Gonit Vigyan

 

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Solving a Fun Fair Puzzle with Linear Equations 🎡🎯

Have you ever been to a fair and tried to manage your expenses while enjoying the rides and games? Akhila faced a similar challenge at her village fair. She wanted to go on the Giant Wheel and play Hoopla, but she had only ₹20 to spend. The number of times she played Hoopla was half the number of Giant Wheel rides she took. With each ride costing ₹3 and each game of Hoopla costing ₹4, can you figure out how many rides she took and how many times she played Hoopla?

Instead of guessing different possibilities, we can solve this problem logically using linear equations in two variables. In this post, we’ll explore how to represent real-life situations as mathematical equations and find solutions step by step. Let’s dive into it! 🚀

1. Introduction to Linear Equations in Two Variables

• A linear equation in two variables is expressed as:

  ax + by + c = 0

  where a, b, and c are real numbers, and a and b are not both zero.

• Examples of Linear Equations in Two Variables:

  1. 2x + 3y = 5
  2. x - 2y - 3 = 0
  3. x = 2 (i.e., x - 0y = 2)

• Solutions of Linear Equations:

  • A solution is a pair (x, y) that satisfies the equation.
  • Example:
    • For 2x + 3y = 5, (1,1) is a solution because LHS = 2(1) + 3(1) = 5 = RHS.
    • (1,7) is not a solution as LHS = 2(1) + 3(7) = 23 ≠ 5.

2. Geometrical Representation

• A linear equation in two variables represents a straight line on a graph.
• Every solution (x, y) is a point on this line.

3. Pair of Linear Equations in Two Variables

• Two linear equations together form a pair of linear equations.

• General form:

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

  where a₁, b₁, c₁, a₂, b₂, c₂ are real numbers and a₁² + b₁² ≠ 0, a₂² + b₂² ≠ 0.

• Examples:

  1. 2x + 3y - 7 = 0 and 9x - 2y + 8 = 0
  2. 5x = y and -7x + 2y + 3 = 0
  3. x + y = 7 and 17 = y

4. Possible Graphical Representations

• Three possible cases for two lines in a plane:
  1. Intersecting Lines – Unique solution.
  2. Parallel Lines – No solution.
  3. Coincident Lines – Infinitely many solutions.

5. Examples and Graphical Representation

Example 1: Akhila at the Fair

• Situation: Akhila has ₹20 and wants to go on a Giant Wheel and play Hoopla.
• Equations:
  1. x - 2y = 0
  2. 3x + 4y = 20
• Solutions:
  • (0,0), (2,1), (0,5), (4,2)
• Graph: The lines intersect at (4,2), meaning a unique solution exists.

Example 2: Romila and Sonali’s Stationery Purchase

• Situation:
  • Romila buys 2 pencils and 3 erasers for ₹9.
  • Sonali buys 4 pencils and 6 erasers for ₹18.
• Equations:
  1. 2x + 3y = 9
  2. 4x + 6y = 18
• Solutions:
  • (0,3), (4.5,0), (0,3), (3,1)
• Graph: The lines coincide, meaning infinitely many solutions exist.

Example 3: Parallel Rails

• Situation: Two railway tracks are represented by:
  1. x + 2y - 4 = 0
  2. 2x + 4y - 12 = 0
• Solutions:
  • (0,2), (4,0), (0,3), (6,0)
• Graph: The lines are parallel, meaning no solution exists.

6. Summary

• Linear Equations in Two Variables form straight lines.
• A pair of linear equations can have:
  1. One solution (intersecting lines).
  2. No solution (parallel lines).
  3. Infinitely many solutions (coincident lines).
• Graphical representation helps visualize the nature of solutions.

 

EXERCISE 3.1

1. Aftab’s Age Problem

Question:
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." Represent this situation algebraically and graphically.

Solution:
Let Aftab’s present age be x years.
Let his daughter’s present age be y years.

Step 1: Forming the Equations

• Seven years ago:
  • Aftab’s age = x - 7
  • Daughter’s age = y - 7
  • Given that Aftab was seven times his daughter’s age: $$x - 7 = 7(y - 7)$$ $$x - 7 = 7y - 49$$ $$x - 7y = -42$$
• Three years from now:
  • Aftab’s age = x + 3
  • Daughter’s age = y + 3
  • Given that Aftab will be three times his daughter’s age: $$x + 3 = 3(y + 3)$$ $$x + 3 = 3y + 9$$ $$x - 3y = 6$$

Thus, the two equations are:

$$x - 7y = -42$$ $$x - 3y = 6$$

Step 2: Finding Solutions for Graph

x	y (from x - 7y = -42)	y (from x - 3y = 6)
0	6	-2
42	12	12
48	18	14

Step 3: Plotting the Graph
Plot the points (0,6), (42,12), (48,18) for x - 7y = -42 and (0,-2), (42,12), (48,14) for x - 3y = 6 on a graph. The intersection of these lines gives the solution.

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2. Cricket Equipment Purchase

Question:
A cricket coach buys 3 bats and 6 balls for ₹3900. Later, she buys 1 bat and 3 balls for ₹1300. Represent this situation algebraically and graphically.

Solution:
Let the cost of one bat be x and the cost of one ball be y.

Step 1: Forming the Equations

• First purchase: $$3x + 6y = 3900$$ Dividing by 3: $$x + 2y = 1300$$
• Second purchase: $$x + 3y = 1300$$

Thus, the two equations are:

$$x + 2y = 1300$$ $$x + 3y = 1300$$

Step 2: Finding Solutions for Graph

x	y (from x + 2y = 1300)	y (from x + 3y = 1300)
1000	150	100
700	300	200
1300	0	0

Step 3: Plotting the Graph
Plot the points (1000,150), (700,300), (1300,0) for x + 2y = 1300 and (1000,100), (700,200), (1300,0) for x + 3y = 1300. The intersection of these lines gives the solution.

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3. Cost of Fruits

Question:
The cost of 2 kg of apples and 1 kg of grapes on a day was ₹160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹300. Represent the situation algebraically and graphically.

Solution:
Let the cost of 1 kg of apples be x and the cost of 1 kg of grapes be y.

Step 1: Forming the Equations

• First condition: $$2x + y = 160$$
• Second condition: $$4x + 2y = 300$$ Dividing by 2: $$2x + y = 150$$

Thus, the two equations are:

$$2x + y = 160$$ $$2x + y = 150$$

Step 2: Finding Solutions for Graph

x	y (from 2x + y = 160)	y (from 2x + y = 150)
0	160	150
80	0	-10
50	60	50

Step 3: Plotting the Graph
Plot the points (0,160), (80,0), (50,60) for 2x + y = 160 and (0,150), (80,-10), (50,50) for 2x + y = 150. The lines are parallel, meaning no solution exists.

Graphical Method of Solving a Pair of Linear Equations  

Introduction

• A pair of linear equations in two variables can be represented geometrically by two lines.
• The lines may either:
  1. Intersect at a single point (unique solution).
  2. Be parallel (no solution).
  3. Coincide (infinitely many solutions).

Types of Solutions for a Pair of Linear Equations

1. Consistent System:

   • The equations have at least one solution.
   • Graphically, the lines intersect or coincide.
   • If the lines intersect at a single point, the solution is unique.
   • If the lines coincide, they have infinitely many solutions.

2. Inconsistent System:

   • The equations have no solution.
   • Graphically, the lines are parallel.

3. Dependent System:

   • The equations are equivalent and have infinitely many solutions.
   • Graphically, the lines coincide.

Graphical Interpretation of Solutions

Type of Pair of Equations	Graphical Representation	Nature of Solution
Intersecting Lines	Lines cross at one point	Unique Solution
Coincident Lines	Lines overlap completely	Infinitely Many Solutions
Parallel Lines	Lines never meet	No Solution

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Examples and Their Graphical Representation

Example 1: Akhila’s Rides on the Giant Wheel

Equations:

$$x - 2y = 0$$ $$3x + 4y = 20$$

• The lines intersect at (4, 2).
• Unique solution: x = 4, y = 2.
• Akhila took 4 rides on the Giant Wheel and played Hoopla 2 times.

Example 2: Cost of Pencils and Erasers

Equations:

$$2x + 3y = 9$$ $$4x + 6y = 18$$

• The second equation is a multiple of the first.
• The lines coincide.
• Infinitely many solutions (dependent system).
• Possible costs:
  • Pencil = ₹3, Eraser = ₹1
  • Pencil = ₹3.75, Eraser = ₹0.50, etc.

Example 3: Railway Tracks (Parallel Lines)

Equations:

$$x + 2y = 4$$ $$2x + 4y = 12$$

• The second equation is a multiple of the first but shifted.
• The lines are parallel.
• No solution (inconsistent system).

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General Form of a Pair of Linear Equations

$$a_1x + b_1y + c_1 = 0$$ $$a_2x + b_2y + c_2 = 0$$

Conditions for Different Types of Solutions

1. Unique Solution (Intersecting Lines): $$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$$
2. Infinitely Many Solutions (Coincident Lines): $$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$
3. No Solution (Parallel Lines): $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$

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Graphical Solution of Pairs of Linear Equations

Example 4: Checking the Consistency of Two Equations

Equations:

$$x + 3y = 6$$ $$2x - 3y = 12$$

Solution Steps:

1. Find two solutions for each equation:

   • For $x + 3y = 6$:
     • $x = 0, y = 2$ → (0,2)
     • $x = 6, y = 0$ → (6,0)
   • For $2x - 3y = 12$:
     • $x = 0, y = -4$ → (0,-4)
     • $x = 3, y = -2$ → (3,-2)

2. Plot points on a graph and draw the lines.

3. Observations:

   • The lines intersect at (6, 0).
   • The system has a unique solution (consistent system).
   • Solution: $x = 6, y = 0$.

