Chapter 3: Pair of Linear Equations in Two Variables
Exercise 3.1
Q1
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.
(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 that of one pen.
(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 ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.
Solution:
(i) Let the number of girls be \(x\) and the number of boys be \(y\).
Total students: \(x + y = 10\)
Number of girls is 4 more than boys: \(x = y + 4 \Rightarrow x - y = 4\)
Plotting these lines on a graph:
- For \(x + y = 10\): Points (5, 5), (4, 6), (6, 4).
- For \(x - y = 4\): Points (4, 0), (5, 1), (6, 2).
The lines intersect at the point (7, 3).
Answer: Number of girls = 7, Number of boys = 3.
(ii) Let the cost of 1 pencil be \(x\) and 1 pen be \(y\).
Case 1: \(5x + 7y = 50\)
Case 2: \(7x + 5y = 46\)
Plotting these lines on a graph:
- For \(5x + 7y = 50\): Points (3, 5), (10, 0).
- For \(7x + 5y = 46\): Points (8, -2), (3, 5).
The lines intersect at (3, 5).
Answer: Cost of one pencil = ₹ 3, Cost of one pen = ₹ 5.
Q2
On comparing the ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2}\) and \(\frac{c_1}{c_2}\), find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) \(5x - 4y + 8 = 0; \quad 7x + 6y - 9 = 0\)
(ii) \(9x + 3y + 12 = 0; \quad 18x + 6y + 24 = 0\)
(iii) \(6x - 3y + 10 = 0; \quad 2x - y + 9 = 0\)
(i) \(5x - 4y + 8 = 0; \quad 7x + 6y - 9 = 0\)
(ii) \(9x + 3y + 12 = 0; \quad 18x + 6y + 24 = 0\)
(iii) \(6x - 3y + 10 = 0; \quad 2x - y + 9 = 0\)
Solution:
(i) \(5x - 4y + 8 = 0\) and \(7x + 6y - 9 = 0\)
\(\frac{a_1}{a_2} = \frac{5}{7}\), \(\frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3}\)
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the lines intersect at a point.
(ii) \(9x + 3y + 12 = 0\) and \(18x + 6y + 24 = 0\)
\(\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}\), \(\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}\), \(\frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}\)
Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the lines are coincident.
(iii) \(6x - 3y + 10 = 0\) and \(2x - y + 9 = 0\)
\(\frac{a_1}{a_2} = \frac{6}{2} = 3\), \(\frac{b_1}{b_2} = \frac{-3}{-1} = 3\), \(\frac{c_1}{c_2} = \frac{10}{9}\)
Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the lines are parallel.
Q3
On comparing the ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2}\) and \(\frac{c_1}{c_2}\), find out whether the following pair of linear equations are consistent, or inconsistent.
(i) \(3x + 2y = 5; \quad 2x - 3y = 7\)
(ii) \(2x - 3y = 8; \quad 4x - 6y = 9\)
(iii) \(\frac{3}{2}x + \frac{5}{3}y = 7; \quad 9x - 10y = 14\)
(iv) \(5x - 3y = 11; \quad -10x + 6y = -22\)
(v) \(\frac{4}{3}x + 2y = 8; \quad 2x + 3y = 12\)
(i) \(3x + 2y = 5; \quad 2x - 3y = 7\)
(ii) \(2x - 3y = 8; \quad 4x - 6y = 9\)
(iii) \(\frac{3}{2}x + \frac{5}{3}y = 7; \quad 9x - 10y = 14\)
(iv) \(5x - 3y = 11; \quad -10x + 6y = -22\)
(v) \(\frac{4}{3}x + 2y = 8; \quad 2x + 3y = 12\)
Solution:
(i) \(\frac{3}{2} \neq \frac{2}{-3}\). Consistent (Intersecting).
(ii) \(\frac{2}{4} = \frac{-3}{-6} = \frac{1}{2}\). \(\frac{c_1}{c_2} = \frac{8}{9}\). Since \(\frac{1}{2} \neq \frac{8}{9}\), Inconsistent (Parallel).
(iii) \(\frac{3/2}{9} = \frac{1}{6}\). \(\frac{5/3}{-10} = \frac{-1}{6}\). Since \(\frac{1}{6} \neq \frac{-1}{6}\), Consistent (Intersecting).
(iv) \(\frac{5}{-10} = -\frac{1}{2}\), \(\frac{-3}{6} = -\frac{1}{2}\), \(\frac{11}{-22} = -\frac{1}{2}\). Consistent (Coincident/Dependent).
