CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Worksheet Set 10

Question. The pair of equations \( x = 2 \) and \( y = -3 \) has
(a) no solution
(b) one solution
(c) infinitely many solutions
(d) two solutions
Answer: (b) one solution
Explanation: Here, a unique solution of each variable of a pair of linear equations is given, therefore, it has one solution to a system of linear equations.

Question. Solution of \( \frac{a^2}{x} - \frac{b^2}{y} = 0 \) and \( \frac{a^2b}{x} + \frac{b^2a}{y} = a + b \) where \( x, y \neq 0 \) is
(a) \( x = -a^2 \) and \( y = -b^2 \)
(b) \( x = a^2 \) and \( y = -b^2 \)
(c) \( x = a^2 \) and \( y = b^2 \)
(d) \( x = -a^2 \) and \( y = b^2 \)
Answer: (c) \( x = a^2 \) and \( y = b^2 \)
Explanation: First equation: \( \frac{a^2}{x} - \frac{b^2}{y} = 0 \) or \( \frac{a^2}{x} = \frac{b^2}{y} \)
Second Equation: \( \frac{a^2b}{x} + \frac{b^2a}{y} = a + b \)
\( \Rightarrow (\frac{b^2}{y}) \times b + \frac{b^2a}{y} = a + b \)
\( \Rightarrow (\frac{b^2}{y}) \times (b + a) = a + b \)
\( \Rightarrow \frac{b^2}{y} = \frac{a+b}{a+b} = 1 \)
\( \Rightarrow y = b^2 \)
\( \frac{a^2}{x} = \frac{b^2}{y} \)
\( \Rightarrow \frac{a^2}{x} = \frac{b^2}{b^2} = 1 \)
\( \Rightarrow x = a^2 \)
Hence \( x=a^2 \) and \( y=b^2 \)

Question. A system of two linear equations in two variables has a unique solution, if their graphs
(a) do not intersect at any point
(b) coincide
(c) cut the x – axis
(d) intersect only at a point
Answer: (d) intersect only at a point
Explanation: Number of solutions of a system of two linear equations in two variables are equal to number of common points between the graphs of given linear equations. If a system has unique solution then their graphs must intersect in only one point.

Question. A system of two linear equations in two variables is inconsistent, if their graphs
(a) intersect only at a point
(b) coincide
(c) do not intersect at any point
(d) cut the x – axis
Answer: (c) do not intersect at any point
Explanation: A system of two linear equations in two variables is inconsistent, if their graphs do not intersect at any point. In this case, a pair of lines represented by the system are parallel to each other. so they do not intersect each other at any point. the system is an inconsistent system of linear equations and the equations are independent.

Question. Every linear equation in two variables has
(a) two solutions
(b) no solution
(c) an infinite number of solutions
(d) one solution
Answer: (c) an infinite number of solutions
Explanation: A linear equation in two variables is of the form, \( ax + by + c = 0 \), where geometrically it does represent a straight line and every point on this graph is a solution for a given linear equation. As a line consists of an infinite number of points, A linear equation has an infinite number of solutions.

Question. Find whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident: \( 2x - 3y + 6 = 0 \); \( 4x - 5y + 2 = 0 \)
Answer: If two lines are parallel then \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) and if intersect then \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Now given equations are \( 2x - 3y + 6 = 0 \)...(i)
\( \Rightarrow a_1 = 2, b_1 = -3, c_1 = 6 \)
And, \( 4x - 5y + 2 = 0 \)...(ii)
\( \Rightarrow a_2 = 4, b_2 = -5, c_2 = 2 \)
Now, \( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \), \( \frac{b_1}{b_2} = \frac{-3}{-5} = \frac{3}{5} \)
\( \frac{c_1}{c_2} = \frac{6}{2} = 3 \)
\( \Rightarrow \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Therefore lines representing the given pair of linear equations intersect each other at a point.

