Linear Equations Notes

Linear Equations Notes : Formulas, Theorems & Examples (Class 10 Maths)

Ch 3 : PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

LESSON OBJECTIVES:

  1. Framing Equations: Learn to create pairs of linear equations in two variables from real-life situations and word problems.
  2. Graphical Solutions: Solve pairs of linear equations graphically and understand the geometric interpretations:
    • Intersecting Lines (Unique Solution)
    • Parallel Lines (No Solution)
    • Coincident Lines (Infinitely Many Solutions)
    • Consistent and Inconsistent Systems
  3. Algebraic Solutions: Master algebraic methods to solve pairs of linear equations:
    • Substitution Method
    • Elimination Method
    • Cross-Multiplication Method

CONTENTS OF LESSON NOTES:

1. LINEAR EQUATIONS:

  • DEFINITION: A linear equation in two variables is an equation that can be written in the form ax + by + c = 0, where:
    • x and y are the variables.
    • a, b, and c are real numbers, and a and b are not both zero simultaneously. (Meaning at least one of ‘a’ or ‘b’ must be non-zero).
    • a is the coefficient of x, b is the coefficient of y, and c is the constant term.
  • WHY “LINEAR”? The term “linear” comes from the fact that when you graph a linear equation in two variables, it always represents a straight line.
  • EXAMPLES:
    • 2x + 3y – 6 = 0 (Here, a=2, b=3, c=-6)
    • x – y = 5 (Can be rewritten as 1x – 1y – 5 = 0, so a=1, b=-1, c=-5)
    • y = 2x + 1 (Can be rewritten as -2x + 1y – 1 = 0, so a=-2, b=1, c=-1)
    • 3x = 7 (Can be rewritten as 3x + 0y – 7 = 0, so a=3, b=0, c=-7. Note ‘b’ can be zero, but ‘a’ is not zero in this case)
    • 2y = -4 (Can be rewritten as 0x + 2y + 4 = 0, so a=0, b=2, c=4. Note ‘a’ can be zero, but ‘b’ is not zero in this case)

2. PAIR OF LINEAR EQUATIONS:

  • DEFINITION: A pair of linear equations in two variables is a set of two linear equations that involve the same two variables (usually x and y).
  • WHY “PAIR”? We need two equations to solve for two unknowns (variables). One linear equation alone has infinitely many solutions.
  • GENERAL FORM: A pair of linear equations in two variables can be represented as:
    • a₁x + b₁y + c₁ = 0 (Equation 1)
    • a₂x + b₂y + c₂ = 0 (Equation 2)
    • Where a₁, b₁, c₁, a₂, b₂, c₂ are real numbers, and a₁ and b₁ are not both zero, and a₂ and b₂ are not both zero.
  • EXAMPLES:
      • Equation 1: x + y = 10
      • Equation 2: x – y = 4
      • Equation 1: 2x – 3y = 7
      • Equation 2: 4x + y = -2

3. GRAPHICAL METHOD OF SOLUTION OF A PAIR OF LINEAR EQUATIONS:

  • GEOMETRIC INTERPRETATION: Each linear equation in a pair represents a straight line on a graph. When we solve a pair of linear equations graphically, we are essentially finding the point(s) of intersection of these two lines.
  • POSSIBLE CASES: When you graph two lines in a plane, there are three possibilities:

(a) INTERSECTING LINES:

    • Graphical Representation: The two lines intersect each other at exactly one point.
    • Number of Solutions: There is exactly one solution (unique solution). The coordinates of the point of intersection (x, y) give the solution to the pair of equations.
    • Consistent Pair: Such a pair of equations is called a consistent pair of equations because it has at least one solution.

graph

(b) COINCIDENT LINES:

    • Graphical Representation: The two lines coincide or overlap each other completely. This means they are essentially the same line.
    • Number of Solutions: There are infinitely many solutions. Every point on the line is a solution to both equations.
    • Dependent and Consistent Pair: Such a pair of equations is called a dependent and consistent pair of equations. “Consistent” because there are solutions, and “dependent” because one equation can be derived from the other (they are essentially the same).
    • coincident_lines

(c) PARALLEL LINES:

    • Graphical Representation: The two lines are parallel to each other. They never intersect.
    • Number of Solutions: There is no solution. There is no point that lies on both lines simultaneously.
    • Inconsistent Pair: Such a pair of equations is called an inconsistent pair of equations because it has no solution.
    • parallel_lines

4. ALGEBRAIC METHODS OF SOLVING A PAIR OF LINEAR EQUATIONS:

(a) SUBSTITUTION METHOD:

  • PROCESS: In this method, we solve for one variable in terms of the other from one equation and then substitute this expression into the second equation to solve for the remaining variable.
  • STEPS:
  1. Step 1: Express one variable in terms of the other: Choose either equation and solve for one variable (say, y) in terms of the other variable (x). Pick the equation that looks simpler to manipulate.
  2. Step 2: Substitute: Substitute this expression for y (in terms of x) into the other This will result in an equation in just one variable (x).
  3. Step 3: Solve for x: Solve the new linear equation in x to find the value of x.
  4. Step 4: Substitute back to find y: Substitute the value of x you just found back into either of the original equations (or the expression for y in terms of x from Step 1) to find the value of y.
  5. Step 5: Check your solution (optional but recommended): Substitute the values of x and y you found into both original equations to verify that they satisfy both equations.
  • EXAMPLE: Solve the pair of equations:
    • x + y = 14 (Equation 1)
    • x – y = 4 (Equation 2)
  1. Step 1: From Equation 1, express y in terms of x: y = 14 – x
  2. Step 2: Substitute this into Equation 2: x – (14 – x) = 4
  3. Step 3: Solve for x: x – 14 + x = 4 => 2x – 14 = 4 => 2x = 18 => x = 9
  4. Step 4: Substitute x = 9 back into y = 14 – x: y = 14 – 9 = 5
  5. Solution: x = 9, y = 5 or (9, 5)

(b) ELIMINATION METHOD:

  • PROCESS: In this method, we eliminate one variable by making its coefficients equal in both equations and then adding or subtracting the equations.
  • STEPS:
  1. Step 1: Make coefficients equal: Multiply one or both equations by suitable non-zero constants so that the coefficients of one of the variables (either x or y) become numerically equal.
  2. Step 2: Eliminate a variable:
    • Add the equations if the coefficients of the variable you want to eliminate have opposite signs.
    • Subtract the equations if the coefficients of the variable you want to eliminate have the same sign.

This will eliminate one variable, resulting in an equation in just one variable.

  1. Step 3: Solve for the remaining variable: Solve the new linear equation in one variable to find its value.
  2. Step 4: Substitute back to find the other variable: Substitute the value you just found back into either of the original equations to find the value of the other variable.
  3. Step 5: Check your solution (optional but recommended): Substitute the values of x and y into both original equations to verify.
  • EXAMPLE: Solve the pair of equations:
    • 3x + 4y = 10 (Equation 1)
    • 2x – 2y = 2 (Equation 2)
  1. Step 1: Make coefficients of y equal: Multiply Equation 2 by 2: 4x – 4y = 4 (New Equation 2′) Now, coefficients of y are +4 and -4 in Equation 1 and New Equation 2′.
  2. Step 2: Eliminate y (by adding): Add Equation 1 and New Equation 2′: (3x + 4y) + (4x – 4y) = 10 + 4 => 7x = 14
  3. Step 3: Solve for x: 7x = 14 => x = 2
  4. Step 4: Substitute x = 2 into Equation 2 (or Equation 1): 2(2) – 2y = 2 => 4 – 2y = 2 => -2y = -2 => y = 1
  5. Solution: x = 2, y = 1 or (2, 1)

(c) CROSS-MULTIPLICATION METHOD:

(Formula-based method, less commonly used but in syllabus)

  • For a pair of linear equations:
  • a₁x + b₁y + c₁ = 0
  • a₂x + b₂y + c₂ = 0
  • The solution is given by:

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

  • To find x: x = (b₁c₂ – b₂c₁) / (a₁b₂ – a₂b₁)
  • To find y: y = (c₁a₂ – c₂a₁) / (a₁b₂ – a₂b₁)
  • Condition for Unique Solution: This method works only when (a₁b₂ – a₂b₁) ≠ 0. If (a₁b₂ – a₂b₁) = 0, then the pair of equations either has no solution or infinitely many solutions, and this method cannot be directly applied to find a unique solution.
  • EXAMPLE: Solve using cross-multiplication:
  • 2x + y = 5 (=> 2x + y – 5 = 0) (Equation 1)
  • 3x + 2y = 8 (=> 3x + 2y – 8 = 0) (Equation 2)
  • Here, a₁=2, b₁=1, c₁=-5, a₂=3, b₂=2, c₂=-8
  • x = (b₁c₂ – b₂c₁) / (a₁b₂ – a₂b₁) = (1*(-8) – 2*(-5)) / (2*2 – 3*1) = (-8 + 10) / (4 – 3) = 2 / 1 = 2
  • y = (c₁a₂ – c₂a₁) / (a₁b₂ – a₂b₁) = ((-5)*3 – (-8)*2) / (2*2 – 3*1) = (-15 + 16) / (4 – 3) = 1 / 1 = 1
  • Solution: x = 2, y = 1 or (2, 1)

5. INTERPRETATION OF THE PAIR OF EQUATIONS (Based on Coefficient Ratios):

  • For a pair of linear equations:
  • a₁x + b₁y + c₁ = 0
  • a₂x + b₂y + c₂ = 0
  • RATIO COMPARISON and INTERPRETATION TABLE:

Ratio Comparison

Graphical Representation

Algebraic Interpretation

Consistency

a₁/a₂ ≠ b₁/b₂

Intersecting Lines

Unique Solution

Consistent

a₁/a₂ = b₁/b₂ = c₁/c₂

Coincident Lines

Infinitely Many Solutions

Dependent & Consistent

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Parallel Lines

No Solution

Inconsistent

  • EXPLANATION of Ratios:
  • a₁/a₂ ≠ b₁/b₂: This condition implies that the slopes of the two lines are different, so they must intersect at one point.
  • a₁/a₂ = b₁/b₂ = c₁/c₂: This condition implies that the two equations are essentially multiples of each other, representing the same line.
  • a₁/a₂ = b₁/b₂ ≠ c₁/c₂: This condition implies that the lines have the same slope (parallel) but different y-intercepts, so they never intersect.

6. EQUATIONS REDUCIBLE TO A PAIR OF LINEAR EQUATIONS:

  • INTRODUCTION: Some equations are not linear in x and y directly, but they can be transformed into linear equations by using suitable substitutions.
  • COMMON TYPE: Equations where variables x and y appear in the denominator, like:
    • a/x + b/y = c (and similar forms)
  • REDUCTION TECHNIQUE: SUBSTITUTION
  1. Step 1: Introduce new variables: Let m = 1/x and n = 1/y.
  2. Step 2: Substitute: Substitute ‘m’ for ‘1/x’ and ‘n’ for ‘1/y’ in the given equations. This will transform the equations into linear equations in terms of ‘m’ and ‘n’.
  3. Step 3: Solve the linear equations in m and n: Solve the new pair of linear equations in ‘m’ and ‘n’ using any algebraic method (substitution, elimination, or cross-multiplication).
  4. Step 4: Substitute back to find x and y: Once you have the values of ‘m’ and ‘n’, substitute back x = 1/m and y = 1/n to find the values of x and y.
  • EXAMPLE: Solve the equations:
    • 2/x + 3/y = 13 (Equation 1)
    • 5/x – 4/y = -2 (Equation 2)

Step 1: Substitution: Let m = 1/x and n = 1/y.

Step 2: Transformed Equations:

  • Equation 1 becomes: 2m + 3n = 13 (Equation 1′)
  • Equation 2 becomes: 5m – 4n = -2 (Equation 2′)

      Step 3: Solve for m and n (using Elimination method here):

  • Multiply Equation 1′ by 4 and Equation 2′ by 3 to equate coefficients of ‘n’:
    • 8m + 12n = 52
    • 15m – 12n = -6
  • Add these equations: 23m = 46 => m = 2
  • Substitute m = 2 into Equation 1′:
  • 2(2) + 3n = 13 => 4 + 3n = 13 =>
  • 3n = 9 => n = 3
  1. Step 4: Substitute back:
    • x = 1/m = 1/2
    • y = 1/n = 1/3
  2. Solution: x = 1/2, y = 1/3