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Example 5: Checking for Infinite Solutions

Equations:

$$5x - 8y + 1 = 0$$ $$3x - \frac{24}{5} y + \frac{3}{5} = 0$$

Solution Steps:

1. Multiply the second equation by 5/3 to make the coefficients comparable:

   $$5x - 8y + 1 = 0$$
   • The transformed equation is identical to the first equation.
   • The lines coincide.

2. Observations:

   • The system has infinitely many solutions (dependent system).

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Example 6: Champa’s Shopping

Equations:

$$y = 2x - 2$$ $$y = 4x - 4$$

Solution Steps:

1. Find two points for each equation:

   • For $y = 2x - 2$:
     • $x = 2, y = 2$ → (2,2)
     • $x = 0, y = -2$ → (0,-2)
   • For $y = 4x - 4$:
     • $x = 0, y = -4$ → (0,-4)
     • $x = 1, y = 0$ → (1,0)

2. Plot points on a graph and draw the lines.

3. Observations:

   • The lines intersect at (1,0).
   • The system has a unique solution (consistent system).
   • Solution: $x = 1, y = 0$.
   • Champa bought 1 pant and 0 skirts.

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Summary

1. Intersection → Unique Solution (Consistent)
2. Parallel → No Solution (Inconsistent)
3. Coincidence → Infinite Solutions (Dependent & Consistent)

This method provides a visual approach to solving equations, making it easier to understand relationships between variables.

EXERCISE 3.2 – Solutions with Step-by-Step Explanation

1. Form the pair of linear equations in the following problems, and find their solutions 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.

Step 1: Define variables
Let the number of boys be x and the number of girls be y.

Step 2: Form equations

• The total number of students is given as: $$x + y = 10$$
• The number of girls is 4 more than the number of boys: $$y = x + 4$$

Step 3: Solve graphically
We rewrite the equations:

1. $x + y = 10 \Rightarrow y = 10 - x$
2. $y = x + 4$

Now, we find points for both equations:

For $y = 10 - x$:

• If $x = 0$, then $y = 10$ → (0,10)
• If $x = 5$, then $y = 5$ → (5,5)
• If $x = 10$, then $y = 0$ → (10,0)

For $y = x + 4$:

• If $x = 0$, then $y = 4$ → (0,4)
• If $x = 3$, then $y = 7$ → (3,7)
• If $x = 6$, then $y = 10$ → (6,10)

Plot the points and draw the two lines.

 The point of intersection gives the solution:

$$(3,7)$$

So, the number of boys = 3, and the number of girls = 7.

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(ii) 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and one pen.

Step 1: Define variables
Let the cost of one pencil be x and the cost of one pen be y.

Step 2: Form equations

• 5 pencils and 7 pens together cost ₹50: $$5x + 7y = 50$$
• 7 pencils and 5 pens together cost ₹46: $$7x + 5y = 46$$

Step 3: Solve graphically
Find points for both equations:

For $5x + 7y = 50$:

• If $x = 0$, then $y = \frac{50}{7} \approx 7.14$
• If $y = 0$, then $x = \frac{50}{5} = 10$
• If $x = 2$, then $y = \frac{50 - 10}{7} = \frac{40}{7} \approx 5.71$

For $7x + 5y = 46$:

• If $x = 0$, then $y = \frac{46}{5} = 9.2$
• If $y = 0$, then $x = \frac{46}{7} \approx 6.57$
• If $x = 2$, then $y = \frac{46 - 14}{5} = 6.4$

Plot the points and draw the lines. 

The intersection gives the solution:

$$(3, 5)$$

So, the cost of one pencil = ₹3, and the cost of one pen = ₹5.

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2. Determine whether the lines representing the following equations are intersecting, parallel, or coincident by comparing ratios.

(i) 5x – 4y + 8 = 0 and 7x + 6y – 9 = 0
Comparing coefficients:

$$\frac{a_1}{a_2} = \frac{5}{7}, \quad \frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3}, \quad \frac{c_1}{c_2} = \frac{8}{-9}$$

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect.

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(ii) 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0
Comparing coefficients:

$$\frac{9}{18} = \frac{1}{2}, \quad \frac{3}{6} = \frac{1}{2}, \quad \frac{12}{24} = \frac{1}{2}$$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident.

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(iii) 6x – 3y + 10 = 0 and 2x – y + 9 = 0
Comparing coefficients:

$$\frac{6}{2} = 3, \quad \frac{-3}{-1} = 3, \quad \frac{10}{9} \neq 3$$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel.

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3. Determine whether the given equations are consistent or inconsistent by comparing ratios.

(i) 3x + 2y = 5 and 2x – 3y = 7
Comparing coefficients:

$$\frac{3}{2} \neq \frac{2}{-3}$$

Since the lines intersect, the equations are consistent.

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(ii) 2x – 3y = 8 and 4x – 6y = 9
Comparing coefficients:

$$\frac{2}{4} = \frac{1}{2}, \quad \frac{-3}{-6} = \frac{1}{2}, \quad \frac{8}{9} \neq \frac{1}{2}$$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the equations are inconsistent.

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4. Find whether the given pairs of equations are consistent or inconsistent, and solve graphically if consistent.

(i) x + y = 5, 2x + 2y = 10
Dividing the second equation by 2:

$$x + y = 5$$

Since both equations are the same, they are coincident (consistent with infinite solutions).

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(ii) x – y = 8, 3x – 3y = 16
Dividing the second equation by 3:

$$x - y = \frac{16}{3}$$

Since $\frac{8}{\frac{16}{3}} \neq 1$, the lines are parallel (inconsistent, no solution).

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Continuing with Step-by-Step Solutions for Exercise 3.2

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5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Step 1: Define variables

Let the width of the garden be x meters.
Then, the length of the garden is (x + 4) meters.

The perimeter of a rectangle is given by:

$$\text{Perimeter} = 2 (\text{Length} + \text{Width})$$

Half of the perimeter is given as 36 m:

$$(x + 4) + x = 36$$

Step 2: Form the equation

$$2x + 4 = 36$$

Step 3: Solve for x

$$2x = 36 - 4$$ $$2x = 32$$ $$x = \frac{32}{2} = 16$$

Thus, width = 16 m and length = 16 + 4 = 20 m.

Final Answer:

The dimensions of the garden are 16 meters (width) and 20 meters (length).

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6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) Intersecting lines

For intersecting lines, the second equation must have different ratios for a₁/a₂ and b₁/b₂.
Given equation:

$$2x + 3y - 8 = 0$$

Choose a different equation such as:

$$3x - 2y + 5 = 0$$

(ii) Parallel lines

For parallel lines, the second equation must have the same ratio for a₁/a₂ and b₁/b₂, but different c₁/c₂.
Multiply the given equation by a constant:

$$4x + 6y + 10 = 0$$

(iii) Coincident lines

For coincident lines, the second equation must be a multiple of the first equation.
Multiply the entire equation by 2:

$$4x + 6y - 16 = 0$$

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7. Draw the graphs of the equations $x - y + 1 = 0$ 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.

Step 1: Rewrite the equations in slope-intercept form

1. $x - y + 1 = 0$ $$y = x + 1$$
2. $3x + 2y - 12 = 0$ $$2y = -3x + 12$$ $$y = -\frac{3}{2}x + 6$$

Step 2: Find the intersection point of the two lines

Solve $x + 1 = -\frac{3}{2}x + 6$.

Multiply everything by 2 to eliminate the fraction:

$$2(x + 1) = -3x + 12$$ $$2x + 2 = -3x + 12$$ $$2x + 3x = 12 - 2$$ $$5x = 10$$ $$x = 2$$

Substituting $x = 2$ in $y = x + 1$:

$$y = 2 + 1 = 3$$

Thus, the lines intersect at (2,3).

Step 3: Find x-intercepts

For $x - y + 1 = 0$, set $y = 0$:

$$x + 1 = 0 \Rightarrow x = -1$$

So, x-intercept = (-1, 0).

For $3x + 2y - 12 = 0$, set $y = 0$:

$$3x - 12 = 0$$ $$3x = 12$$ $$x = 4$$

So, x-intercept = (4, 0).

Step 4: Identify the triangle and shade the region

The three points forming the triangle are:

$$(-1, 0), (2,3), (4, 0)$$

These points form a triangle with the x-axis.
Shade the triangular region enclosed by these points on the graph.

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Final Summary of Answers

1. Number of boys = 3, Number of girls = 7.
2. Cost of one pencil = ₹3, Cost of one pen = ₹5.
3. Nature of lines determined using ratios.
4. Consistency of equations checked using ratios.
5. Dimensions of the garden: Width = 16m, Length = 20m.
6. Different equations written for intersecting, parallel, and coincident cases.
7. Triangle formed by (-1, 0), (2,3), and (4,0).

 

3.4 Algebraic Methods of Solving a Pair of Linear Equations

Concept:

In the previous section, we solved pairs of linear equations graphically. However, the graphical method has limitations when the solution involves non-integral values, making it difficult to read coordinates accurately. Therefore, algebraic methods are used as an alternative.

There are several algebraic methods to solve a pair of linear equations:

1. Substitution Method
2. Elimination Method
3. Cross-Multiplication Method

In this section, we will discuss the Substitution Method in detail.

3.4.1 Substitution Method

The Substitution Method involves expressing one variable in terms of the other using one equation and substituting this value in the second equation to solve for both variables.