(v) \(\frac{4/3}{2} = \frac{2}{3}\), \(\frac{2}{3} = \frac{2}{3}\), \(\frac{8}{12} = \frac{2}{3}\). Consistent (Coincident/Dependent).
Q4
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) \(x + y = 5, \quad 2x + 2y = 10\)
(ii) \(x - y = 8, \quad 3x - 3y = 16\)
(iii) \(2x + y - 6 = 0, \quad 4x - 2y - 4 = 0\)
(iv) \(2x - 2y - 2 = 0, \quad 4x - 4y - 5 = 0\)
(i) \(x + y = 5, \quad 2x + 2y = 10\)
(ii) \(x - y = 8, \quad 3x - 3y = 16\)
(iii) \(2x + y - 6 = 0, \quad 4x - 2y - 4 = 0\)
(iv) \(2x - 2y - 2 = 0, \quad 4x - 4y - 5 = 0\)
Solution:
(i) \( \frac{1}{2} = \frac{1}{2} = \frac{5}{10} \). Consistent (Coincident). The lines overlap; infinitely many solutions.
(ii) \( \frac{1}{3} = \frac{-1}{-3} \neq \frac{8}{16} \). Inconsistent (Parallel). No solution.
(iii) \( \frac{2}{4} \neq \frac{1}{-2} \). Consistent (Intersecting).
Graphing \( 2x + y = 6 \) (Points: 0,6 and 3,0) and \( 4x - 2y = 4 \) (Points: 1,0 and 0,-2).
Solution: \( x = 2, y = 2 \).
(iv) \( \frac{2}{4} = \frac{-2}{-4} \neq \frac{-2}{-5} \). Inconsistent (Parallel). No solution.
Q5
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.
Solution:
Let length be \( l \) and width be \( w \).
1. Length is 4 m more than width: \( l = w + 4 \) or \( l - w = 4 \).
2. Half perimeter is 36: \( \frac{1}{2} \times 2(l + w) = 36 \Rightarrow l + w = 36 \).
Solving graphically or algebraically:
Substituting \( l = w + 4 \) into \( l + w = 36 \):
\( (w + 4) + w = 36 \Rightarrow 2w = 32 \Rightarrow w = 16 \).
\( l = 16 + 4 = 20 \).
Answer: Length = 20 m, Width = 16 m.
Q6
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
(ii) parallel lines
(iii) coincident lines
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Solution:
Given equation: \( 2x + 3y - 8 = 0 \)
(i) Intersecting lines: We need \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\).
Example: \( 3x + 2y - 7 = 0 \) (Here \(\frac{2}{3} \neq \frac{3}{2}\)).
(ii) Parallel lines: We need \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).
Example: \( 4x + 6y - 12 = 0 \) (Here \(\frac{2}{4} = \frac{3}{6} \neq \frac{-8}{-12}\)).
(iii) Coincident lines: We need \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\).
Example: \( 6x + 9y - 24 = 0 \) (Here all ratios are \(\frac{1}{3}\)).
Q7
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.
Solution:
Equation 1: \( x - y = -1 \)
Points: (0, 1), (-1, 0), (2, 3).
Equation 2: \( 3x + 2y = 12 \)
Points: (0, 6), (4, 0), (2, 3).
Graph:
The lines intersect at (2, 3).
The lines meet the x-axis at (-1, 0) and (4, 0).
Vertices of the triangle: (2, 3), (-1, 0), and (4, 0).
Exercise 3.2
Q1
Solve the following pair of linear equations by the substitution method.
(i) \( x + y = 14 \); \( x - y = 4 \)
(ii) \( s - t = 3 \); \( \frac{s}{3} + \frac{t}{2} = 6 \)
(iii) \( 3x - y = 3 \); \( 9x - 3y = 9 \)
(iv) \( 0.2x + 0.3y = 1.3 \); \( 0.4x + 0.5y = 2.3 \)
(v) \( \sqrt{2}x + \sqrt{3}y = 0 \); \( \sqrt{3}x - \sqrt{8}y = 0 \)
(vi) \( \frac{3x}{2} - \frac{5y}{3} = -2 \); \( \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \)
(i) \( x + y = 14 \); \( x - y = 4 \)
(ii) \( s - t = 3 \); \( \frac{s}{3} + \frac{t}{2} = 6 \)
(iii) \( 3x - y = 3 \); \( 9x - 3y = 9 \)
(iv) \( 0.2x + 0.3y = 1.3 \); \( 0.4x + 0.5y = 2.3 \)
(v) \( \sqrt{2}x + \sqrt{3}y = 0 \); \( \sqrt{3}x - \sqrt{8}y = 0 \)
(vi) \( \frac{3x}{2} - \frac{5y}{3} = -2 \); \( \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \)
Solution:
(i) From \( x - y = 4 \), we get \( x = y + 4 \).