Question. The larger of two supplementary angles exceeds the smaller by 20 degrees. Find the angles.
Answer: Suppose larger angle = \( x^\circ \) and smaller angle = \( y^\circ \)
Now \( x - y = 20 \)...(i)
\( x + y = 180 \)...(ii)
Adding equation (i) and (ii)
\( \Rightarrow 2x = 200 \)
\( \Rightarrow x = 100 \)
\( \Rightarrow y = 80 \)

Question. Find the value of k for which the pair of linear equations \( 4x + 6y - 1 = 0 \) and \( 2x + ky - 7 = 0 \) represents parallel lines.
Answer: Given pair of linear equations are \( 2x - ky + 3 = 0 \); \( 4x + 6y - 5 = 0 \). Since the given equations is consistent, we have \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Here \( a_1 = 2, a_2 = 4, b_1 = -k, b_2 = 6 \)
\( \Rightarrow \frac{2}{4} \neq \frac{-k}{6} \)
\( \Rightarrow -k \neq \frac{2}{4} \times 6 = 3 \)
Hence k is a real number except 3.

Question. Which type of solution will equations \( x + 2y = 4 \) and \( 2x + y = 5 \) have?
Answer: Since, \( a_1 = 1, b_1 = 2, c_1 = -4 \)
\( a_2 = 2, b_2 = 1, c_2 = -5 \)
\( \therefore \frac{a_1}{a_2} = \frac{1}{2} \) and \( \frac{b_1}{b_2} = \frac{2}{1} \)
\( \Rightarrow \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
\( \therefore \) System of equations has unique solution.

Question. Without drawing the graph, find out whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident: \( 18x - 7y = 24 \); \( \frac{9}{5}x - \frac{7}{10}y = \frac{9}{10} \)
Answer: The condition for two lines to be parallel is that \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Given lines are \( 18x - 7y = 24 \). Here \( a_1=18, b_1=-7, c_1=-24 \) and
\( \frac{9}{5}x - \frac{7}{10}y = \frac{9}{10} \) or \( \frac{18x - 7y}{10} = \frac{9}{10} \)
\( \Rightarrow 18x - 7y = 9 \). Here, \( a_2=18, b_2=-7, c_2=-9 \)
Therefore, \( \frac{a_1}{a_2} = \frac{18}{18} = 1 \) and \( \frac{b_1}{b_2} = \frac{-7}{-7} = 1 \)
and \( \frac{c_1}{c_2} = \frac{-24}{-9} = \frac{8}{3} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Hence the two lines are parallel.

Question. Two lines are given to be parallel. The equation of one of the lines is \( 4x + 3y = 14 \), then find the equation of the second line.
Answer: Equation of a line \( ax + by + c = 0 \) parallel to given line is \( ax + by + k = 0 \), here k is any real number. The equation of one line is \( 4x + 3y = 14 \).
\( a_1 = 4, b_1 = 3 \) and \( c_1 = -14 \)
Two lines are given to be parallel. So, No solutions and the pair of linear equations is inconsistent.
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
or \( \frac{4}{a_2} = \frac{3}{b_2} \neq \frac{14}{c_2} \)
\( \Rightarrow \frac{a_2}{b_2} = \frac{4}{3} = \frac{12}{9} \)
Hence, one of the possible, second parallel line is \( 12x + 9y = 5 \).

Question. The difference between two numbers is 26 and one number is three times the other. Find the numbers.
Answer: Let the two numbers be \( x \) and \( y \) (\( x > y \))
We are given that, The difference between two numbers is 26.
\( x - y = 26 \)...(i)
And one number is three times the other.
\( x = 3y \)...(ii)
On substituting the value x from eqn. (ii) in eqn. (i), we get
3y - y = 26
2y = 26
\( \therefore y = 13 \)
Put y = 13 in (ii) we get
x = 3(13) = 39
Hence, the two numbers are 39 and 13.

Question. Draw the graph of the equation \( 3x + 2y = 12 \). Also, find the co-ordinates of the points where the line meets the x-axis and the y-axis.
Answer: \( 3x + 2y = 12 \Rightarrow y = \frac{12-3x}{2} \)
Table:
x: 0, 2, 4
y: 6, 3, 0
Steps:
i. Given equation.
ii. Write y in terms of x.
iii. Complete the table.
iv. Plot the points A(0, 6), B(2, 3) and C(4, 0) on the graph paper.
v. Join the points.
The Line meets the x-axis at (4, 0) and the y-axis at (0, 6).

