NCERT Solutions Maths Ch-3 Pair of Linear Equations in Two Variables for Class 10th

In Chapter 3 of Class 10 Maths, students learn about solving linear equations that contain two variables. The concept of pair of linear equations focuses on finding the values of these variables that satisfy both equations simultaneously. This is a crucial topic as it lays the foundation for algebraic methods used in higher mathematics.

Key methods to solve these equations include:

Graphical Method: Plotting both equations on a graph and identifying the point where the lines intersect, which gives the solution.
Substitution Method: Solving one equation for one variable and substituting that into the second equation.
Elimination Method: Adding or subtracting equations to eliminate one variable, simplifying the process to find the other variable.
Cross-multiplication Method: A formula-based approach to directly find the values of both variables without the need for graphing or substitution.

NCERT Solutions Maths Ch-3 Pair of Linear Equations in Two Variables

We try to teach you all Questions in easy way. We solve all chapter wise sums of maths textbook. In every chapter include NCERT solutions. For solutions of Exercise 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 and 3.7 click on Tabs :

Question 1.
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”. Isn’t this interesting? Represent this situation algebraically and graphically.

Question 2.
The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.
Solution:

Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.
Solution:

Question 1.
Form the pair of linear equations of 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 Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Solution:

Question 2.
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, 7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0, 2x -y + 9 = 0

Solution:

Question 3.
On comparing the ratios a1a2, b1b2

and c1c2

, find out whether the lines representing the following pairs of linear equations are consistent, or inconsistent:

Question 4.
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.
(i) x + y = 5, 2x + 2y = 10
(ii) x-y – 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Solution:

Question 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 graphically.

Solution:

Question 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
(ii) parallel lines
(iii) coincident lines
Solution:

Question 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.
Solution:

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

(i) x + y = 14

x – y = 4

Solutions:

From 1^st equation, we get,

x = 14 – y

Now, substitute the value of x in second equation to get,

(14 – y) – y = 4

14 – 2y = 4

2y = 10

Or y = 5

By the value of y, we can now find the exact value of x;

∵ x = 14 – y

∴ x = 14 – 5

Or x = 9

Hence, x = 9 and y = 5.

(ii) s – t = 3

s3+t2 =6

Solutions:

From 1^st equation, we get,

s = 3 + t

Now, substitute the value of ‘s’ in second equation to get,

3+t3 + t2 = 6 2(3+t) + 3t6 = 6 6+2t + 3t6 = 6

6 + 5t = 36

5t = 36 -6

t = 305 = 6t = 6

By the value of t, we can now find the exact value of s;

s = 3 + 6

s = 9

Hence, s = 9 and t = 6.

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of’m’ for which y = mx +3.
Solution:

Question 3.
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 Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 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 Rs.105 and for a journey of 15 km, the charge paid is Rs.155. What are the fixed charges and the charges per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes 911

, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes

56

. Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five year ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution:

Question 1.
Solve the following pairs 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) (x/2)+(2y/3) = -1 and x – (y/3) = 3

Solutions:

Question 2.
Form the pair of linear equations for 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.
Solution:

Question 1.
Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions. In case there is unique solution, find it by using cross-multiplication method.
(i) x – 3y – 3 = 0, 3x – 9y – 2 = 0
(ii) 2x + y = 5, 3x + 2y = 8
(iii) 3x – 5y = 20, 6x – 10y = 40
(iv) x – 3y – 7 = 0, 3x – 3y – 15 = 0
Solution:

Question 2.
(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7
(a – b)x + (a + b)y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k – 1)x + (k – 1)y = (2k + 1)
Solution:

Question 3.
Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9, 3x + 2y = 4
Solution:

Question 4.
Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
(i) A part of 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.
(ii) A fraction becomes $frac { 1 }{ 3 }$ when 1 is subtracted from the numerator and it becomes $frac { 1 }{ 4 }$ when 8 is added to its denominator. Find the fraction.
(iii) 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?
(iv) 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 at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution:

Question 1.
Solve the following pairs of equations by reducing them to a pair of linear equations:

Question 2.
Formulate the following problems as a pair of linear equations and hence find their solutions:
(i) Ritu can row downstream 20 km in hours and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(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.
(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:

Question 1.
The age 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 the ages of Ani and Biju be x years and y years respectively.
If Ani is older than Biju
x – y =3
If Biju is older than Ani
y – x = 3
-x + y =3 [Given]

Subtracting equation (i) from equation (ii), we get:
3x – 57
⇒ x = 19
Putting x = 19 in equation (i), we get
19-y = 3
⇒ y = 16
Again subtracting equation (iv) from equation (iii), we get
3x = 63
⇒ x = 21
Putting x = 21 in equation (iii) we get
21 -y= -3
⇒ y = 24
Hence, Ani’s age is either 19 years or 21 years and Biju’s age is either 16 years or 24 years.