Substitute in \( x + y = 14 \):
\( (y + 4) + y = 14 \Rightarrow 2y = 10 \Rightarrow y = 5 \).
\( x = 5 + 4 = 9 \). Answer: \( x = 9, y = 5 \).
(ii) From \( s - t = 3 \), \( s = t + 3 \).
Substitute in \( \frac{s}{3} + \frac{t}{2} = 6 \) (Multiply by 6: \( 2s + 3t = 36 \)).
\( 2(t + 3) + 3t = 36 \Rightarrow 2t + 6 + 3t = 36 \Rightarrow 5t = 30 \Rightarrow t = 6 \).
\( s = 6 + 3 = 9 \). Answer: \( s = 9, t = 6 \).
(iii) From \( 3x - y = 3 \), \( y = 3x - 3 \).
Substitute in \( 9x - 3y = 9 \):
\( 9x - 3(3x - 3) = 9 \Rightarrow 9x - 9x + 9 = 9 \Rightarrow 9 = 9 \).
This statement is always true. The equations are coincident. Answer: Infinitely many solutions.
(iv) Multiply by 10: \( 2x + 3y = 13 \) and \( 4x + 5y = 23 \).
From eq 1: \( 2x = 13 - 3y \).
Substitute in eq 2: \( 2(13 - 3y) + 5y = 23 \Rightarrow 26 - 6y + 5y = 23 \Rightarrow -y = -3 \Rightarrow y = 3 \).
\( 2x = 13 - 3(3) = 4 \Rightarrow x = 2 \). Answer: \( x = 2, y = 3 \).
(v) From \( \sqrt{2}x + \sqrt{3}y = 0 \), \( x = -\frac{\sqrt{3}}{\sqrt{2}}y \).
Substitute in \( \sqrt{3}x - \sqrt{8}y = 0 \):
\( \sqrt{3}(-\frac{\sqrt{3}}{\sqrt{2}}y) - 2\sqrt{2}y = 0 \Rightarrow y(-\frac{3}{\sqrt{2}} - 2\sqrt{2}) = 0 \Rightarrow y = 0 \).
\( x = 0 \). Answer: \( x = 0, y = 0 \).
(vi) Simplify equations:
1. \( \frac{3x}{2} - \frac{5y}{3} = -2 \Rightarrow 9x - 10y = -12 \).
2. \( \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \Rightarrow 2x + 3y = 13 \).
From eq 2: \( 2x = 13 - 3y \Rightarrow x = \frac{13-3y}{2} \).
Substitute in eq 1: \( 9(\frac{13-3y}{2}) - 10y = -12 \).
\( 117 - 27y - 20y = -24 \Rightarrow -47y = -141 \Rightarrow y = 3 \).
\( x = \frac{13 - 9}{2} = 2 \). Answer: \( x = 2, y = 3 \).
Q2
Solve \( 2x + 3y = 11 \) and \( 2x - 4y = -24 \) and hence find the value of 'm' for which \( y = mx + 3 \).
Solution:
Equation 1: \( 2x + 3y = 11 \)
Equation 2: \( 2x - 4y = -24 \)
Subtract eq 2 from eq 1: \( (2x - 2x) + (3y - (-4y)) = 11 - (-24) \).
\( 7y = 35 \Rightarrow y = 5 \).
Substitute \( y = 5 \) in eq 1: \( 2x + 15 = 11 \Rightarrow 2x = -4 \Rightarrow x = -2 \).
Solution: \( x = -2, y = 5 \).
Find m: \( y = mx + 3 \Rightarrow 5 = m(-2) + 3 \).
\( 2 = -2m \Rightarrow m = -1 \).
Answer: \( m = -1 \).
Q3
Form the pair of linear equations for the following problems and find their solution by substitution method.
(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 ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 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 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 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 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?
(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 ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 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 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 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 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?
Solution:
(i) Let numbers be \(x\) and \(y\) (\(x > y\)).
\(x - y = 26\) and \(x = 3y\).
Substitute \(x = 3y\) in first eq: \(3y - y = 26 \Rightarrow 2y = 26 \Rightarrow y = 13\).