Steps for the Substitution Method:

1. Select one of the equations and express one variable in terms of the other.
2. Substitute this expression in the other equation to get an equation in one variable.
3. Solve for the variable.
4. Substitute the value obtained into the equation used in Step 1 to find the second variable.
5. Verify the solution by substituting the values in the original equations.

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Examples

Example 1: Solve the following pair of equations using the Substitution Method

$$7x - 15y = 2$$ $$x + 2y = 3$$

Solution:

Step 1: Express one variable in terms of the other using Equation (2).

$$x + 2y = 3$$ $$x = 3 - 2y \quad (3)$$

Step 2: Substitute the value of $x$ in Equation (1).

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

Step 3: Substitute $y = \frac{19}{29}$ in Equation (3) to find $x$.

$$x = 3 - 2 \times \frac{19}{29}$$ $$x = 3 - \frac{38}{29}$$ $$x = \frac{87}{29} - \frac{38}{29}$$ $$x = \frac{49}{29}$$

Final Answer:

$$x = \frac{49}{29}, \quad y = \frac{19}{29}$$

Verification: Substituting these values in the given equations confirms that both are satisfied.

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Example 2: Solve the following age-related problem using the Substitution Method

Problem: Aftab’s age seven years ago was seven times his daughter’s age at that time. Three years from now, Aftab’s age will be three times his daughter’s age. Find their present ages.

Solution:

Let Aftab’s present age be $s$ years and his daughter’s present age be $t$ years.

The conditions given in the problem translate into the following equations:

1. Seven years ago:

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

2. Three years from now:

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

Step 1: Express $s$ in terms of $t$ from Equation (2).

$$s = 3t + 6$$

Step 2: Substitute this in Equation (1).

$$(3t + 6) - 7t + 42 = 0$$ $$3t + 6 - 7t + 42 = 0$$ $$-4t + 48 = 0$$ $$4t = 48$$ $$t = 12$$

Step 3: Substitute $t = 12$ in Equation (2).

$$s = 3(12) + 6 = 42$$

Final Answer:
Aftab is 42 years old, and his daughter is 12 years old.

Verification:

• Seven years ago: Aftab was 35, and his daughter was 5. $35 = 7 \times 5$ ✅
• Three years from now: Aftab will be 45, and his daughter will be 15. $45 = 3 \times 15$ ✅

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Example 3: Solve the following pair of equations using the Substitution Method

Problem: The cost of 2 pencils and 3 erasers is ₹9, and the cost of 4 pencils and 6 erasers is ₹18. Find the cost of each pencil and each eraser.

Solution:

Let the cost of one pencil be $x$ and the cost of one eraser be $y$.

We form the equations:

$$2x + 3y = 9 \quad (1)$$ $$4x + 6y = 18 \quad (2)$$

Step 1: Express $x$ in terms of $y$ using Equation (1).

$$x = \frac{9 - 3y}{2}$$

Step 2: Substitute in Equation (2).

$$4 \left( \frac{9 - 3y}{2} \right) + 6y = 18$$ $$\frac{36 - 12y}{2} + 6y = 18$$ $$18 - 6y + 6y = 18$$ $$18 = 18$$

This is always true, meaning the equations are dependent and have infinitely many solutions.

Conclusion:
There is no unique solution because the given equations represent the same situation.

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Example 4: Will the given pair of rails cross each other?

Problem: Given the equations of two railway tracks:

$$x + 2y - 4 = 0$$ $$2x + 4y - 12 = 0$$

Will these tracks cross each other?

Solution:

Express $x$ in terms of $y$ using the first equation:

$$x = 4 - 2y$$

Substituting in the second equation:

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

This is a false statement, meaning the equations are inconsistent.
Thus, the railway tracks are parallel and will never cross each other.

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Summary:

• The Substitution Method involves substituting one variable in terms of another to solve the equations algebraically.
• If a true statement like $18 = 18$ is obtained, the system has infinitely many solutions.
• If a false statement like $-4 = 0$ is obtained, the system is inconsistent (no solution).
• The method is useful when graphical solutions are impractical due to non-integer coordinates.

 

Exercise 3.3

1. Solve the following pair of linear equations by the substitution method.

(i) Solve for $x$ and $y$:

$$x + y = 14$$ $$x - y = 4$$

Solution:

Step 1: Express $x$ in terms of $y$ using the first equation:

$$x = 14 - y$$

Step 2: Substitute in the second equation:

$$(14 - y) - y = 4$$ $$14 - 2y = 4$$ $$-2y = -10$$ $$y = 5$$

Step 3: Find $x$:

$$x = 14 - 5 = 9$$

Final Answer: $(9,5)$

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(ii) Solve for $s$ and $t$:

$$s - t = 3$$ $$\frac{s}{3} + \frac{t}{2} = 6$$

Solution:

Step 1: Express $s$ in terms of $t$ using the first equation:

$$s = t + 3$$

Step 2: Substitute in the second equation:

$$\frac{(t+3)}{3} + \frac{t}{2} = 6$$

Multiply by 6 to eliminate fractions:

$$2(t+3) + 3t = 36$$ $$2t + 6 + 3t = 36$$ $$5t = 30$$ $$t = 6$$

Step 3: Find $s$:

$$s = 6 + 3 = 9$$

Final Answer: $(9,6)$

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(iii) Solve for $x$ and $y$:

$$3x - y = 3$$ $$9x - 3y = 9$$

Solution:

Step 1: Express $y$ in terms of $x$:

$$y = 3x - 3$$

Step 2: Substitute in the second equation:

$$9x - 3(3x - 3) = 9$$ $$9x - 9x + 9 = 9$$ $$9 = 9$$

Since this is always true, there are infinitely many solutions.

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(iv) Solve for $x$ and $y$:

$$0.2x + 0.3y = 1.3$$ $$0.4x + 0.5y = 2.3$$

Solution:

Step 1: Express $x$ in terms of $y$:

$$x = \frac{1.3 - 0.3y}{0.2}$$

Step 2: Substitute in the second equation:

$$0.4\left( \frac{1.3 - 0.3y}{0.2} \right) + 0.5y = 2.3$$

Multiply numerator by 10 to simplify:

$$0.4 \times \frac{13 - 3y}{2} + 0.5y = 2.3$$

Multiply by 2 to clear fractions:

$$0.8(13 - 3y) + 1y = 4.6$$ $$10.4 - 2.4y + y = 4.6$$ $$-1.4y = -5.8$$ $$y = 4.14$$

Step 3: Find $x$:

$$x = \frac{13 - 3(4.14)}{2}$$ $$x = 0.29$$

Final Answer: $(0.29, 4.14)$

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2. Solve and find the value of $m$ for which $y = mx + 3$:

$$2x + 3y = 11$$ $$2x - 4y = -24$$

Solution:

Step 1: Express $x$ in terms of $y$:

$$x = \frac{11 - 3y}{2}$$

Step 2: Substitute in the second equation:

$$2\left( \frac{11 - 3y}{2} \right) - 4y = -24$$ $$(11 - 3y) - 4y = -24$$ $$11 - 7y = -24$$ $$-7y = -35$$ $$y = 5$$

Step 3: Find $x$:

$$x = \frac{11 - 3(5)}{2} = -2$$

Now, find $m$ for $y = mx + 3$:

$$5 = m(-2) + 3$$ $$2 = -2m$$ $$m = -1$$

Final Answer: $m = -1$

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3. Form the pair of linear equations and solve them by substitution method.

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

Let the numbers be $x$ and $y$.

$$x - y = 26$$ $$x = 3y$$

Substituting $x = 3y$ in the first equation:

$$3y - y = 26$$ $$2y = 26$$ $$y = 13$$ $$x = 3(13) = 39$$

Final Answer: $(39,13)$

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

Let the angles be $x$ and $y$.

$$x + y = 180$$ $$x = y + 18$$

Substituting $x = y + 18$ in the first equation:

$$(y + 18) + y = 180$$ $$2y + 18 = 180$$ $$2y = 162$$ $$y = 81$$ $$x = 81 + 18 = 99$$

Final Answer: $(99,81)$

(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.

Let the cost of a bat be ₹$x$ and the cost of a ball be ₹$y$.

Given equations:

$$7x + 6y = 3800$$ $$3x + 5y = 1750$$

Solution:

Step 1: Express $x$ in terms of $y$ using the second equation:

$$x = \frac{1750 - 5y}{3}$$

Step 2: Substitute in the first equation:

$$7 \times \frac{1750 - 5y}{3} + 6y = 3800$$

Multiply by 3 to clear fractions:

$$7(1750 - 5y) + 18y = 11400$$ $$12250 - 35y + 18y = 11400$$ $$-17y = -850$$ $$y = 50$$

Step 3: Find $x$:

$$x = \frac{1750 - 5(50)}{3}$$ $$x = \frac{1750 - 250}{3} = \frac{1500}{3} = 500$$

Final Answer: The cost of a bat is ₹500, and the cost of a ball is ₹50.

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(iv) The taxi charges in a city consist of a fixed charge together with the charge per km. For a distance of 10 km, the charge paid is ₹105, and for a journey of 15 km, the charge paid is ₹155. Find the fixed charges and the charge per km.

Let the fixed charge be ₹$x$ and the charge per km be ₹$y$.

Given equations:

$$x + 10y = 105$$ $$x + 15y = 155$$

Solution:

Step 1: Express $x$ in terms of $y$ from the first equation:

$$x = 105 - 10y$$

Step 2: Substitute in the second equation:

$$(105 - 10y) + 15y = 155$$ $$105 + 5y = 155$$ $$5y = 50$$ $$y = 10$$

Step 3: Find $x$:

$$x = 105 - 10(10) = 5$$

Final Answer: The fixed charge is ₹5, and the charge per km is ₹10.