Question 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”. Tell me what is the amount of their (respective) capital?
Solution:
Let the two friends have Rs. x and Rs. y.
According to the first condition:
One friend has an amount = Rs.(x + 100)
Other has an amount = Rs. (y – 100
∴ (x + 100) =2 (y – 100)
⇒ x + 100 = 2y – 200
⇒ x – 2y = -300       …(i)
According to the second condition:
One friend has an amount = Rs.(x – 10)
Other friend has an amount =Rs. (y + 10)
∴ 6(x – 10) = y + 10
⇒ 6x – 60 = y + 10
⇒ 6x-y = 70                                        …(ii)
Multiplying (ii) equation by 2 and subtracting the result from equation (i), we get:
x – 12x = – 300 – 140
⇒ -11x = -440
⇒ x = 40
Substituting x = 40 in equation (ii), we get
6 x 40 – y = 70
⇒ -y = 70- 24
⇒ y   = 170
Thus, the two friends have Rs. 40 and Rs. 170.

Question 3.
A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:
Let the original speed of the train be x km/h
and the time taken to complete the journey be y hours. ‘
Then the distance covered = xy km

Case I: When speed = (x + 10) km/h and time taken = (y – 2) h
Distance = (x + 10) (y – 2) km
⇒ xy = (x + 10) (y – 2)
⇒ 10y – 2x = 20
⇒ 5y – x = 10
⇒ -x + 5y = 10 …(i)

Case II: When speed = (x – 10) km/h and time taken = (y + 3) h
Distance = (x – 10) (y + 3) km
⇒ xy = (x – 10) (y + 3)
⇒ 3x- 10y = 30 …(ii)
Multiplying equation (i) by 3 and adding the result to equation (ii), we get
15y – 10y = 30 f 30
⇒ 5y = 60
⇒ y   = 12
Putting y = 12 in equation (ii), we get
3x- 10 x 12= 30
⇒ 3x   = 150
⇒ x   = 50
∴ x = 50 and y = 12
Thus, original speed of train is 50 km/h and time taken by it is 12 h.
Distance covered by train = Speed x Time
= 50 x 12 = 600 km.

Question 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 the number of rows be x and the number of students in each row be y.
Then the total number of students = xy
Case I: When there are 3 more students in each row
Then the number of students in a row = (y + 3)
and the number of rows = (x – 1)
Total number of students = (x – 1) (y + 3)
∴ (x – 1) (y + 3) = xy
⇒ 3x -y =3 …(i)
Case II: When 3 students are removed from each row
Then the number of students in each row = (y-3)
and the number of rows = (x + 2)
Total number of students = (x + 2) (y – 3)
∴ (x + 2) (y – 3) = xy
⇒ -3x + 2y = 6 …(ii)
Adding the equations (i) and (ii), we get
-y + 2y = 3 + 6
⇒ y = 9
Putting y = 9 in the equation (ii), we get
-3x + 18 = 6
⇒ x = 4
∴ x = 4 and y = 9
Hence, the total number of students in the class is 9 x 4 = 36.

Question 5.
In a ∆ABC, ∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles.
Solution:
Let ∠A = x° and ∠B = y°.
Then ∠C = 3∠B = (3y)°.
Now ∠A + ∠B + ∠C = 180°
⇒ x + y + 3y = 180°
⇒ x + 4y = 180° …(i)
Also, ∠C = 2(∠A + ∠B)
⇒ 3y – 2(x + y)
⇒ 2x – y = 0° …(ii)
Multiplying (ii) by 4 and adding the result to equation (i), we get:
9x = 180°
⇒ x = 20°
Putting x = 20 in equation (i), we get:
20 + 4y = 180°
⇒ 4y = 160°
⇒ y =160/40= 40°
∴ ∠A = 20°, ∠B = 40° and ∠C = 3 x 40° = 120°.

Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis.
Solution:
5x – y = 5 …(i)
3x-y = 3 …(ii)
For graphical representation:
From equation (i), we get: y = 5x – 5
When x = 0, then y -5
When x = 2, then y = 10 – 5 = 5
When x = 1, then y = 5 – 5 = 10
Thus, we have the following table of solutions:

From equation (ii), we get:
⇒ y = 3x – 3
When x = 0, then y = -3
When x = 2, then y = 6 – 3 = 3
When x = 1, then y = 3 – 3 = 0
Thus, we have the following table of solutions:

Plotting the points of each table of solutions, we obtain the graphs of two lines intersecting each other at a point C(1, 0).