Question. A man invested an amount at 12% per annum simple interest and another amount at 10% per annum simple interest. He received an annual interest of Rs 2600. But, if he had interchanged the amounts invested, he would have received Rs 140 less. What amounts did he invest at the different rates?
Answer: Let the amount invested at 12% be Rs x and that invested at 10% be Rs y.
Then, total annual interest = \( (\frac{x \times 12 \times 1}{100} + \frac{y \times 10 \times 1}{100}) = (\frac{6x+5y}{50}) \)
\( \therefore \frac{6x+5y}{50} = 2600 \Rightarrow 6x + 5y = 130000 \).......(i)
Again, the amount invested at 12% is Rs y and that invested at 10% is Rs x.
Total annual interest at the new rates = \( (\frac{y \times 12 \times 1}{100} + \frac{x \times 10 \times 1}{100}) = (\frac{6y+5x}{50}) \)
But, interest received at the new rates = Rs. (2600 – 140) = Rs. 2460.
\( \therefore \frac{6y+5x}{50} = 2460 \Rightarrow 5x + 6y = 123000 \)........(ii)
Adding (i) and (ii), we get
11x + 11y = 253000
\( \Rightarrow 11(x + y) = 253000 \Rightarrow x + y = 23000 \). ... (iii)
Subtracting (ii) from (i), we get
x - y = 7000. ...(iv)
Adding (iii) and (iv), we get 2x = 30000 \( \Rightarrow \) x = 15000.
Putting x = 15000 in (i), we get
15000 + y = 23000 \( \Rightarrow \) y = 23000 - 15000 = 8000
\( \therefore x = 15000 \) and \( y = 8000 \).
Hence, the amount at 12% is Rs 15000 and that at 10% is Rs 8000.

Question. Find the values of a and b for which \( 2x + 3y = 7, 2ax + (a + b)y = 28 \) has an infinite number of solutions.
Answer: We know that, if a system of linear equations \( a_1x + b_1y + c_1 = 0 \), \( a_2x + b_2y + c_2 = 0 \) has infinite number of solutions, then \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Given that, \( 2x + 3y = 7 \), \( 2ax + (a + b)y = 28 \) have an infinite number of solutions.
\( \Rightarrow 2x + 3y - 7 = 0 \), \( 2ax + (a + b)y - 28 = 0 \)
Since, the pair of lines have an infinite number of solutions,
so, \( \frac{2}{2a} = \frac{3}{a+b} = \frac{-7}{-28} \)
\( \Rightarrow \frac{1}{a} = \frac{3}{a+b} = \frac{7}{28} \)
a = 4
and a + b = 3a \( \Rightarrow \) 4 + b = 12 \( \Rightarrow \) b = 8
Hence, a = 4 and b = 8.

Question. Solve: \( \frac{3}{x+y} + \frac{2}{x-y} = 2 \) and \( \frac{9}{x+y} - \frac{4}{x-y} = 1 \).
Answer: According to question the given system of equations are as
\( \frac{3}{x+y} + \frac{2}{x-y} = 2 \)
and \( \frac{9}{x+y} - \frac{4}{x-y} = 1 \)
Put \( \frac{1}{x+y} = u \) and \( \frac{1}{x-y} = v \)
So, given system of equations become
3u + 2v = 2................(i)
and 9u - 4v = 1..........(ii)
Multiply equation (i) by 2
\( \Rightarrow 6u + 4v = 4 \)........(iii)
Adding equation (ii) and equation (iii) we get
9u - 4v + 6u + 4v = 1 + 4
\( \Rightarrow 15u = 5 \)
\( \Rightarrow u = \frac{1}{3} \)
Substitute the value of \( u = \frac{1}{3} \) in equation (i), we get \( v = \frac{1}{2} \)
\( \Rightarrow \frac{1}{x+y} = \frac{1}{3} \) and \( \frac{1}{x-y} = \frac{1}{2} \)
\( \Rightarrow x + y = 3 \)...........(iv)
and x - y = 2........(v)
Adding (iv) and (v), we get
2x = 5 \( \Rightarrow x = \frac{5}{2} \)
Substituting \( x = \frac{5}{2} \) in (iv), we get \( y = \frac{1}{2} \)
Hence, \( x = \frac{5}{2} \) and \( y = \frac{1}{2} \)