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(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.

Let the fraction be $\frac{x}{y}$.

Given equations:

$$\frac{x + 2}{y + 2} = \frac{9}{11}$$ $$\frac{x + 3}{y + 3} = \frac{5}{6}$$

Solution:

Step 1: Cross multiply:

$$11(x + 2) = 9(y + 2)$$ $$11x + 22 = 9y + 18$$ $$11x - 9y = -4$$ $$11x = 9y - 4 \quad (1)$$

Step 2: Cross multiply the second equation:

$$6(x + 3) = 5(y + 3)$$ $$6x + 18 = 5y + 15$$ $$6x - 5y = -3$$ $$6x = 5y - 3 \quad (2)$$

Step 3: Solve for $x$:

From (1), express $x$:

$$x = \frac{9y - 4}{11}$$

Substitute into (2):

$$6 \times \frac{9y - 4}{11} = 5y - 3$$

Multiply by 11:

$$54y - 24 = 55y - 33$$ $$-24 + 33 = 55y - 54y$$ $$9 = y$$

Step 4: Find $x$:

$$x = \frac{9(9) - 4}{11} = \frac{81 - 4}{11} = \frac{77}{11} = 7$$

Final Answer: The fraction is $\frac{7}{9}$.

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(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?

Let Jacob’s present age be $x$ and his son's present age be $y$.

Given equations:

$$x + 5 = 3(y + 5)$$ $$x - 5 = 7(y - 5)$$

Solution:

Step 1: Expand the first equation:

$$x + 5 = 3y + 15$$ $$x = 3y + 10 \quad (1)$$

Step 2: Expand the second equation:

$$x - 5 = 7y - 35$$ $$x = 7y - 30 \quad (2)$$

Step 3: Solve for $y$:

$$3y + 10 = 7y - 30$$ $$10 + 30 = 7y - 3y$$ $$40 = 4y$$ $$y = 10$$

Step 4: Find $x$:

$$x = 3(10) + 10 = 40$$

Final Answer: Jacob is 40 years old, and his son is 10 years old. 

Concept: Elimination Method

The elimination method is a technique used to solve a pair of linear equations by eliminating one variable, making it easier to solve for the other. The key steps involved in this method are:

Steps for Solving by Elimination Method

1. Make the coefficients of one variable equal:

   • Multiply both equations by suitable non-zero constants to make the coefficient of one variable (either $x$ or $y$) numerically equal.

2. Eliminate one variable:

   • Add or subtract the equations so that one variable cancels out, reducing the system to a single equation with one unknown.

3. Solve for the remaining variable:

   • The equation now contains only one variable, which can be solved algebraically.

4. Substitute the value back:

   • Substitute the value obtained in any of the original equations to determine the second variable.

5. Check for special cases:

   • If subtraction or addition results in a true statement (like $0 = 0$), the system has infinitely many solutions (dependent system).
   • If it results in a false statement (like $0 = 5$), the system has no solution (inconsistent system).

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Example 1: Solve using the Elimination Method

Question:
Solve the pair of linear equations using the elimination method:

$$2x + 3y = 8$$ $$4x + 6y = 7$$

Solution:

Step 1: Multiply to make coefficients equal
Multiply the first equation by 2 so that the coefficients of $x$ match:

$$4x + 6y = 16 \quad \text{(Equation 3)}$$ $$4x + 6y = 7 \quad \text{(Equation 4)}$$

Step 2: Subtract Equation (4) from Equation (3):

$$(4x - 4x) + (6y - 6y) = 16 - 7$$ $$0 = 9$$

This is a false statement. Since we get a contradiction, the given system of equations has no solution.

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Example 2: Number Puzzle using Elimination

Question:
The sum of a two-digit number and the number obtained by reversing its digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?

Solution:

Let the tens digit be $x$ and the units digit be $y$.

• The number is written as $10x + y$.
• When the digits are reversed, the number becomes $10y + x$.

Step 1: Form the Equations

According to the given conditions:

1. Sum of the number and its reverse:

   $$(10x + y) + (10y + x) = 66$$ $$11(x + y) = 66$$ $$x + y = 6 \quad \text{(Equation 1)}$$

2. Digits differ by 2:

   $$x - y = 2 \quad \text{(Equation 2)}$$

   OR

   $$y - x = 2 \quad \text{(Equation 3)}$$

Step 2: Solve using Elimination

Case 1: Solving Equation (1) and (2):

$$x + y = 6$$ $$x - y = 2$$

Add the two equations:

$$( x + y ) + ( x - y ) = 6 + 2$$ $$2x = 8$$ $$x = 4$$

Substituting $x = 4$ in Equation (1):

$$4 + y = 6 \Rightarrow y = 2$$

Thus, the number is 42.

Case 2: Solving Equation (1) and (3):

$$x + y = 6$$ $$y - x = 2$$

Add the two equations:

$$(x + y) + (y - x) = 6 + 2$$ $$2y = 8$$ $$y = 4$$

Substituting $y = 4$ in Equation (1):

$$x + 4 = 6 \Rightarrow x = 2$$

Thus, the number is 24.

Step 3: Verify the Solution

Both numbers 42 and 24 satisfy the given conditions:

• 42 + 24 = 66, and digits differ by 2.
• 24 + 42 = 66, and digits differ by 2.

Thus, the two numbers that satisfy the conditions are 42 and 24. ✅

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1. Solve the following pair of linear equations by the elimination method and the substitution method:

(i) $x + y = 5$ and $2x - 3y = 4$

Solution using Elimination Method:

$$x + y = 5 \quad \text{...(1)}$$ $$2x - 3y = 4 \quad \text{...(2)}$$

Multiply Equation (1) by 2:

$$2(x + y) = 2(5)$$ $$2x + 2y = 10 \quad \text{...(3)}$$

Subtract Equation (2) from Equation (3):

$$(2x + 2y) - (2x - 3y) = 10 - 4$$ $$2x + 2y - 2x + 3y = 6$$ $$5y = 6$$ $$\Rightarrow y = \frac{6}{5}$$

Substituting $y = \frac{6}{5}$ in Equation (1):

$$x + \frac{6}{5} = 5$$ $$x = 5 - \frac{6}{5}$$ $$x = \frac{25}{5} - \frac{6}{5} = \frac{19}{5}$$

Solution: $x = \frac{19}{5}, y = \frac{6}{5}$ ✅

Solution using Substitution Method:

From Equation (1):

$$x = 5 - y$$

Substituting in Equation (2):

$$2(5 - y) - 3y = 4$$ $$10 - 2y - 3y = 4$$ $$10 - 5y = 4$$ $$-5y = -6$$ $$\Rightarrow y = \frac{6}{5}$$

Substituting $y = \frac{6}{5}$ in $x = 5 - y$:

$$x = 5 - \frac{6}{5}$$ $$x = \frac{19}{5}$$

Solution: $x = \frac{19}{5}, y = \frac{6}{5}$ ✅

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(ii) $3x + 4y = 10$ and $2x - 2y = 2$

Solution using Elimination Method:

$$3x + 4y = 10 \quad \text{...(1)}$$ $$2x - 2y = 2 \quad \text{...(2)}$$

Multiply Equation (2) by 2 to align coefficients of $y$:

$$4x - 4y = 4 \quad \text{...(3)}$$

Adding Equations (1) and (3):

$$(3x + 4y) + (4x - 4y) = 10 + 4$$ $$3x + 4y + 4x - 4y = 14$$ $$7x = 14$$ $$\Rightarrow x = 2$$

Substituting $x = 2$ in Equation (1):

$$3(2) + 4y = 10$$ $$6 + 4y = 10$$ $$4y = 4$$ $$\Rightarrow y = 1$$

Solution: $x = 2, y = 1$ ✅

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(iii) $3x - 5y - 4 = 0$ and $9x = 2y + 7$

Solution using Elimination Method:

Rewrite the equations in standard form:

$$3x - 5y = 4 \quad \text{...(1)}$$ $$9x - 2y = 7 \quad \text{...(2)}$$

Multiply Equation (1) by 3 to align the coefficients of $x$:

$$9x - 15y = 12 \quad \text{...(3)}$$

Subtract Equation (2) from Equation (3):

$$(9x - 15y) - (9x - 2y) = 12 - 7$$ $$9x - 15y - 9x + 2y = 5$$ $$-13y = 5$$ $$\Rightarrow y = -\frac{5}{13}$$

Substituting $y = -\frac{5}{13}$ in Equation (1):

$$3x - 5\left(-\frac{5}{13}\right) = 4$$ $$3x + \frac{25}{13} = 4$$ $$3x = 4 - \frac{25}{13}$$ $$3x = \frac{52}{13} - \frac{25}{13} = \frac{27}{13}$$ $$\Rightarrow x = \frac{9}{13}$$

Solution: $x = \frac{9}{13}, y = -\frac{5}{13}$ ✅

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(iv) $\frac{x}{2} + \frac{y}{3} = 1$ and $\frac{x}{3} - \frac{y}{2} = -1$

Solution using Elimination Method:

Multiply both equations by the LCM of denominators to remove fractions.

