Comprehensive Study Notes on Pair of Linear Equations in Two Variables

Real-World Application and Equation Formulation

• Situations involving two unknown quantities can be represented algebraically using linear equations in two variables.

• Akhila's Village Fair Example:

  • Activities: Giant Wheel rides and playing Hoopla (a game of throwing a ring onto items at a stall; if the ring covers an item completely, the player wins it).

  • Conditions given:

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

  • Cost per Giant Wheel ride: $\text{` } 3$

  • Cost per game of Hoopla: $\text{` } 4$

  • Total amount spent: $\text{` } 20$

  • Variable representation:

  • Let $x$ represent the number of rides on the Giant Wheel.

  • Let $y$ represent the number of times she played Hoopla.

  • Algebraic equations formed:

  • $y = \frac{1}{2}x$

  • $3x + 4y = 20$

General Form and Classification of Linear Equations

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

  • $a_1x + b_1y + c_1 = 0$

  • $a_2x + b_2y + c_2 = 0$

  • Where $a_1, b_1, c_1, a_2, b_2, c_2$ are real numbers.

• Consistent Pair of Linear Equations: A pair of linear equations in two variables that has at least one solution.

• Inconsistent Pair of Linear Equations: A pair of linear equations in two variables that has no solution.

• Dependent Pair of Linear Equations: A pair of linear equations that are equivalent and have infinitely many distinct common solutions. A dependent pair of linear equations is always consistent.

Graphical Method and Ratio Criteria

• The geometric behavior of lines representing a pair of linear equations and their solution conditions are categorized by comparing coefficient ratios:

• Intersecting Lines:

  • Condition: $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$

  • Graphical representation: The lines intersect at a single unique point.

  • Algebraic interpretation: Exactly one unique solution.

  • System consistency: Consistent pair.

  • Example: $x - 2y = 0$ and $3x + 4y - 20 = 0$

  • Coefficients: $a_1 = 1, b_1 = -2, c_1 = 0$ and $a_2 = 3, b_2 = 4, c_2 = -20$

  • Ratios: $\frac{a_1}{a_2} = \frac{1}{3}$, $\frac{b_1}{b_2} = \frac{-2}{4} = \frac{-1}{2}$, $\frac{c_1}{c_2} = \frac{0}{-20} = 0$

  • Comparison: $\frac{1}{3} \neq \frac{-1}{2}$

• Coincident Lines:

  • Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

  • Graphical representation: The lines coincide (overlap completely).

  • Algebraic interpretation: Infinitely many solutions.

  • System consistency: Dependent (consistent) pair.

  • Example: $2x + 3y - 9 = 0$ and $4x + 6y - 18 = 0$

  • Coefficients: $a_1 = 2, b_1 = 3, c_1 = -9$ and $a_2 = 4, b_2 = 6, c_2 = -18$

  • Ratios: $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{-9}{-18} = \frac{1}{2}$

  • Comparison: $\frac{1}{2} = \frac{1}{2} = \frac{1}{2}$

• Parallel Lines:

  • Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

  • Graphical representation: The lines are parallel and never meet.

  • Algebraic interpretation: No solution.

  • System consistency: Inconsistent pair.

  • Example: $x + 2y - 4 = 0$ and $2x + 4y - 12 = 0$

  • Coefficients: $a_1 = 1, b_1 = 2, c_1 = -4$ and $a_2 = 2, b_2 = 4, c_2 = -12$

  • Ratios: $\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{c_1}{c_2} = \frac{-4}{-12} = \frac{1}{3}$

  • Comparison: $\frac{1}{2} = \frac{1}{2} \neq \frac{1}{3}$

Detailed Worked Graphical Examples

• Example 1:

  • Problem: Check graphically whether $x + 3y = 6$ (1) and $2x - 3y = 12$ (2) is consistent. If so, solve graphically.

  • Solution Procedure:

  • Table values for $x + 3y = 6 \implies y = \frac{6 - x}{3}$:

    • When $x = 0$, $y = 2 \implies A(0, 2)$

    • When $x = 6$, $y = 0 \implies B(6, 0)$

  • Table values for $2x - 3y = 12 \implies y = \frac{2x - 12}{3}$:

    • When $x = 0$, $y = -4 \implies P(0, -4)$

    • When $x = 3$, $y = -2 \implies Q(3, -2)$

  • Plot points $A(0, 2)$, $B(6, 0)$, $P(0, -4)$, and $Q(3, -2)$ and draw lines $AB$ and $PQ$

  • Point $B(6, 0)$ is common to both lines $AB$ and $PQ$

  • Solution: $x = 6$ and $y = 0$. The pair of equations is consistent.