Question. Solve the following system of equation by elimination method: \( \frac{x}{2} - \frac{y}{5} = 4 \) and \( \frac{x}{7} + \frac{y}{15} = 3 \)
Answer: The given system of equation is
\( \frac{x}{2} - \frac{y}{5} = 4 \)...(1)
\( \frac{x}{7} + \frac{y}{15} = 3 \)...(2)
Multiplying equation (2) by 3, we get
\( \frac{3x}{7} + \frac{y}{5} = 9 \)...(3)
Adding equation (1) and equation (3), we get
\( \frac{x}{2} + \frac{3x}{7} = 13 \)
\( \Rightarrow \frac{13}{14}x = 13 \Rightarrow x = \frac{13 \times 14}{13} = 14 \)
Substituting this value of x in equation (2), we get
\( \frac{14}{7} + \frac{y}{15} = 3 \)
\( \Rightarrow 2 + \frac{y}{15} = 3 \Rightarrow \frac{y}{15} = 3 - 2 \)
\( \Rightarrow \frac{y}{15} = 1 \Rightarrow y = 15 \)
So, the solution of the given system of equations is x = 14, y = 15
Verification; Substituting x = 14, y = 15.
We find that both the equations (1) and (2) are satisfied as shown below;
\( \frac{x}{2} - \frac{y}{5} = \frac{14}{2} - \frac{15}{5} = 7 - 3 = 4 \)
\( \frac{x}{7} + \frac{y}{15} = \frac{14}{7} + \frac{15}{15} = 2 + 1 = 3 \)
Hence, the solution is correct.

Question. A boat goes 12 km upstream and 40 km downstream in 8 hours. It can go 16 km upstream and 32 km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.
Answer: Let the speed of the boat in still water be x km/hr and speed of the stream be y km/hr.
Then, Speed of the boat while going upstream = (x - y) km/hr
Speed of the boat while going downstream = (x + y) km/hr
Also we know that, time taken to cover 'd' Km with speed 's' Km/hr is \( \frac{d}{s} \)
Hence, Time taken by the boat to cover 12 km upstream = \( \frac{12}{x-y} \) hrs
And, Time taken by the boat to cover 40 km downstream = \( \frac{40}{x+y} \) hrs
According to the question, Total time taken = 8 hrs
\( \therefore \frac{12}{x-y} + \frac{40}{x+y} = 8 \).........(1)
Again, time taken by the boat to cover 16 km upstream = \( \frac{16}{x-y} \)
And, Time taken by the boat to cover 32 km downstream = \( \frac{32}{x+y} \)
According to the question, Total time taken = 8 hrs
\( \therefore \frac{16}{(x-y)} + \frac{32}{(x+y)} = 8 \).........(2)
Putting \( \frac{1}{(x-y)} = u \) and \( \frac{1}{(x+y)} = v \) in equation (1) & equation (2), so that we may get linear equations in the variables u & v as following :-
12u + 40v = 8
\( \Rightarrow \) 3u + 10v = 2........(3)
and 16u + 32v = 8
\( \Rightarrow \) 2u + 4v = 1.........(4)
Multiplying equation (3) by 4 and equation (4) by 10, we get ;
12u + 40v = 8..........(5)
20u + 40v = 10........(6)
Subtracting equation (5) from equation (6), we get
8u = 2 \( \Rightarrow u = \frac{1}{4} \)
Putting \( u = \frac{1}{4} \) in equation (3), we get
\( 3 \times \frac{1}{4} + 10v = 2 \Rightarrow 10v = \frac{5}{4} \Rightarrow v = \frac{1}{8} \)
\( u = \frac{1}{4} \Rightarrow \frac{1}{x-y} = \frac{1}{4} \Rightarrow x - y = 4 \).....(7)
\( v = \frac{1}{8} \Rightarrow \frac{1}{x+y} = \frac{1}{8} \Rightarrow x + y = 8 \)......(8)
On adding (7) and (8), we get
2x = 12 \( \Rightarrow \) x = 6
Putting x = 6 in (8), we get
6 + y = 8 \( \Rightarrow \) y = 8 - 6 = 2
\( \therefore x = 6, y = 2 \)
Hence, the speed of the boat in still water = 6 km/hr and speed of the stream = 2 km/hr