\(x = 3(13) = 39\). Numbers are 39 and 13.
(ii) Let angles be \(x\) and \(y\) (\(x > y\)).
\(x + y = 180\) (Supplementary) and \(x = y + 18\).
\((y + 18) + y = 180 \Rightarrow 2y = 162 \Rightarrow y = 81^\circ\).
\(x = 81 + 18 = 99^\circ\).
(iii) Let bat cost \(x\) and ball cost \(y\).
\(7x + 6y = 3800\) ... (1)
\(3x + 5y = 1750\) ... (2)
From (2): \(3x = 1750 - 5y \Rightarrow x = \frac{1750 - 5y}{3}\).
Substitute in (1): \(7(\frac{1750 - 5y}{3}) + 6y = 3800\).
\(12250 - 35y + 18y = 11400 \Rightarrow -17y = -850 \Rightarrow y = 50\).
\(x = \frac{1750 - 250}{3} = 500\). Bat: ₹ 500, Ball: ₹ 50.
(iv) Let fixed charge \(x\) and per km charge \(y\).
\(x + 10y = 105\) ... (1)
\(x + 15y = 155\) ... (2)
From (1), \(x = 105 - 10y\). Substitute in (2):
\((105 - 10y) + 15y = 155 \Rightarrow 5y = 50 \Rightarrow y = 10\).
\(x = 105 - 100 = 5\).
Fixed: ₹ 5, Per km: ₹ 10.
Cost for 25 km: \(x + 25y = 5 + 250 = \) ₹ 255.
(v) Fraction \(\frac{x}{y}\).
\(\frac{x+2}{y+2} = \frac{9}{11} \Rightarrow 11x - 9y = -4\).
\(\frac{x+3}{y+3} = \frac{5}{6} \Rightarrow 6x - 5y = -3\).
Solve: From eq 2, \(x = \frac{5y - 3}{6}\). Substitute into eq 1.
\(11(\frac{5y-3}{6}) - 9y = -4 \Rightarrow 55y - 33 - 54y = -24 \Rightarrow y = 9\).
\(x = \frac{45 - 3}{6} = 7\). Fraction
Exercise 3.3
Q1
Solve the following pair of linear equations by the elimination method and the substitution method:
(i) \( x + y = 5 \) and \( 2x - 3y = 4 \)
(ii) \( 3x + 4y = 10 \) and \( 2x - 2y = 2 \)
(iii) \( 3x - 5y - 4 = 0 \) and \( 9x = 2y + 7 \)
(iv) \( \frac{x}{2} + \frac{2y}{3} = -1 \) and \( x - \frac{y}{3} = 3 \)
(i) \( x + y = 5 \) and \( 2x - 3y = 4 \)
(ii) \( 3x + 4y = 10 \) and \( 2x - 2y = 2 \)
(iii) \( 3x - 5y - 4 = 0 \) and \( 9x = 2y + 7 \)
(iv) \( \frac{x}{2} + \frac{2y}{3} = -1 \) and \( x - \frac{y}{3} = 3 \)
Solution:
(i) \( x + y = 5 \) ... (1) and \( 2x - 3y = 4 \) ... (2)
Multiply eq (1) by 3: \( 3x + 3y = 15 \) ... (3)
Add eq (2) and (3): \( (2x - 3y) + (3x + 3y) = 4 + 15 \Rightarrow 5x = 19 \Rightarrow x = \frac{19}{5} \).
Substitute \( x \) in eq (1): \( \frac{19}{5} + y = 5 \Rightarrow y = 5 - \frac{19}{5} = \frac{6}{5} \).
Answer: \( x = \frac{19}{5}, y = \frac{6}{5} \).
(ii) \( 3x + 4y = 10 \) ... (1) and \( 2x - 2y = 2 \) ... (2)
From eq (2), divide by 2: \( x - y = 1 \Rightarrow x = y + 1 \).
Substitute in eq (1): \( 3(y + 1) + 4y = 10 \Rightarrow 3y + 3 + 4y = 10 \Rightarrow 7y = 7 \Rightarrow y = 1 \).
\( x = 1 + 1 = 2 \).
Answer: \( x = 2, y = 1 \).
(iii) \( 3x - 5y = 4 \) ... (1) and \( 9x - 2y = 7 \) ... (2)
Multiply eq (1) by 3: \( 9x - 15y = 12 \) ... (3)
Subtract eq (3) from eq (2): \( (9x - 2y) - (9x - 15y) = 7 - 12 \Rightarrow 13y = -5 \Rightarrow y = -\frac{5}{13} \).