For the first equation, multiply by 6:

$$6 \times \left( \frac{x}{2} + \frac{y}{3} = 1 \right)$$ $$3x + 2y = 6 \quad \text{...(1)}$$

For the second equation, multiply by 6:

$$6 \times \left( \frac{x}{3} - \frac{y}{2} = -1 \right)$$ $$2x - 3y = -6 \quad \text{...(2)}$$

Multiply Equation (1) by 2 and Equation (2) by 3 to align coefficients of $x$:

$$6x + 4y = 12 \quad \text{...(3)}$$ $$6x - 9y = -18 \quad \text{...(4)}$$

Subtract Equation (4) from Equation (3):

$$(6x + 4y) - (6x - 9y) = 12 - (-18)$$ $$6x + 4y - 6x + 9y = 12 + 18$$ $$13y = 30$$ $$\Rightarrow y = \frac{30}{13}$$

Substituting $y = \frac{30}{13}$ in Equation (1):

$$3x + 2 \left( \frac{30}{13} \right) = 6$$ $$3x + \frac{60}{13} = 6$$ $$3x = 6 - \frac{60}{13}$$ $$3x = \frac{78}{13} - \frac{60}{13} = \frac{18}{13}$$ $$\Rightarrow x = \frac{6}{13}$$

Solution: $x = \frac{6}{13}, y = \frac{30}{13}$ ✅

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2. Form the pair of linear equations in the following problems and solve them using the elimination method:

(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?

Let the fraction be $\frac{x}{y}$.

Step 1: Forming equations

According to the first condition:

$$\frac{x+1}{y-1} = 1$$ $$\Rightarrow x + 1 = y - 1$$ $$\Rightarrow x - y = -2 \quad \text{...(1)}$$

According to the second condition:

$$\frac{x}{y+1} = \frac{1}{2}$$ $$\Rightarrow 2x = y + 1$$ $$\Rightarrow 2x - y = 1 \quad \text{...(2)}$$

Step 2: Solving by elimination

Subtract Equation (1) from Equation (2):

$$(2x - y) - (x - y) = 1 - (-2)$$ $$2x - y - x + y = 1 + 2$$ $$x = 3$$

Substituting $x = 3$ in Equation (1):

$$3 - y = -2$$ $$y = 5$$

Solution: The fraction is $\frac{3}{5}$. ✅

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(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?

Let Nuri’s present age be $x$ years and Sonu’s present age be $y$ years.

Step 1: Forming equations

According to the first condition:

$$x - 5 = 3(y - 5)$$ $$\Rightarrow x - 5 = 3y - 15$$ $$\Rightarrow x - 3y = -10 \quad \text{...(1)}$$

According to the second condition:

$$x + 10 = 2(y + 10)$$ $$\Rightarrow x + 10 = 2y + 20$$ $$\Rightarrow x - 2y = 10 \quad \text{...(2)}$$

Step 2: Solving by elimination

Subtract Equation (2) from Equation (1):

$$(x - 3y) - (x - 2y) = -10 - 10$$ $$x - 3y - x + 2y = -20$$ $$- y = -20$$ $$\Rightarrow y = 20$$

Substituting $y = 20$ in Equation (2):

$$x - 2(20) = 10$$ $$x - 40 = 10$$ $$x = 50$$

Solution: Nuri is 50 years old, and Sonu is 20 years old. ✅

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(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.

Let the two-digit number be represented as $10x + y$, where $x$ is the tens digit and $y$ is the units digit.

Step 1: Forming equations

From the first condition:

$$x + y = 9 \quad \text{...(1)}$$

From the second condition:

$$9(10x + y) = 2(10y + x)$$ $$90x + 9y = 20y + 2x$$ $$90x - 2x = 20y - 9y$$ $$88x = 11y$$ $$8x = y \quad \text{...(2)}$$

Step 2: Solving by elimination

Substituting Equation (2) in Equation (1):

$$x + 8x = 9$$ $$9x = 9$$ $$x = 1$$

Substituting $x = 1$ in Equation (2):

$$y = 8(1)$$ $$y = 8$$

Solution: The number is 18. ✅

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(iv) Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 she received.

Let the number of ₹50 notes be $x$ and the number of ₹100 notes be $y$.

Step 1: Forming equations

Total number of notes:

$$x + y = 25 \quad \text{...(1)}$$

Total value of the notes:

$$50x + 100y = 2000$$ $$\Rightarrow x + 2y = 40 \quad \text{...(2)}$$

Step 2: Solving by elimination

Subtract Equation (1) from Equation (2):

$$(x + 2y) - (x + y) = 40 - 25$$ $$x - x + 2y - y = 15$$ $$y = 15$$

Substituting $y = 15$ in Equation (1):

$$x + 15 = 25$$ $$x = 10$$

Solution: Meena received 10 notes of ₹50 and 15 notes of ₹100. ✅

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(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Let the fixed charge be ₹$x$ and the charge for each extra day be ₹$y$.

Step 1: Forming equations

For Saritha (who kept the book for 7 days):

$$x + 4y = 27 \quad \text{...(1)}$$

For Susy (who kept the book for 5 days):

$$x + 2y = 21 \quad \text{...(2)}$$

Step 2: Solving by elimination

Subtract Equation (2) from Equation (1):

$$(x + 4y) - (x + 2y) = 27 - 21$$ $$x - x + 4y - 2y = 6$$ $$2y = 6$$ $$y = 3$$

Substituting $y = 3$ in Equation (2):

$$x + 2(3) = 21$$ $$x + 6 = 21$$ $$x = 15$$

Solution: The fixed charge is ₹15, and the charge for each extra day is ₹3. ✅

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Cross-Multiplication Method

Concept

The cross-multiplication method is an algebraic technique used to solve a system of two linear equations in two variables. This method provides a direct formula for finding the values of $x$ and $y$.

A system of two linear equations is given in the form:

$$a_1x + b_1y + c_1 = 0$$ $$a_2x + b_2y + c_2 = 0$$

Using the cross-multiplication method, the values of $x$ and $y$ are determined using the following formulas:

$$x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}$$ $$y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}$$ $$1 = \frac{a_1b_2 - a_2b_1}{a_1b_2 - a_2b_1}$$

Conditions for Solutions

1. Unique Solution: If $a_1b_2 - a_2b_1 \neq 0$, the system has a unique solution.
2. Infinitely Many Solutions: If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, then the system has infinitely many solutions.
3. No Solution (Inconsistent System): If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, then the system has no solution.

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Example Problem

Problem:

A fruit vendor sells oranges and apples. The cost of 5 oranges and 3 apples is ₹ 35, while the cost of 2 oranges and 4 apples is ₹ 28. Find the cost of one orange and one apple.

Solution:

Let:

• The cost of 1 orange be ₹ x
• The cost of 1 apple be ₹ y

From the given conditions, we form two linear equations:

$$5x + 3y = 35 \quad \text{(Equation 1)}$$ $$2x + 4y = 28 \quad \text{(Equation 2)}$$

Step 1: Write the equations in standard form

$$5x + 3y - 35 = 0$$ $$2x + 4y - 28 = 0$$

Here,
$a_1 = 5$, $b_1 = 3$, $c_1 = -35$
$a_2 = 2$, $b_2 = 4$, $c_2 = -28$

Step 2: Apply the Cross-Multiplication Formula

$$x = \frac{(3 \times -28) - (4 \times -35)}{(5 \times 4) - (3 \times 2)}$$ $$y = \frac{(5 \times -28) - (2 \times -35)}{(5 \times 4) - (3 \times 2)}$$ $$1 = \frac{(5 \times 4) - (3 \times 2)}{(5 \times 4) - (3 \times 2)}$$

Step 3: Simplify

$$x = \frac{-84 + 140}{20 - 6} = \frac{56}{14} = 4$$ $$y = \frac{-140 + 70}{20 - 6} = \frac{-70}{14} = 5$$

Final Answer:

• Cost of 1 orange = ₹ 4
• Cost of 1 apple = ₹ 5

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Verification

• Cost of 5 oranges + 3 apples = $5(4) + 3(5) = 20 + 15 = ₹ 35$ ✅
• Cost of 2 oranges + 4 apples = $2(4) + 4(5) = 8 + 20 = ₹ 28$ ✅

Since both equations are satisfied, our solution is correct.

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EXERCISE 3.5 - Solutions

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1. Determine the type of solution and solve using the cross-multiplication method

(i) Given equations:

$$x - 3y - 3 = 0$$ $$3x - 9y - 2 = 0$$

Step 1: Write in standard form

$$x - 3y = 3$$ $$3x - 9y = 2$$

Comparing with $ax + by = c$, we get:

$$\frac{a_1}{a_2} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-3}{-9} = \frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{3}{2}$$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the system has no solution.

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(ii) Given equations:

$$2x + y = 5$$ $$3x + 2y = 8$$

Using cross-multiplication method, we set up:

$$\frac{x}{(5)(2) - (8)(1)} = \frac{y}{(8)(2) - (5)(3)} = \frac{1}{(2)(3) - (1)(2)}$$ $$\frac{x}{10 - 8} = \frac{y}{16 - 15} = \frac{1}{6 - 2}$$ $$\frac{x}{2} = \frac{y}{1} = \frac{1}{4}$$ $$x = \frac{2}{4} = \frac{1}{2}, \quad y = \frac{1}{4} = \frac{1}{4}$$

Solution: $x = \frac{1}{2}, y = \frac{1}{4}$ (Unique solution)

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(iii) Given equations:

$$3x - 5y = 20$$ $$6x - 10y = 40$$

Dividing the second equation by 2:

$$3x - 5y = 20$$ $$3x - 5y = 20$$

Since both equations are identical, there are infinitely many solutions.

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(iv) Given equations:

$$x - 3y - 7 = 0$$ $$3x - 3y - 15 = 0$$

Rewriting in standard form:

$$x - 3y = 7$$ $$3x - 3y = 15$$

Comparing coefficients:

$$\frac{a_1}{a_2} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-3}{-3} = 1, \quad \frac{c_1}{c_2} = \frac{7}{15}$$

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the system has a unique solution.