• Example 2:

  • Problem: Determine graphically whether $5x - 8y + 1 = 0$ (1) and $3x - \frac{24}{5}y + \frac{3}{5} = 0$ (2) has no solution, unique solution, or infinitely many solutions.

  • Solution Procedure:

  • Multiply Equation (2) by $\frac{5}{3}$:

    • $\frac{5}{3} \left(3x - \frac{24}{5}y + \frac{3}{5}\right) = 0 \implies 5x - 8y + 1 = 0$

  • Since this matches Equation (1) exactly, the two equations represent coincident lines.

  • Conclusion: The system has infinitely many solutions.

• Example 3:

  • Problem: Champa bought pants ($x$) and skirts ($y$). She told her friends: "The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased." Find how many pants and skirts she bought.

  • Equations formed:

  • $y = 2x - 2$ (1)

  • $y = 4x - 4$ (2)

  • Table values for $y = 2x - 2$:

  • When $x = 2$, $y = 2 \implies (2, 2)$

  • When $x = 0$, $y = -2 \implies (0, -2)$

  • Table values for $y = 4x - 4$:

  • When $x = 0$, $y = -4 \implies (0, -4)$

  • When $x = 1$, $y = 0 \implies (1, 0)$

  • Plotting and intersection: The lines intersect at $(1, 0)$.

  • Solution: $x = 1$ and $y = 0$. Champa bought 1 pair of pants and 0 skirts.

Exercise 3.1 Problems

• Problem 1: Form linear equations and solve graphically:

  • (i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

  • (ii) 5 pencils and 7 pens together cost $\text{` } 50$, whereas 7 pencils and 5 pens together cost $\text{` } 46$. Find the cost of one pencil and that of one pen.

• Problem 2: Compare ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}$ to check if lines intersect at a point, are parallel, or coincident:

  • (i) $5x - 4y + 8 = 0$ and $7x + 6y - 9 = 0$

  • (ii) $9x + 3y + 12 = 0$ and $18x + 6y + 24 = 0$

  • (iii) $6x - 3y + 10 = 0$ and $2x - y + 9 = 0$

• Problem 3: Compare ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}$ to check if the linear pairs are consistent or inconsistent:

  • (i) $3x + 2y = 5$ ; $2x - 3y = 7$

  • (ii) $2x - 3y = 8$ ; $4x - 6y = 9$

  • (iii) $\frac{3}{2}x + \frac{5}{3}y = 7$ ; $9x - 10y = 14$

  • (iv) $5x - 3y = 11$ ; $-10x + 6y = -22$

  • (v) $\frac{4}{3}x + 2y = 8$ ; $2x + 3y = 12$

• Problem 4: Determine consistency; if consistent, obtain solution graphically:

  • (i) $x + y = 5$, $2x + 2y = 10$

  • (ii) $x - y = 8$, $3x - 3y = 16$

  • (iii) $2x + y - 6 = 0$, $4x - 2y - 4 = 0$

  • (iv) $2x - 2y - 2 = 0$, $4x - 4y - 5 = 0$

• Problem 5: Half the perimeter of a rectangular garden, whose length is $4\,\text{m}$ more than its width, is $36\,\text{m}$. Find the dimensions of the garden.

• Problem 6: Given the linear equation $2x + 3y - 8 = 0$, write another linear equation in two variables such that the geometric representation formed is:

  • (i) Intersecting lines

  • (ii) Parallel lines

  • (iii) Coincident lines

• Problem 7: Draw graphs of $x - y + 1 = 0$ and $3x + 2y - 12 = 0$. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Need for Algebraic Methods

• Graphical solutions are inconvenient when coordinate values are non-integral fractions or irrational expressions like $\left(\sqrt{3}, 2\sqrt{7}\right)$, $(-1.75, 3.3)$, or $\left(\frac{4}{13}, \frac{1}{19}\right)$, which lead to estimation errors when reading points on graph paper.

Substitution Method

• Method Steps:

  • Step 1: Pick either equation and write one variable in terms of the other (e.g., $y$ in terms of $x$).

  • Step 2: Substitute this expression into the other equation to obtain a single-variable equation, then solve for that variable.

  • If a true statement with no variable is obtained (e.g., $18 = 18$), the pair has infinitely many solutions.

  • If a false statement with no variable is obtained (e.g., $-4 = 0$), the pair is inconsistent (no solution).

  • Step 3: Substitute the calculated value into the equation from Step 1 to solve for the remaining variable.