Question. A man travels 370 km, partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.
Answer: Let the speed of the train be x km/hr and that of the car be y km/hr.
Case I Distance covered by train = 250 km.
Distance covered by car = (370 - 250) km = 120 km.
Time taken to cover 250 km by train = \( \frac{250}{x} \) hours
Time taken to cover 120 km by car = \( \frac{120}{y} \) hours
Total time taken = 4 hours
\( \therefore \frac{250}{x} + \frac{120}{y} = 4 \Rightarrow \frac{125}{x} + \frac{60}{y} = 2 \)......(i)
Case II Distance covered by train = 130 km.
Distance covered by car = (370 - 130) km = 240 km.
Time taken to cover 130 km by train = \( \frac{130}{x} \) hours
Time taken to cover 240 km by car = \( \frac{240}{y} \) hours
Total time taken = \( 4 \frac{18}{60} \) hours = \( 4 \frac{3}{10} \) hours = \( \frac{43}{10} \) hours
\( \therefore \frac{130}{x} + \frac{240}{y} = \frac{43}{10} \Rightarrow \frac{1300}{x} + \frac{2400}{y} = 43 \)......(ii)
Putting \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \), equations (i) and (ii) become
125u + 60v = 2 ... (iii) and 1300u + 2400v = 43 ... (iv)
On multiplying (iii) by 40 and subtracting (iv) from the result, we get
5000u - 1300u = 80 - 43 \( \Rightarrow \) 3700u = 37
\( \Rightarrow u = \frac{37}{3700} = \frac{1}{100} \Rightarrow \frac{1}{x} = \frac{1}{100} \Rightarrow x = 100 \)
Putting \( u = \frac{1}{100} \) in (iv), we get
\( (1300 \times \frac{1}{100}) + 2400v = 43 \Rightarrow 2400v = 43 - 13 = 30 \)
\( \Rightarrow v = \frac{30}{2400} = \frac{1}{80} \Rightarrow \frac{1}{y} = \frac{1}{80} \Rightarrow y = 80 \)
\( \therefore x = 100 \) and y = 80.
Hence, the speed of the train is 100 km/hr and that of the car is 80 km/hr

Question. Solve for x and y : \( \frac{8}{2x-3y} + \frac{21}{2x+3y} = 11 \); \( \frac{5}{2x-3y} + \frac{7}{2x+3y} = 6 \), (\( 2x - 3y \neq 0, 2x + 3y \neq 0 \))
Answer: Given equations are
\( \frac{8}{2x-3y} + \frac{21}{2x+3y} = 11 \)...(i)
\( \frac{5}{2x-3y} + \frac{7}{2x+3y} = 6 \)....(ii)
Putting \( \frac{1}{2x-3y} = A \) and \( \frac{1}{2x+3y} = B \) in equation (i) & (ii) so that we may get the pair of linear equations in variables A & B as following :-
8A + 21B = 11 ...(iii). and 5A + 7B = 6...(iv)
Multiplying eq. (iv) by 3 & then subtracting eq. (iii) from it, we get ;
15A + 21B = 18
8A + 21B = 11
----------------
7A = 7
\( \Rightarrow A = 1 \)
Substituting A = 1 in eq. (iii) ,
8 × 1 + 21B = 11
\( \Rightarrow \) 21B = 3
\( \Rightarrow B = \frac{1}{7} \)
Since, A = 1
\( \Rightarrow \frac{1}{2x-3y} = 1 \)
\( \Rightarrow 2x - 3y = 1 \)...(vi)
Where \( B = \frac{1}{7} \)
\( \Rightarrow \frac{1}{2x+3y} = \frac{1}{7} \)
\( \Rightarrow 2x + 3y = 7 \)...(vii)
Adding (vi) and (vii), we get
2x – 3y = 1
2x + 3y = 7
----------------
4x = 8
\( \Rightarrow x = 2 \)
Substituting x = 2 in eq.(vi),
2 × 2 - 3y = 1
\( \Rightarrow -3y = -3 \Rightarrow y = 1 \)
\( \therefore x = 2, y = 1 \).

Question. The area of the triangle formed by \( y = x \), \( x = 6 \) and \( y = 0 \) is 
(a) 18 sq. units
(b) 72 sq. units
(c) 36 sq. units
(d) 9 sq. units
Answer: (a) 18 sq. units

Question. The system of linear equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) has a unique solution if 
(a) \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
(b) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
(c) None of the options
(d) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Answer: (a) \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)

Question. The difference between two numbers is 26 and one number is three times the other. The numbers are 
(a) 39 and 13
(b) 30 and 10
(c) 36 and 12
(d) 36 and 10
Answer: (a) 39 and 13

Question. If \( 2x - 3y = 11 \) and \( (a + b)x - (a + b - 3)y = 4a + b \) has infinite number of solutions, then 
(a) \( a = -9 \) and \( b = 3 \)
(b) \( a = -9 \) and \( b = -3 \)
(c) \( a = 9 \) and \( b = 3 \)
(d) \( a = 9 \) and \( b = -3 \)
Answer: (a) \( a = -9 \) and \( b = 3 \)