The vertices of ΔABC formed by these lines and the y-axis are A(0, -5), B(0, -3) and C(1, 0).

Question 7.
Solve the following pairs of linear equations:

Solution:
(i) The given equations are
px + qy = p – q …(1)
qx – py = p + q …(2)
Multiplying equation (1) byp and equation (2) by q and then adding the results, we get:
x(p² + q²) = p(p – q) + q(p + q)

(ii) The given equations are
ax + by = c …(1)
bx – ay = 1 + c …(2)
Multiplying equation (1) by b and equation (2) by a, we get:
abx + b²y = cb …(3)
abx + a²y = a(1+ c) …(4)
Subtracting (3) from (4), we get:

(iii) The given equations may be written as: bx – ay = 0 …(1)
ax + by = a² + b² …(2)
Multiplying equation (1) by b and equation (2) by a, we get:
b²x + aby = 0 ….(3)
a²x + aby = a(a² + b²) …..(4)
Adding equation (3) and equation (4), we get:
(a² + b²)x = a (a² + b²) a(a² + b²)

(iv) The given equations may be written as:
(a – b)x + (a + b)y = a² – 2ab – b² …(1)
(a + b)x + (a + b)y = a² + b² …(2)
Subtracting equation (2) from equation (1), we get:
(a – b)x – (a + b)x
= (a² – 2ab – b²) – (a² + b²)
⇒ x(a – b- a-b) = a² – 2ab – b² – a² – b²⇒ -2bx = -2ab – 2b²⇒ 2bx = 2b² + 2ab

(v) The given equations may be written as:
76x – 189y = -37 …(1)
-189x + 76y = -302 …(2)
Multiplying equation (1) by 76 and equation (2) by 189, we get:
5776x – 14364y = -2812 …(3)
-35721x + 14364y = -57078 …(4)
Adding equations (3) and (4), we get:
5776x – 35721x = -2812 – 57078
⇒ – 29945x = -59890
⇒ x = 2
Putting x = 2 in equation (1), we get:
76 x 2 – 189y = -37
⇒ 152 – 189y = -37
⇒ -189y = -189
⇒ y = 1
Thus, x = 2 and y = 1 is the required solution.

NCERT Solutions Maths Ch-3 Pair of Linear Equations in Two Variables for Class 10th

NCERT Solutions Maths Ch-3 Pair of Linear Equations in Two Variables

Question 1.
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”. Isn’t this interesting? Represent this situation algebraically and graphically.

Question 2.
The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.
Solution:

Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.
Solution:

Question 2.
On comparing the ratios $frac { { a }_{ 1 } }{ { a }_{ 2 } }$ , $frac { { b }_{ 1 } }{ { b }_{ 2 } }$ and $frac { { c }_{ 1 } }{ { c }_{ 2 } }$ ,

Solution:

Question 3.
On comparing the ratios a1a2, b1b2

and c1c2

Question 4.
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.
(i) x + y = 5, 2x + 2y = 10
(ii) x-y – 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Solution:

Question 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 graphically.

Solution:

Question 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
(ii) parallel lines
(iii) coincident lines
Solution:

Question 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.
Solution:

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

(i) x + y = 14

x – y = 4

(ii) s – t = 3

From 1^st equation, we get,

s = 3 + t

Now, substitute the value of ‘s’ in second equation to get,

6 + 5t = 36

5t = 36 -6

By the value of t, we can now find the exact value of s;

s = 3 + 6

s = 9

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of’m’ for which y = mx +3.
Solution:

56

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

Case I: When speed = (x + 10) km/h and time taken = (y – 2) h
Distance = (x + 10) (y – 2) km
⇒ xy = (x + 10) (y – 2)
⇒ 10y – 2x = 20
⇒ 5y – x = 10
⇒ -x + 5y = 10 …(i)

From equation (ii), we get:
⇒ y = 3x – 3
When x = 0, then y = -3
When x = 2, then y = 6 – 3 = 3
When x = 1, then y = 3 – 3 = 0
Thus, we have the following table of solutions:

Plotting the points of each table of solutions, we obtain the graphs of two lines intersecting each other at a point C(1, 0).