• Example 4:

  • Solve $7x - 15y = 2$ (1) and $x + 2y = 3$ (2).

  • Step 1: From Equation (2), $x = 3 - 2y$ (3).

  • Step 2: Substitute into Equation (1):

  • $7(3 - 2y) - 15y = 2 \implies 21 - 14y - 15y = 2 \implies -29y = -19 \implies y = \frac{19}{29}$

  • Step 3: Substitute $y = \frac{19}{29}$ into Equation (3):

  • $x = 3 - 2\left(\frac{19}{29}\right) = \frac{87 - 38}{29} = \frac{49}{29}$

  • Solution: $x = \frac{49}{29}$, $y = \frac{19}{29}$.

• Example 5:

  • Problem: Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." Solve algebraically by substitution.

  • Variable definitions: Let Aftab's age be $s$ and daughter's age be $t$ in years.

  • Equations:

  • $s - 7 = 7(t - 7) \implies s - 7t + 42 = 0$ (1)

  • $s + 3 = 3(t + 3) \implies s - 3t = 6$ (2)

  • Solution:

  • From (2), $s = 3t + 6$

  • Substitute into (1): $(3t + 6) - 7t + 42 = 0 \implies -4t + 48 = 0 \implies 4t = 48 \implies t = 12$

  • Substitute $t = 12$ into (2): $s = 3(12) + 6 = 42$

  • Result: Aftab is 42 years old and his daughter is 12 years old.

• Example 6:

  • Cost of 2 pencils and 3 erasers is $\text{` } 9$ ($2x + 3y = 9$) (1). Cost of 4 pencils and 6 erasers is $\text{` } 18$ ($4x + 6y = 18$) (2).

  • Express $x$ from (1): $x = \frac{9 - 3y}{2}$ (3).

  • Substitute into (2):

  • $4\left(\frac{9 - 3y}{2}\right) + 6y = 18 \implies 18 - 6y + 6y = 18 \implies 18 = 18$

  • Conclusion: This statement is true for all values of $y$. Both equations are identical; there are infinitely many common solutions, and a unique individual cost cannot be determined.

• Example 7:

  • Two rails are given by $x + 2y - 4 = 0$ (1) and $2x + 4y - 12 = 0$ (2). Will the rails cross each other?

  • From (1): $x = 4 - 2y$.

  • Substitute into (2):

  • $2(4 - 2y) + 4y - 12 = 0 \implies 8 - 4y + 4y - 12 = 0 \implies -4 = 0$

  • Conclusion: False statement. Equations have no common solution, meaning the two rails will not cross each other.

Exercise 3.2 Problems

• Problem 1: Solve the following pairs by substitution method:

  • (i) $x + y = 14$ and $x - y = 4$

  • (ii) $s - t = 3$ and $\frac{s}{3} + \frac{t}{2} = 6$

  • (iii) $3x - y = 3$ and $9x - 3y = 9$

  • (iv) $0.2x + 0.3y = 1.3$ and $0.4x + 0.5y = 2.3$

  • (v) $\sqrt{2}x + \sqrt{3}y = 0$ and $\sqrt{3}x - \sqrt{8}y = 0$

  • (vi) $\frac{3}{2}x - \frac{5}{3}y = -2$ and $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

• Problem 2: Solve $2x + 3y = 11$ and $2x - 4y = -24$, and find the value of '$m$' for which $y = mx + 3$

• Problem 3: Form linear equations and solve by substitution:

  • (i) The difference between two numbers is 26 and one number is three times the other. Find them.

  • (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

  • (iii) The coach of a cricket team buys 7 bats and 6 balls for $\text{` } 3800$. Later, she buys 3 bats and 5 balls for $\text{` } 1750$. Find the cost of each bat and each ball.

  • (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of $10\,\text{km}$, the charge paid is $\text{` } 105$ and for a journey of $15\,\text{km}$, the charge paid is $\text{` } 155$. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of $25\,\text{km}$?

  • (v) A fraction becomes $\frac{9}{11}$, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes $\frac{5}{6}$. Find the fraction.

  • (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?

Elimination Method

• Method Steps:

  • Step 1: Multiply both equations by suitable non-zero constants to make the numerical coefficients of one variable ($x$ or $y$) equal.

  • Step 2: Add or subtract one equation from the other so that one variable gets eliminated.

  • If a single-variable equation is obtained, proceed to Step 3.

  • If a true statement with no variable is obtained, the pair has infinitely many solutions.

  • If a false statement with no variable is obtained, the pair has no solution (inconsistent).

