = 5
    Substitute in (i):
    5x + 10 = 20 → x = 2

✅ Answer: Small = 2L, Large = 5L
🧩 Topic 4: Number Puzzles (Quick Solving)
🔍 Example 1: Sum and Difference

Problem: Two numbers sum to 28, and difference is 12. Find them.

Step-by-step:

    x + y = 28, x − y = 12
    → Add: 2x = 40 → x = 20
    → Subtract: 2y = 16 → y = 8

✅ Answer: 20 and 8
🔍 Example 2: Digit Problem

Problem: A two-digit number has digits adding to 11. Reversing digits increases it by 27.

Step-by-step:

    Let tens digit = x, units = y
    → Number = 10x + y
    → Reversed = 10y + x
    → x + y = 11
    → 10y + x = 10x + y + 27
    → Simplify: 9y − 9x = 27 → y − x = 3

    x + y = 11, y − x = 3
    Add: 2y = 14 → y = 7 → x = 4
    → Number = 47

✅ Answer: 47
🔍 Example 3: Triangle Sides

Problem: Difference of two angles = 20°, and third is 90°. Find all angles.

Step-by-step:

    Let angles be x and x + 20
    → Third = 90°
    → x + x + 20 + 90 = 180 → 2x = 70 → x = 35
    → Other = 55

✅ Answer: 35°, 55°, 90°





🔍 Summary

This chapter introduces students to the concept of solving problems involving two unknowns using pairs of linear equations. Through relatable word problems—like pricing, measurements, and number puzzles—students learn to set up and solve equations either algebraically or through logical reasoning. It highlights different strategies: mental math, substitution, elimination, and how real-life problems can be mathematically modeled. The chapter encourages flexible thinking, showing that the same problem can often be solved in more than one way. The use of Computer Algebra Systems (CAS) like GeoGebra is briefly introduced to solve these equations digitally. By the end of the chapter, students understand how to form and solve linear equation pairs and apply this knowledge to practical and abstract problems.
📌 Capsule Notes
🧮 Basic Concepts

    Many real-life problems involve finding two unknown values.

    Such problems can be solved by forming two equations with two variables.

💡 Approaches to Solve

    Mental math: Logical deduction without algebra.

    Substitution: Replace one variable with an expression from another equation.

    Elimination: Add or subtract equations to eliminate one variable.

    Trial and error: Useful in simple scenarios.

🧾 Algebraic Representation

    Assign variables (e.g., x = price of chair, y = price of table).

    Form two equations based on the given information.

    Solve the pair using substitution or elimination.

💻 Digital Tools

    GeoGebra, Maxima, and SageMath can solve equation pairs using the CAS (Computer Algebra System).

    Example GeoGebra command:
    Solve({5x + 2y = 20, 2x + 3y = 19}, {x, y})

❓ Questions with Answers (Q&A)
🟡 MCQs

    What type of system is used when solving two equations with two variables?

        a) Linear equation

        b) Simultaneous equations ✅

        c) Quadratic equation

        d) Inequality system

    If x + y = 10 and x − y = 2, what is x?

        a) 5

        b) 6 ✅

        c) 8

        d) 4

🟢 Short Answer

    What is the method of elimination in solving equations?
    Ans: Eliminating one variable by adding or subtracting the equations.

    Define substitution method.
    Ans: Replacing one variable in an equation with its equivalent expression from another.

    What is the solution to the equations:
    2x + 3y = 12 and x + y = 5?
    Ans: x = 3, y = 2

    Solve:
    3x + 4y = 66 and 6x + 3y = 72
    Ans: x = 6, y = 12

🔵 Application-Based

    A bag and slippers cost ₹1100. Bag costs ₹300 more. Find their prices.
    Ans: Slippers = ₹400, Bag = ₹700

    The sum of two numbers is 26, difference is 4. Find the numbers.
    Ans: 15 and 11

    A person invested ₹100000 in two schemes at 6% and 7%. Total interest is ₹6750. Find the investment in each.
    Ans: ₹25000 at 6%, ₹75000 at 7%

    Speed at t = 1s is 5 m/s and at t = 5s is 13 m/s. Find initial speed and acceleration.
    Ans: Initial speed u = 3 m/s, acceleration a = 2 m/s²

📚 Definitions & Key Terms
Term    Definition
Linear Equation    An equation of the first degree involving one or more variables. Example: 2x + 3y = 10
Simultaneous Equations    A set of equations with multiple variables solved together to find a common solution.
Substitution Method    Solving one equation for one variable and replacing it in the other.
Elimination Method    Adding or subtracting equations to eliminate one variable.
CAS (Computer Algebra System)    Software used to solve algebraic equations digitally. Examples: GeoGebra, Maxima.
🧠 Main Points to Remember (For Revision)

    🔹 2 equations with 2 unknowns → use substitution or elimination.

    🔹 Always assign variables clearly based on the problem.

    🔹 Check your solution by substituting into both equations.

    🔹 Sum + Difference of numbers → useful shortcut:

        x + y = A, x − y = B →

            x = (A + B)/2

            y = (A − B)/2

    🔹 Parallel lines, proportions, and geometry can also involve pairs of equations.

    🔹 Equations involving rates (e.g., speed, interest) often lead to two-variable systems.

📂 Topic-Wise Breakdown
🧩 1. Introduction to Equation Pairs

    Real-life examples like bead counts, table-chair pricing.

    Logical reasoning & visualization.

🧩 2. Algebraic Methods

    Represent problems using x and y.

    Two main solving methods: substitution & elimination.

🧩 3. Practical Applications

    Age problems, investment returns, speed-time-distance, geometric measurements.

🧩 4. Visual Problem Solving

    Tables, diagrams, proportional lines.

    Extending equations to geometry problems.

🧩 5. Using Technology (CAS Tools)

    GeoGebra's CAS helps solve equation pairs.

    Command: Solve({eq1, eq2}, {x, y})