Question. A system of two linear equations in two variables is dependent consistent, if their graphs 
(a) do not intersect at any point
(b) cut the x – axis
(c) intersect only at a point
(d) coincide
Answer: (d) coincide

Question. Find the value of k so that the following system of equations has no solution: \( 3x - y - 5 = 0 \), \( 6x - 2y + k = 0 \)  
Answer: Given system of equations is \( 3x - y - 5 = 0 \) and \( 6x - 2y + k = 0 \)
Here \( a_1 = 3, b_1 = -1, c_1 = -5 \) and \( a_2 = 6, b_2 = -2, c_2 = k \)
For no solution \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \implies \frac{3}{6} = \frac{-1}{-2} \neq \frac{-5}{k} \)
\( \implies \frac{1}{2} \neq \frac{-5}{k} \)
\( \implies k \neq -10 \)

Question. For what value of a the following pair of linear equation has infinitely many solution? (1) \( ax - 3y = 1 \), \( -12x + ay = 2 \)
Answer: For infinitely many solutions, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \implies \frac{a}{-12} = \frac{-3}{a} = \frac{1}{2} \)
\( \implies \frac{a}{-12} = \frac{-3}{a} \) and \( \frac{-3}{a} = \frac{1}{2} \)
\( \implies a^2 = 36 \) and \( a = -6 \)
\( \implies a = \pm 6 \) and \( a = -6 \)
For \( a = -6 \), pair of given linear equations has infinitely many solutions.

Question. For what value of k, the following pair of linear equations has infinitely many solutions? (1) \( 10x + 5y - (k - 5) = 0 \), \( 20x + 10y - k = 0 \)
Answer: The system of linear equations has infinite number of solutions if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \frac{a_1}{a_2} = \frac{10}{20} = \frac{1}{2} \)
\( \frac{b_1}{b_2} = \frac{5}{10} = \frac{1}{2} \)
\( \frac{c_1}{c_2} = \frac{-(k-5)}{-k} = \frac{k-5}{k} \)
The system has infinite number of solutions,
\( \implies \frac{1}{2} = \frac{1}{2} = \frac{k-5}{k} \)
\( \implies k = 2k - 10 \)
\( \implies k = 10 \)

Question. The equation \( ax^n + by^n + c = 0 \) represents a straight line if 
Answer: n = 1

Question. Find whether the pair of linear equations \( y = 0 \) and \( y = -5 \) has no solution, unique solution or infinitely many solutions 
Answer: Since, given variable y has different values so, The pair of equations \( y = 0 \) and \( y = -5 \) has no solution. Both the lines \( y = 0 \) and \( y = -5 \) are parallel to x-axis. Hence they do not have solution.

Question. A and B each have a certain number of mangoes. A says to B, "if you give 30 of your mangoes, I will have twice as many as left with you." B replies, "if you give me 10, I will have thrice as many as left with you." How many mangoes does each have? 
Answer: Assume A has x mangoes and B has y mangoes.
As per given condition if B gives 30 then A will have twice as many as left by B.
\( (x + 30) = 2(y - 30) \)
\( \implies x + 30 = 2y - 60 \)
\( \implies x = 2y - 90....(1) \)
And if A gives 10 mangoes to B, then B will have thrice as many as left by A.
\( 3(x - 10) = (y + 10) \)
\( \implies 3x = y + 40 ....(2) \)
\( \implies 3(2y - 90) = y + 40 \) (using x from Eq. 1)
\( \implies 6y - 270 = y + 40 \)
\( \implies 6y - y = 310 \)
\( \implies y = 62 \)
Put \( y = 62 \) in (1)
So, \( x = 124 - 90 = 34 \)
Therefore A has 34 mangoes and B has 62 mangoes.

Question. Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received.   
Answer: Let the number of Rs. 50 notes and Rs. 100 notes be x and y respectively.
According to given condition,
Meena got 25 notes in all.
\( \implies x + y = 25 ...........(i) \)
and Meena withdraw Rs. 2000.
\( \implies 50x + 100y = 2000 ............(ii) \)
Multiplying equation (i) by 50, we obtain:
\( 50x + 50y = 1250 ............. (iii) \)
Subtracting equation (iii) from equation (ii), we obtain:
\( (50x + 100y) - (50x + 50y) = 2000 - 1250 \)
\( 50x + 100y - 50x - 50y = 750 \)
\( 50y = 750 \)
\( y = 15 \)
Substituting the value of y in equation (i), we obtain:
\( x = 10 \)
Hence, Meena received 10 notes of Rs. 50 and 15 notes of Rs. 100.