  • Step 3: Solve the single-variable equation obtained.

  • Step 4: Substitute this value into either original equation to calculate the second variable.

• Example 8:

  • Problem: Ratio of monthly incomes of two persons is $9 : 7$ and ratio of expenditures is $4 : 3$. Each saves $\text{` } 2000$ per month. Find monthly incomes.

  • Equations: Incomes are $\text{` } 9x$ and $\text{` } 7x$; expenditures are $\text{` } 4y$ and $\text{` } 3y$.

  • $9x - 4y = 2000$ (1)

  • $7x - 3y = 2000$ (2)

  • Step 1: Multiply (1) by 3 and (2) by 4:

  • $27x - 12y = 6000$ (3)

  • $28x - 12y = 8000$ (4)

  • Step 2: Subtract (3) from (4):

  • $(28x - 27x) - (12y - 12y) = 8000 - 6000 \implies x = 2000$

  • Step 3 & 4: Substitute $x = 2000$ into (1):

  • $9(2000) - 4y = 2000 \implies 18000 - 4y = 2000 \implies 4y = 16000 \implies y = 4000$

  • Monthly Incomes:

  • First person: $9 \times 2000 = \text{` } 18000$

  • Second person: $7 \times 2000 = \text{` } 14000$

  • Verification: Income ratio $18000 : 14000 = 9 : 7$. Expenditure ratio (18000 - 2000) : (14000 - 2000) = 16000 : 12000 = 4 : 3$.\n\n- **Example 9**:\n - Solve 2x + 3y = 8$(1) and$4x + 6y = 7 (2) by elimination.\n - Step 1: Multiply Equation (1) by 2:\n - 4x + 6y = 16 (3)\n - Step 2: Subtract Equation (2) from Equation (3):\n - (4x - 4x) + (6y - 6y) = 16 - 7 \implies 0 = 9\n - Conclusion: False statement. The system of equations has no solution.\n\n- **Example 10**:\n - Problem: The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits differ by 2, find the number. How many such numbers are there?\n - Variable definitions: Let ten's digit be x$and unit's digit be$y$. Original number =$10x + y$. Reversed number =$10y + x.\n - First condition:\n - (10x + y) + (10y + x) = 66 \implies 11(x + y) = 66 \implies x + y = 6 (1)\n - Second condition (digits differ by 2):\n - Either x - y = 2$(2) OR$y - x = 2 (3)\n - Solving Case 1 (Equations 1 & 2):\n - (x + y) + (x - y) = 6 + 2 \implies 2x = 8 \implies x = 4\n - y = 6 - 4 = 2\n - Number: 42\n - Solving Case 2 (Equations 1 & 3):\n - (x + y) + (y - x) = 6 + 2 \implies 2y = 8 \implies y = 4\n - x = 6 - 4 = 2\n - Number: 24\n - Result: There are two such numbers: 42 and 24.\n\n# Exercise 3.3 Problems\n\n- **Problem 1**: Solve the following linear pairs using both elimination and substitution methods:\n - (i) x + y = 5$and$2x - 3y = 4\n - (ii) 3x + 4y = 10$and$2x - 2y = 2\n - (iii) 3x - 5y - 4 = 0$and$9x = 2y + 7\n - (iv) \frac{x}{2} + \frac{2y}{3} = -1$and$x - \frac{y}{3} = 3\n\n- **Problem 2**: Form linear equations and solve by elimination method:\n - (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \frac{1}{2} if we only add 1 to the denominator. What is the fraction?\n - (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?\n - (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.\n - (iv) Meena went to a bank to withdraw \text{} 2000$. She asked the cashier to give her$\text{ } 50$and$\text{} 100$notes only. Meena got 25 notes in all. Find how many notes of$\text{ } 50$and$\text{} 100 she received.\n - (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid \text{ } 27$for a book kept for seven days, while Susy paid$\text{` } 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.\n\n# Chapter Summary\n\n- A pair of linear equations in two variables can be solved via:\n - Graphical Method\n - Algebraic Methods (Substitution Method, Elimination Method)\n\n- Ratio comparisons for a_1x + b_1y + c_1 = 0$and$a_2x + b_2y + c_2 = 0:\n - Intersecting lines: \frac{a_1}{a_2} \neq \frac{b_1}{b_2} (Consistent system with a unique solution).\n - Parallel lines: \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} (Inconsistent system with no solution).\n - Coincident lines: \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$ (Dependent and consistent system with infinitely many solutions).

• Equations that are non-linear at start can often be altered/reduced into linear equations in two variables to enable standard solution methods.