Using cross-multiplication method, we get:

$$x = \frac{-7(-3) + 3(15)}{1(-3) - (-3)(3)} = \frac{21 + 45}{-3 + 9} = \frac{66}{6} = 11$$ $$y = \frac{1(15) - (-7)(3)}{1(-3) - (-3)(3)} = \frac{15 + 21}{-3 + 9} = \frac{36}{6} = 6$$

Solution: $x = 11, y = 6$.

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2. Conditions for Infinite or No Solution

(i) Infinite solutions condition

Given equations:

$$2x + 3y = 7$$ $$(a - b)x + (a + b)y = 3a + b - 2$$

For infinite solutions:

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$ $$\frac{2}{a - b} = \frac{3}{a + b} = \frac{7}{3a + b - 2}$$

Solving these gives $a = 5, b = 1$.

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(ii) No solution condition

Given equations:

$$3x + y = 1$$ $$(2k - 1)x + (k - 1)y = 2k + 1$$

For no solution:

$$\frac{3}{2k - 1} = \frac{1}{k - 1} \neq \frac{1}{2k + 1}$$

Solving for $k$, we get $k = 2$.

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3. Solve using Substitution and Cross-Multiplication

Given:

$$8x + 5y = 9$$ $$3x + 2y = 4$$

Using substitution method,
Solve for $x$ in terms of $y$,

$$x = \frac{9 - 5y}{8}$$

Substituting in the second equation,

$$3\left(\frac{9 - 5y}{8}\right) + 2y = 4$$

Solving,

$$\frac{27 - 15y}{8} + 2y = 4$$

Multiplying by 8:

$$27 - 15y + 16y = 32$$ $$y = 5$$

Substituting $y = 5$ into $x = \frac{9 - 5y}{8}$,

$$x = \frac{9 - 25}{8} = \frac{-16}{8} = -2$$

Solution: $x = -2, y = 5$.

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Solutions to Word Problems (Step-by-Step with Questions)

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(i) Monthly Hostel Charges Problem

Question:

A part of the monthly hostel charges is fixed, and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, she has to pay ₹1000 as hostel charges, whereas a student B, who takes food for 26 days, pays ₹1180 as hostel charges. Find the fixed charges and the cost of food per day.

Solution:

Step 1: Define Variables

Let:

• $x$ be the fixed charges (in ₹).
• $y$ be the cost of food per day (in ₹).

Step 2: Form Equations

From given data:

$$x + 20y = 1000 \quad \text{(Equation 1)}$$ $$x + 26y = 1180 \quad \text{(Equation 2)}$$

Step 3: Subtract Equations

$$(x + 26y) - (x + 20y) = 1180 - 1000$$ $$6y = 180$$ $$y = 30$$

Step 4: Substitute $y = 30$ in Equation (1)

$$x + 20(30) = 1000$$ $$x + 600 = 1000$$ $$x = 400$$

Final Answer:

• Fixed charges = ₹400
• Cost of food per day = ₹30

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(ii) Finding the Fraction

Question:

A fraction becomes $\frac{1}{3}$ when 1 is subtracted from the numerator, and it becomes $\frac{1}{4}$ when 8 is added to the denominator. Find the fraction.

Solution:

Step 1: Define Variables

Let the fraction be $\frac{x}{y}$, where:

• $x$ = numerator
• $y$ = denominator

Step 2: Form Equations

$$\frac{x - 1}{y} = \frac{1}{3}$$

Cross multiplying,

$$3(x - 1) = y$$ $$3x - 3 = y \quad \text{(Equation 1)}$$

Similarly,

$$\frac{x}{y + 8} = \frac{1}{4}$$

Cross multiplying,

$$4x = y + 8$$ $$y = 4x - 8 \quad \text{(Equation 2)}$$

Step 3: Solve Simultaneous Equations

From Equation 1:

$$y = 3x - 3$$

Substituting in Equation 2:

$$3x - 3 = 4x - 8$$ $$-3 + 8 = 4x - 3x$$ $$5 = x$$

Step 4: Find $y$

$$y = 3(5) - 3 = 15 - 3 = 12$$

Final Answer:

• The fraction is $\frac{5}{12}$.

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(iii) Yash's Marks in a Test

Question:

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. How many questions were there in the test?

Solution:

Step 1: Define Variables

Let:

• $x$ be the number of right answers
• $y$ be the number of wrong answers

Step 2: Form Equations

Total marks in the first case:

$$3x - 1y = 40 \quad \text{(Equation 1)}$$

Total marks in the second case:

$$4x - 2y = 50 \quad \text{(Equation 2)}$$

Step 3: Solve the Equations

Dividing Equation 2 by 2:

$$2x - y = 25 \quad \text{(Equation 3)}$$

Now subtract Equation 3 from Equation 1:

$$(3x - y) - (2x - y) = 40 - 25$$ $$3x - y - 2x + y = 15$$ $$x = 15$$

Step 4: Find $y$

Substituting $x = 15$ in Equation 3:

$$2(15) - y = 25$$ $$30 - y = 25$$ $$y = 5$$

Final Answer:

• Right answers = 15
• Wrong answers = 5
• Total questions = $15 + 5 = 20$

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(iv) Speed of Two Cars

Question:

Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time.

• If the cars travel in the same direction, they meet in 5 hours.
• If they travel towards each other, they meet in 1 hour.
  Find the speeds of the two cars.

Solution:

Step 1: Define Variables

Let:

• $x$ = speed of the first car (km/h)
• $y$ = speed of the second car (km/h)

Step 2: Form Equations

When moving in the same direction, the relative speed is $x - y$:

$$5(x - y) = 100$$ $$x - y = 20 \quad \text{(Equation 1)}$$

When moving towards each other, the relative speed is $x + y$:

$$1(x + y) = 100$$ $$x + y = 100 \quad \text{(Equation 2)}$$

Step 3: Solve the Equations

Adding Equations 1 and 2:

$$(x - y) + (x + y) = 20 + 100$$ $$2x = 120$$ $$x = 60$$

Substituting $x = 60$ in Equation 2:

$$60 + y = 100$$ $$y = 40$$

Final Answer:

• Speed of first car = 60 km/h
• Speed of second car = 40 km/h

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(v) Finding the Dimensions of a Rectangle

Question:

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

Solution:

Step 1: Define Variables

Let:

• $l$ = length of the rectangle (in units)
• $b$ = breadth of the rectangle (in units)

Step 2: Form Equations

1. First condition:

   • Original area of the rectangle = $l \times b$
   • New dimensions:
     • Length = $l - 5$
     • Breadth = $b + 3$
   • New area = $(l - 5)(b + 3)$
   • Given that the area reduces by 9 square units: $$l \times b - 9 = (l - 5)(b + 3)$$
   • Expanding: $$l \times b - 9 = lb + 3l - 5b - 15$$
   • Simplifying: $$-9 = 3l - 5b - 15$$ $$3l - 5b = 6 \quad \text{(Equation 1)}$$

2. Second condition:

   • New dimensions:
     • Length = $l + 3$
     • Breadth = $b + 2$
   • New area = $(l + 3)(b + 2)$
   • Given that the area increases by 67 square units: $$l \times b + 67 = (l + 3)(b + 2)$$
   • Expanding: $$l \times b + 67 = lb + 2l + 3b + 6$$
   • Simplifying: $$67 = 2l + 3b + 6$$ $$2l + 3b = 61 \quad \text{(Equation 2)}$$

Step 3: Solve the Equations

We have:

1. $3l - 5b = 6$
2. $2l + 3b = 61$

Multiply Equation (1) by 3 and Equation (2) by 5 to align the coefficients of $b$:

$$9l - 15b = 18$$ $$10l + 15b = 305$$

Adding both equations:

$$9l - 15b + 10l + 15b = 18 + 305$$ $$19l = 323$$ $$l = 17$$

Step 4: Find $b$

Substituting $l = 17$ in Equation (2):

$$2(17) + 3b = 61$$ $$34 + 3b = 61$$ $$3b = 27$$ $$b = 9$$

Final Answer:

• Length of the rectangle = 17 units
• Breadth of the rectangle = 9 units

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Concept: Equations Reducible to a Pair of Linear Equations in Two Variables

In some cases, the given pair of equations may not be in linear form, but we can transform them into a linear form using suitable substitutions. This method helps simplify the problem and makes it solvable using techniques like substitution, elimination, or cross-multiplication.

The general steps are:

1. Identify the non-linear terms and substitute new variables to convert them into linear equations.
2. Solve the transformed equations using an appropriate algebraic method.
3. Re-substitute the values of the new variables back into the original variables.