Question. Solve the system of equations: \( x - 2y = 0, 3x + 4y = 20 \). 
Answer: The given equations are
\( x - 2y = 0 ...........(i) \)
\( 3x + 4y = 20..........(ii) \)
Multiply (i) by 2, we get
\( 2x - 4y = 0 ......... (iii) \)
Add (ii) and (iii), we get
\( 5x = 20 \)
\( \implies x = 4 \)
Put \( x = 4 \) in (i), we get
\( 4 - 2y = 0 \)
\( \implies 2y = 4 \)
\( \implies y = 2 \)
So, \( x = 4 \) and \( y = 2 \) is the solution of given equations.

Question. 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 year. Find the ages of Ani and Biju. 
Answer: Let the ages of Ani and Biju be x yr. and y yr, respectively.
According to the given condition,
\( x - y = \pm 3 ..(i) \)
Also, age of Ani's father Dharam = \( 2x \) years
And age of Biju's sister = \( \frac{y}{2} \) years
According to the given condition,
\( 2x - \frac{y}{2} = 30 \)
or, \( 4x - y = 60 .....(ii) \)
Case I : When \( x - y = 3 ......(iii) \)
On subtracting eqn. (iii) from eqn. (ii),
\( 3x = 57 \)
\( \therefore x = 19 \) years
On putting \( x = 19 \) in eqn. (iii),
\( 19 - y = 3 \)
\( \therefore y = 16 \) years
Case II : When \( x - y = -3 ...(iv) \)
On subtracting eqn. (iv) from eqn. (ii),
\( 3x = 60 + 3 \)
\( 3x = 63 \)
\( \therefore x = 21 \) years
On putting \( x = 21 \) in eqn. (iv), we get
\( 21 - y = -3 \)
\( \therefore y = 24 \) years
Hence, Ani's age = 19 years or 21 years.
Biju age = 16 years or 24 years

Question. Find the value of k for which the system of equations \( x + 2y = 5, 3x + ky - 15 = 0 \) has no solution. 
Answer: The given system of equations is
\( x + 2y - 5 = 0 ...(i) \)
\( 3x + ky - 15 = 0 ... (ii) \)
These equations are of the form \( a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0 \), where \( a_1 = 1, b_1 = 2, c_1 = -5 \) and \( a_2 = 3, b_2 = k, c_2 = -15 \)
\( \therefore \frac{a_1}{a_2} = \frac{1}{3}, \frac{b_1}{b_2} = \frac{2}{k} \) and \( \frac{c_1}{c_2} = \frac{-5}{-15} = \frac{1}{3} \)
Let the given system of equations have no solution.
Then, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \implies \frac{1}{3} = \frac{2}{k} \neq \frac{1}{3} \)
\( \implies \frac{1}{3} = \frac{2}{k} \) and \( \frac{2}{k} \neq \frac{1}{3} \)
\( \implies k = 6 \) and \( k \neq 6 \), which is impossible.
Hence, there is no value of k for which the given system of equations has no solution.

Question. Solve for x and y: \( \frac{2}{x} + \frac{3}{y} = 13, \frac{5}{x} - \frac{4}{y} = -2 \) 
Answer: Putting \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \), the given equations become
\( 2u + 3v = 13 .... (i) \)
\( 5u - 4v = -2 .......(ii) \)
Multiplying (i) by 4 and (ii) by 3 and adding the results, we get
\( 8u + 15u = 52 - 6 \)
\( \implies 23u = 46 \)
\( \implies u = \frac{46}{23} = 2 \)
Putting \( u = 2 \) in (i), we get
\( (2 \times 2) + 3v = 13 \implies 3v = 13 - 4 = 9 \implies v = 3 \).
Now, \( u = 2 \implies \frac{1}{x} = 2 \implies 2x = 1 \implies x = \frac{1}{2} \)
And, \( v = 3 \implies \frac{1}{y} = 3 \implies 3y = 1 \implies y = \frac{1}{3} \)
Hence, \( x = \frac{1}{2} \) and \( y = \frac{1}{3} \)