Substitute \( y \) in eq (1): \( 3x - 5(-\frac{5}{13}) = 4 \Rightarrow 3x + \frac{25}{13} = 4 \Rightarrow 3x = \frac{52 - 25}{13} = \frac{27}{13} \Rightarrow x = \frac{9}{13} \).
Answer: \( x = \frac{9}{13}, y = -\frac{5}{13} \).
(iv) \( \frac{x}{2} + \frac{2y}{3} = -1 \Rightarrow 3x + 4y = -6 \) ... (1)
\( x - \frac{y}{3} = 3 \Rightarrow 3x - y = 9 \) ... (2)
Subtract eq (2) from eq (1): \( (3x + 4y) - (3x - y) = -6 - 9 \Rightarrow 5y = -15 \Rightarrow y = -3 \).
Substitute \( y \) in eq (2): \( 3x - (-3) = 9 \Rightarrow 3x = 6 \Rightarrow x = 2 \).
Answer: \( x = 2, y = -3 \).
Q2
Form the pair of linear equations in the following problems, and find their solutions (if they exist) by 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?
(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?
(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.
(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.
(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.
(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?
(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?
(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.
(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.
(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.
Solution:
(i) Let the fraction be \( \frac{x}{y} \).
Case 1: \( \frac{x+1}{y-1} = 1 \Rightarrow x + 1 = y - 1 \Rightarrow x - y = -2 \) ... (1)
Case 2: \( \frac{x}{y+1} = \frac{1}{2} \Rightarrow 2x = y + 1 \Rightarrow 2x - y = 1 \) ... (2)
Subtract eq (1) from eq (2): \( (2x - y) - (x - y) = 1 - (-2) \Rightarrow x = 3 \).
Substitute \( x \) in eq (1): \( 3 - y = -2 \Rightarrow y = 5 \).
Answer: The fraction is \( \frac{3}{5} \).
(ii) Let Nuri's age be \( x \) and Sonu's age be \( y \).
5 years ago: \( (x - 5) = 3(y - 5) \Rightarrow x - 3y = -10 \) ... (1)
10 years later: \( (x + 10) = 2(y + 10) \Rightarrow x - 2y = 10 \) ... (2)
Subtract eq (1) from eq (2): \( (x - 2y) - (x - 3y) = 10 - (-10) \Rightarrow y = 20 \).
Substitute \( y \) in eq (2): \( x - 40 = 10 \Rightarrow x = 50 \).
Answer: Nuri is 50 years old, Sonu is 20 years old.
(iii) Let the number be \( 10x + y \) (digits \( x, y \)).
Sum of digits: \( x + y = 9 \) ... (1)
Reversed number: \( 10y + x \).
Condition: \( 9(10x + y) = 2(10y + x) \Rightarrow 90x + 9y = 20y + 2x \Rightarrow 88x - 11y = 0 \Rightarrow 8x - y = 0 \) ... (2)
Add eq (1) and (2): \( (x + y) + (8x - y) = 9 + 0 \Rightarrow 9x = 9 \Rightarrow x = 1 \).
Substitute \( x \) in eq (1): \( 1 + y = 9 \Rightarrow y = 8 \).
Answer: The number is 18.
(iv) Let notes of ₹ 50 be \( x \) and ₹ 100 be \( y \).
Total notes: \( x + y = 25 \) ... (1)
Total value: \( 50x + 100y = 2000 \Rightarrow x + 2y = 40 \) ... (2)
Subtract eq (1) from eq (2): \( (x + 2y) - (x + y) = 40 - 25 \Rightarrow y = 15 \).
Substitute \( y \) in eq (1): \( x + 15 = 25 \Rightarrow x = 10 \).
Answer: 10 notes of ₹ 50 and 15 notes of ₹ 100.
(v) Let fixed charge be \( x \) and charge per day be \( y \).
Saritha (7 days = 3 fixed + 4 extra): \( x + 4y = 27 \) ... (1)
Susy (5 days = 3 fixed + 2 extra): \( x + 2y = 21 \) ... (2)
Subtract eq (2) from eq (1): \( 2y = 6 \Rightarrow y = 3 \).
Substitute \( y \) in eq (2): \( x + 6 = 21 \Rightarrow x = 15 \).
Answer: Fixed charge = ₹ 15, Charge per extra day = ₹ 3.