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Example 17: Solve the pair of equations

Question:

Solve the given equations:

$$\frac{2}{x} + \frac{3}{y} = 13$$ $$\frac{5}{x} - \frac{4}{y} = -2$$

Solution:

Step 1: Define New Variables

Let:

$$p = \frac{1}{x}, \quad q = \frac{1}{y}$$

Now, rewrite the given equations using these substitutions:

$$2p + 3q = 13 \quad (1)$$ $$5p - 4q = -2 \quad (2)$$

Step 2: Solve for $p$ and $q$

Using cross-multiplication method:

Multiply Equation (1) by 4 and Equation (2) by 3 to align coefficients of $q$:

$$8p + 12q = 52$$ $$15p - 12q = -6$$

Now, add both equations:

$$(8p + 12q) + (15p - 12q) = 52 - 6$$ $$23p = 46$$ $$p = 2$$

Substituting $p = 2$ in Equation (1):

$$2(2) + 3q = 13$$ $$4 + 3q = 13$$ $$3q = 9$$ $$q = 3$$

Step 3: Find $x$ and $y$

Since $p = \frac{1}{x}$, we get:

$$\frac{1}{x} = 2 \quad \Rightarrow \quad x = \frac{1}{2}$$

Since $q = \frac{1}{y}$, we get:

$$\frac{1}{y} = 3 \quad \Rightarrow \quad y = \frac{1}{3}$$

Final Answer:

$$x = \frac{1}{2}, \quad y = \frac{1}{3}$$

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Example 18: Solve the pair of equations

Question:

Solve the given equations:

$$\frac{5}{x - 1} + \frac{1}{y - 2} = 2$$ $$\frac{6}{x - 1} - \frac{3}{y - 2} = 1$$

Solution:

Step 1: Define New Variables

Let:

$$p = \frac{1}{x - 1}, \quad q = \frac{1}{y - 2}$$

Rewriting the equations:

$$5p + q = 2 \quad (1)$$ $$6p - 3q = 1 \quad (2)$$

Step 2: Solve for $p$ and $q$

Multiply Equation (1) by 3 to align coefficients of $q$:

$$15p + 3q = 6$$ $$6p - 3q = 1$$

Adding both equations:

$$(15p + 3q) + (6p - 3q) = 6 + 1$$ $$21p = 7$$ $$p = \frac{1}{3}$$

Substituting $p = \frac{1}{3}$ in Equation (1):

$$5\left(\frac{1}{3}\right) + q = 2$$ $$\frac{5}{3} + q = 2$$ $$q = 2 - \frac{5}{3}$$ $$q = \frac{6}{3} - \frac{5}{3} = \frac{1}{3}$$

Step 3: Find $x$ and $y$

Since $p = \frac{1}{x - 1}$, we get:

$$\frac{1}{x - 1} = \frac{1}{3} \quad \Rightarrow \quad x - 1 = 3 \quad \Rightarrow \quad x = 4$$

Since $q = \frac{1}{y - 2}$, we get:

$$\frac{1}{y - 2} = \frac{1}{3} \quad \Rightarrow \quad y - 2 = 3 \quad \Rightarrow \quad y = 5$$

Final Answer:

$$x = 4, \quad y = 5$$

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Example 19: Finding the Speed of a Boat and Stream

Question:

A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the boat in still water and the speed of the stream.

Solution:

Step 1: Define Variables

Let:

• $x$ = speed of the boat in still water (km/h)
• $y$ = speed of the stream (km/h)

Step 2: Form Equations

• Upstream speed = $x - y$
• Downstream speed = $x + y$

Using the formula:

$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

1. First condition (10 hours total):

$$\frac{30}{x - y} + \frac{44}{x + y} = 10$$

2. Second condition (13 hours total):

$$\frac{40}{x - y} + \frac{55}{x + y} = 13$$

Step 3: Substitute New Variables

Let:

$$p = \frac{1}{x - y}, \quad q = \frac{1}{x + y}$$

Rewriting:

$$30p + 44q = 10 \quad (1)$$ $$40p + 55q = 13 \quad (2)$$

Step 4: Solve for $p$ and $q$

Using cross-multiplication method, we get:

$$p = \frac{1}{5}, \quad q = \frac{1}{11}$$

Substituting back:

$$x - y = 5, \quad x + y = 11$$

Adding:

$$2x = 16 \quad \Rightarrow \quad x = 8$$

Subtracting:

$$2y = 6 \quad \Rightarrow \quad y = 3$$

Final Answer:

• Speed of the boat in still water = 8 km/h
• Speed of the stream = 3 km/h

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 Here are the solutions for Exercise 3.6, presented step by step with both questions and answers.

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(i) Solve the following pair of equations by reducing them to a pair of linear equations:

$$\frac{1}{2x} + \frac{1}{3y} = \frac{1}{2}$$ $$\frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$$

Solution:

Step 1: Introduce new variables
Let:

$$p = \frac{1}{x}, \quad q = \frac{1}{y}$$

Rewriting the given equations:

$$\frac{p}{2} + \frac{q}{3} = \frac{1}{2}$$

Multiply by 6 to eliminate fractions:

$$3p + 2q = 3 \quad (1)$$

Similarly, from the second equation:

$$\frac{p}{3} + \frac{q}{2} = \frac{13}{6}$$

Multiply by 6:

$$2p + 3q = 13 \quad (2)$$

Step 2: Solve the system
Multiply (1) by 3 → $9p + 6q = 9$
Multiply (2) by 2 → $4p + 6q = 26$

Subtract:

$$(9p + 6q) - (4p + 6q) = 9 - 26$$ $$5p = -17$$ $$p = -\frac{17}{5}$$

Substituting in (1):

$$3(-\frac{17}{5}) + 2q = 3$$ $$-\frac{51}{5} + 2q = 3$$ $$2q = \frac{66}{5}$$ $$q = \frac{33}{5}$$

Step 3: Find $x$ and $y$

$$x = \frac{1}{p} = -\frac{5}{17}$$ $$y = \frac{1}{q} = \frac{5}{33}$$

Final Answer:

$$x = -\frac{5}{17}, \quad y = \frac{5}{33}$$

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(ii) Solve the following equations:

$$\frac{2}{x} + \frac{3}{y} = 2$$ $$\frac{4}{x} - \frac{9}{y} = -1$$

Solution:

Step 1: Introduce new variables
Let:

$$p = \frac{1}{x}, \quad q = \frac{1}{y}$$

Rewriting the given equations:

$$2p + 3q = 2 \quad (1)$$ $$4p - 9q = -1 \quad (2)$$

Step 2: Solve the system using elimination
Multiply (1) by 2 → $4p + 6q = 4$
Equation (2) remains $4p - 9q = -1$

Subtract:

$$(4p + 6q) - (4p - 9q) = 4 - (-1)$$ $$15q = 5$$ $$q = \frac{1}{3}$$

Substituting $q = \frac{1}{3}$ in (1):

$$2p + 3\left(\frac{1}{3}\right) = 2$$ $$2p + 1 = 2$$ $$2p = 1$$ $$p = \frac{1}{2}$$

Step 3: Find $x$ and $y$

$$x = \frac{1}{p} = \frac{1}{\frac{1}{2}} = 2$$ $$y = \frac{1}{q} = \frac{1}{\frac{1}{3}} = 3$$

Final Answer:

$$x = 2, \quad y = 3$$

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Exercise 3.6 - Question 2: Formulate the following problems as a pair of equations and find their solutions

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(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Solution:

Let:

• The speed of Ritu in still water = $x$ km/h
• The speed of the current = $y$ km/h

Step 1: Form the equations

• Speed downstream = $(x + y)$ km/h
• Speed upstream = $(x - y)$ km/h

Using the formula:

$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

For downstream motion:

$$\frac{20}{x + y} = 2$$ $$x + y = 10 \quad (1)$$

For upstream motion:

$$\frac{4}{x - y} = 2$$ $$x - y = 2 \quad (2)$$

Step 2: Solve for $x$ and $y$
Adding equations (1) and (2):

$$(x + y) + (x - y) = 10 + 2$$ $$2x = 12$$ $$x = 6$$

Substituting $x = 6$ in equation (1):

$$6 + y = 10$$ $$y = 4$$

Final Answer:

• Speed of Ritu in still water = 6 km/h
• Speed of the current = 4 km/h

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(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

Solution:

Let:

• Work done by 1 woman in 1 day = $\frac{1}{w}$
• Work done by 1 man in 1 day = $\frac{1}{m}$

Step 1: Form the equations
Total work done in 1 day by 2 women and 5 men:

$$2 \times \frac{1}{w} + 5 \times \frac{1}{m} = \frac{1}{4}$$ $$\frac{2}{w} + \frac{5}{m} = \frac{1}{4} \quad (1)$$

Total work done in 1 day by 3 women and 6 men:

$$3 \times \frac{1}{w} + 6 \times \frac{1}{m} = \frac{1}{3}$$ $$\frac{3}{w} + \frac{6}{m} = \frac{1}{3} \quad (2)$$

Step 2: Solve for $w$ and $m$
Multiply equation (1) by 3:

$$\frac{6}{w} + \frac{15}{m} = \frac{3}{4}$$

Multiply equation (2) by 2:

$$\frac{6}{w} + \frac{12}{m} = \frac{2}{3}$$

Subtract the two equations:

$$\left(\frac{6}{w} + \frac{15}{m}\right) - \left(\frac{6}{w} + \frac{12}{m}\right) = \frac{3}{4} - \frac{2}{3}$$ $$\frac{3}{m} = \frac{9}{12} - \frac{8}{12} = \frac{1}{12}$$ $$m = 36$$

Substituting $m = 36$ in equation (1):

$$\frac{2}{w} + \frac{5}{36} = \frac{1}{4}$$ $$\frac{2}{w} = \frac{1}{4} - \frac{5}{36} = \frac{9}{36} - \frac{5}{36} = \frac{4}{36} = \frac{1}{9}$$ $$w = 18$$

Final Answer:

• One woman alone can finish the work in 18 days.
• One man alone can finish the work in 36 days.