Question. Represent the following pair of equations graphically and write the coordinates of points where the lines intersect y-axis: \( x + 3y = 6, 2x - 3y = 12 \)  
Answer: The given systems of equations are:
\( x + 3y = 6 \) and \( 2x - 3y = 12 \)
Now, \( x + 3y = 6 \implies y = \frac{6-x}{3} \)
Table for equation \( x + 3y = 6 \)
x: 0, 3
y: 2, 1
Now, \( 2x - 3y = 12 \implies y = \frac{2x-12}{3} \)
Table for equation \( 2x - 3y = 12 \)
x: 0, 6
y: -4, 0
Clearly the two lines meet y-axis at B(0, 2) and C(0, -4) respectively.
Hence the required coordinates are (0,2) and (0, -4)

Question. The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and the denominator both are decreased by 1, the numerator becomes half the denominator. Determine the fraction.  
Answer: Let the numerator and the denominator of the fraction be x and y respectively.
Then the fraction is \( \frac{x}{y} \).
Given, The sum of the numerator and the denominator of the fraction is 3 less than the twice of the denominator.
Thus, we have
\( x + y = 2y - 3 \)
\( \implies x + y - 2y + 3 = 0 \)
\( \implies x - y + 3 = 0 ..............(1) \)
Also given, If the numerator and the denominator both are decreased by 1, the numerator becomes half the denominator. Thus, we have
\( x - 1 = \frac{1}{2}(y - 1) \)
\( \implies 2(x - 1) = (y - 1) \)
\( \implies 2x - 2 = (y - 1) \)
\( \implies 2x - y - 1 = 0.......................(2) \)
So, we have formed two linear equations in x & y as following:-
\( x - y + 3 = 0 \)
\( 2x - y - 1 = 0 \)
By using cross-multiplication method, we have \( x = 4, y = 7 \)
Hence, The fraction is \( \frac{x}{y} = \frac{4}{7} \)

Question. Solve the following system of equations: (4) \( x - y + z = 4 \), \( x - 2y - 2z = 9 \), \( 2x + y + 3z = 1 \)
Answer: We have,
\( x - y + z = 4 ...(i) \)
\( x - 2y - 2z = 9 ...(ii) \)
\( 2x + y + 3z = 1 ...(iii) \)
From equation (i), we get \( z = 4 - x + y \implies z = -x + y + 4 \)
Substituting the value of z in equation (ii),
\( x - 2y - 2(-x + y + 4) = 9 \)
\( \implies x - 2y + 2x - 2y - 8 = 9 \)
\( \implies 3x - 4y = 9 + 8 \)
\( \implies 3x - 4y = 17 ...(iv) \)
Substituting the value of z in equation (iii), we get
\( 2x + y + 3(-x + y + 4) = 1 \)
\( \implies 2x + y - 3x + 3y + 12 = 1 \)
\( \implies -x + 4y = 1 - 12 \)
\( \implies -x + 4y = -11 ...(v) \)
Adding equations (iv) and (v), we get
\( 3x - x - 4y + 4y = 17 - 11 \)
\( \implies 2x = 6 \)
\( \implies x = \frac{6}{2} = 3 \)
Putting \( x = 3 \) in equation (iv), we get
\( 3 \times 3 - 4y = 17 \implies 9 - 4y = 17 \implies -4y = 17 - 9 \implies -4y = 8 \implies y = \frac{8}{-4} = -2 \)
Putting \( x = 3 \) and \( y = -2 \) in \( z = -x + y + 4 \), we get
\( z = -3 - 2 + 4 \implies z = -5 + 4 \implies z = -1 \)
Hence, solution of the giving system of equation is \( x = 3, y = -2, z = -1 \).

Question. Solve the following system of equations graphically: \( x - 2y = 5 \), \( 3x - 6y = 15 \) 
Answer: Given equations, \( x - 2y = 5 \) and \( 3x - 6y = 15 \)
Now, \( x - 2y = 5 \implies x = 2y + 5 \)
When \( y = -1 \) then, \( x = 3 \)
When \( y = 0 \) then, \( x = 5 \)
Table for line \( x - 2y = 5 \):
x: 3, 5
y: -1, 0
Now, \( 3x - 6y = 15 \implies x = \frac{15+6y}{3} \)
When \( y = 0 \), then \( x = 5 \)
When \( y = -1 \), then \( x = 3 \)
Table for line \( 3x - 6y = 15 \):
x: 5, 3
y: 0, -1
On plotting the points, both tables represent the same line. Hence, Given equations have infinitely many solutions.