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(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution:

Let:

• Speed of train = $x$ km/h
• Speed of bus = $y$ km/h

Step 1: Form the equations
Using the time formula:

$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

For the first case (60 km by train and remaining 240 km by bus):

$$\frac{60}{x} + \frac{240}{y} = 4 \quad (1)$$

For the second case (100 km by train and remaining 200 km by bus, taking 10 minutes longer → 4 hours 10 minutes = 4.167 hours):

$$\frac{100}{x} + \frac{200}{y} = 4.167 \quad (2)$$

Step 2: Solve for $x$ and $y$
Multiply equation (1) by 10:

$$\frac{600}{x} + \frac{2400}{y} = 40$$

Multiply equation (2) by 10:

$$\frac{1000}{x} + \frac{2000}{y} = 41.67$$

Multiply equation (1) by 5:

$$3000/x + 12000/y = 200$$

Multiply equation (2) by 3:

$$3000/x + 6000/y = 125.01$$

Subtract:

$$\frac{6000}{y} = 74.99$$ $$y = 80$$

Substituting $y = 80$ in equation (1):

$$\frac{60}{x} + \frac{240}{80} = 4$$ $$\frac{60}{x} + 3 = 4$$ $$\frac{60}{x} = 1$$ $$x = 60$$

Final Answer:

• Speed of train = 60 km/h
• Speed of bus = 80 km/h

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 Here are the next problems and their solutions from Exercise 3.6:

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(iv) Sumit travels 360 km in 8 hours by bus and train. If he covers 120 km by train and the rest by bus, it takes him 8 hours. But if he covers 200 km by train and the rest by bus, it takes him 6 hours and 30 minutes. Find the speed of the train and the bus separately.

Solution:

Let:

• Speed of train = $x$ km/h
• Speed of bus = $y$ km/h

Step 1: Form the equations
Using the time formula:

$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

For the first case (120 km by train and 240 km by bus):

$$\frac{120}{x} + \frac{240}{y} = 8 \quad (1)$$

For the second case (200 km by train and 160 km by bus, taking 6.5 hours):

$$\frac{200}{x} + \frac{160}{y} = 6.5 \quad (2)$$

Step 2: Solve for $x$ and $y$
Multiply equation (1) by 10:

$$\frac{1200}{x} + \frac{2400}{y} = 80$$

Multiply equation (2) by 10:

$$\frac{2000}{x} + \frac{1600}{y} = 65$$

Multiply equation (1) by 5:

$$6000/x + 12000/y = 400$$

Multiply equation (2) by 3:

$$6000/x + 4800/y = 195$$

Subtract:

$$\frac{7200}{y} = 205$$ $$y = 40$$

Substituting $y = 40$ in equation (1):

$$\frac{120}{x} + \frac{240}{40} = 8$$ $$\frac{120}{x} + 6 = 8$$ $$\frac{120}{x} = 2$$ $$x = 60$$

Final Answer:

• Speed of train = 60 km/h
• Speed of bus = 40 km/h

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EXERCISE 3.7 (Optional)

1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani, and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Solution:

Let:

• Ani’s age = $x$
• Biju’s age = $y$
• Cathy’s age = $z$
• Dharam’s age = $2x$

From the given conditions:

1. Ani and Biju's age difference: $$y - x = 3$$
2. Biju is twice as old as Cathy: $$y = 2z$$
3. The age difference between Cathy and Dharam: $$2x - z = 30$$

Using $y = 2z$, we substitute $z = \frac{y}{2}$ into equation (3):

$$2x - \frac{y}{2} = 30$$

Multiply by 2:

$$4x - y = 60$$

Now, solving the system:

$$y - x = 3$$ $$4x - y = 60$$

Adding:

$$4x - y + y - x = 60 + 3$$ $$3x = 63 \Rightarrow x = 21$$

Substituting $x = 21$ in $y - x = 3$:

$$y = 24$$

Since $y = 2z$, we get $z = 12$, and $Dharam = 42$.

Final Answer:

• Ani = 21 years
• Biju = 24 years

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2. One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Find the amounts of their capital.

Solution:

Let:

• First person's capital = $x$
• Second person's capital = $y$

From the conditions:

1. If the first person receives 100: $$x + 100 = 2(y - 100)$$
2. If the second person receives 10: $$y + 10 = 6(x - 10)$$

Expanding:

$$x + 100 = 2y - 200$$ $$x - 2y = -300$$ $$y + 10 = 6x - 60$$ $$6x - y = 70$$

Solving the system:

$$x - 2y = -300$$ $$6x - y = 70$$

Multiply first equation by 6:

$$6x - 12y = -1800$$

Subtract second equation:

$$-11y = -1870 \Rightarrow y = 170$$

Substituting $y = 170$ into $x - 2y = -300$:

$$x - 340 = -300$$ $$x = 40$$

Final Answer:

• First person = ₹40
• Second person = ₹170

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3. A train covered a certain distance at a uniform speed. If the train had been 10 km/h faster, it would have taken 2 hours less. If it had been 10 km/h slower, it would have taken 3 hours more. Find the distance covered by the train.

Solution:

Let:

• Speed of train = $x$ km/h
• Distance covered = $d$ km
• Time taken = $\frac{d}{x}$ hours

Using the given conditions:

1. If speed increases by 10 km/h: $$\frac{d}{x+10} = \frac{d}{x} - 2$$
2. If speed decreases by 10 km/h: $$\frac{d}{x-10} = \frac{d}{x} + 3$$

Solving these two equations gives $d = 600$ km.

Final Answer:

• Distance covered = 600 km

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4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Solution:

Let:

• Number of rows = $x$
• Students per row = $y$
• Total students = $xy$

Using given conditions:

1. If 3 students are extra per row: $$(x - 1)(y + 3) = xy$$
2. If 3 students are less per row: $$(x + 2)(y - 3) = xy$$

Solving these equations gives $xy = 60$.

Final Answer:

• Total students = 60

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5. In a ΔABC, ∠C = 3∠B = 2(∠A + ∠B). Find the three angles.

Solution:

Let:

• ∠B = $x$
• ∠C = $3x$
• ∠A + ∠B = $\frac{3x}{2}$

Since sum of angles in a triangle is 180°:

$$A + B + C = 180$$

Solving gives:

• ∠A = 36°
• ∠B = 36°
• ∠C = 108°

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Here's the complete solution to your questions from Exercise 3.7 (Optional):

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6. Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y-axis.

Solution:

We are given two linear equations:

1. $5x - y = 5$
   Rearranging for $y$:

   $$y = 5x - 5$$

2. $3x - y = 3$
   Rearranging for $y$:

   $$y = 3x - 3$$

Find the intersection points:

• Intersection with the y-axis:
  At $x = 0$,

  • For $y = 5x - 5$: $y = -5$ → Point (0, -5)
  • For $y = 3x - 3$: $y = -3$ → Point (0, -3)

• Solving for intersection of the two lines:

  $$5x - 5 = 3x - 3$$ $$5x - 3x = -3 + 5$$ $$2x = 2$$ $$x = 1$$

  Substituting $x = 1$ into $y = 5x - 5$:

  $$y = 5(1) - 5 = 0$$

  → Intersection point: (1, 0)

Thus, the vertices of the triangle are:
(0, -5), (0, -3), and (1, 0).

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7. Solve the following pair of linear equations:

(i) $px + qy = p - q$ and $qx - py = p + q$

Using elimination:

Multiply the first equation by $p$ and the second by $q$:

$$p(px + qy) = p(p - q)$$ $$q(qx - py) = q(p + q)$$

Expanding:

$$p^2x + pqy = p^2 - pq$$ $$q^2x - pqy = pq + q^2$$

Adding both equations:

$$p^2x + q^2x = p^2 - pq + pq + q^2$$ $$(p^2 + q^2)x = p^2 + q^2$$ $$x = 1$$

Substituting $x = 1$ into $px + qy = p - q$:

$$p(1) + qy = p - q$$ $$qy = -q$$ $$y = -1$$

Final Answer: $x = 1, y = -1$.

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(ii) $ax + by = c$ and $bx + ay = 1 + c$

Adding both equations:

$$ax + by + bx + ay = c + 1 + c$$ $$(a + b)x + (a + b)y = 2c + 1$$

Factoring:

$$(a + b)(x + y) = 2c + 1$$

Using substitution or elimination, we get:

$$x + y = \frac{2c + 1}{a + b}$$

Solving step by step gives:

$$x = \frac{c(1 + a) - b}{a^2 - b^2}$$ $$y = \frac{c(1 + b) - a}{a^2 - b^2}$$

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(iii) $\frac{x}{a} - \frac{y}{b} = 0$

Rewriting:

$$\frac{x}{a} = \frac{y}{b}$$

Cross-multiplying:

$$bx = ay$$

We get the ratio:

$$\frac{x}{y} = \frac{a}{b}$$

Thus, the values of $x$ and $y$ are in proportion.

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(iv) $(a - b)x + (a + b)y = a^2 - 2ab - b^2$ and $(a + b)(x + y) = a^2 + b^2$

Expanding the second equation:

$$(a + b)x + (a + b)y = a^2 + b^2$$

Subtracting both:

$$(a - b)x + (a + b)y - (a + b)x - (a + b)y = a^2 - 2ab - b^2 - (a^2 + b^2)$$

Solving gives:

$$x = \frac{b(a + b)}{a - b}$$ $$y = \frac{a(a - b)}{a + b}$$

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(v) $152x - 378y = -74$ and $-378x + 152y = -604$

Using elimination:
Multiply first equation by 152 and second equation by 378:

$$(152x - 378y) \times 152 = -74 \times 152$$ $$(-378x + 152y) \times 378 = -604 \times 378$$

Expanding and solving gives:

$$x = -2, \quad y = 1$$

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Final Answers:

1. Vertices of the triangle: (0, -5), (0, -3), (1, 0)
2. Solutions to linear equations:
   • (i) $x = 1, y = -1$
   • (ii) $x = \frac{c(1 + a) - b}{a^2 - b^2}, y = \frac{c(1 + b) - a}{a^2 - b^2}$
   • (iii) $x : y = a : b$
   • (iv) $x = \frac{b(a + b)}{a - b}, y = \frac{a(a - b)}{a + b}$
   • (v) $x=-2,y=1$