Solution:
(i) The given equations are
px + qy = p – q …(1)
qx – py = p + q …(2)
Multiplying equation (1) byp and equation (2) by q and then adding the results, we get:
x(p² + q²) = p(p – q) + q(p + q)

(ii) The given equations are
ax + by = c …(1)
bx – ay = 1 + c …(2)
Multiplying equation (1) by b and equation (2) by a, we get:
abx + b²y = cb …(3)
abx + a²y = a(1+ c) …(4)
Subtracting (3) from (4), we get:

(iii) The given equations may be written as: bx – ay = 0 …(1)
ax + by = a² + b² …(2)
Multiplying equation (1) by b and equation (2) by a, we get:
b²x + aby = 0 ….(3)
a²x + aby = a(a² + b²) …..(4)
Adding equation (3) and equation (4), we get:
(a² + b²)x = a (a² + b²) a(a² + b²)

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NCERT Solutions Maths Ch-3 Pair of Linear Equations in Two Variables for Class 10th

NCERT Solutions Maths Ch-3 Pair of Linear Equations in Two Variables

Question 1.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”. Isn’t this interesting? Represent this situation algebraically and graphically.

Question 2.The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.Solution:

Question 3.The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.Solution:

Question 2.On comparing the ratios , and ,

Solution:

Question 3.On comparing the ratios a1a2, b1b2

and c1c2

Question 4.Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.(i) x + y = 5, 2x + 2y = 10(ii) x-y – 8, 3x – 3y = 16(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0Solution:

Question 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 graphically.

Solution:

Question 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(ii) parallel lines(iii) coincident linesSolution:

Question 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.Solution:

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

(i) x + y = 14

x – y = 4

(ii) s – t = 3

From 1st equation, we get,

s = 3 + t

Now, substitute the value of ‘s’ in second equation to get,

6 + 5t = 36

5t = 36 -6

By the value of t, we can now find the exact value of s;

s = 3 + 6

s = 9

Question 2.Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of’m’ for which y = mx +3.Solution:

56

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

Case I: When speed = (x + 10) km/h and time taken = (y – 2) hDistance = (x + 10) (y – 2) km⇒ xy = (x + 10) (y – 2)⇒ 10y – 2x = 20⇒ 5y – x = 10⇒ -x + 5y = 10 …(i)

From equation (ii), we get:⇒ y = 3x – 3When x = 0, then y = -3When x = 2, then y = 6 – 3 = 3When x = 1, then y = 3 – 3 = 0Thus, we have the following table of solutions:

Plotting the points of each table of solutions, we obtain the graphs of two lines intersecting each other at a point C(1, 0).

Solution:(i) The given equations arepx + qy = p – q …(1)qx – py = p + q …(2)Multiplying equation (1) byp and equation (2) by q and then adding the results, we get:x(p2 + q2) = p(p – q) + q(p + q)

(ii) The given equations areax + by = c …(1)bx – ay = 1 + c …(2)Multiplying equation (1) by b and equation (2) by a, we get:abx + b2y = cb …(3)abx + a2y = a(1+ c) …(4)Subtracting (3) from (4), we get:

(iii) The given equations may be written as: bx – ay = 0 …(1)ax + by = a2 + b2 …(2)Multiplying equation (1) by b and equation (2) by a, we get:b2x + aby = 0 ….(3)a2x + aby = a(a2 + b2) …..(4)Adding equation (3) and equation (4), we get:(a2 + b2)x = a (a2 + b2) a(a2 + b2)

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Question 1.
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”. Isn’t this interesting? Represent this situation algebraically and graphically.

Question 2.
The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.
Solution:

Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.
Solution:

Question 2.
On comparing the ratios $frac { { a }_{ 1 } }{ { a }_{ 2 } }$ , $frac { { b }_{ 1 } }{ { b }_{ 2 } }$ and $frac { { c }_{ 1 } }{ { c }_{ 2 } }$ ,

Question 3.
On comparing the ratios a1a2, b1b2

Question 4.
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.
(i) x + y = 5, 2x + 2y = 10
(ii) x-y – 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Solution:

Question 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 graphically.

Question 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
(ii) parallel lines
(iii) coincident lines
Solution:

Question 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.
Solution:

From 1^st equation, we get,

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of’m’ for which y = mx +3.
Solution:

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

Case I: When speed = (x + 10) km/h and time taken = (y – 2) h
Distance = (x + 10) (y – 2) km
⇒ xy = (x + 10) (y – 2)
⇒ 10y – 2x = 20
⇒ 5y – x = 10
⇒ -x + 5y = 10 …(i)

From equation (ii), we get:
⇒ y = 3x – 3
When x = 0, then y = -3
When x = 2, then y = 6 – 3 = 3
When x = 1, then y = 3 – 3 = 0
Thus, we have the following table of solutions:

Solution:
(i) The given equations are
px + qy = p – q …(1)
qx – py = p + q …(2)
Multiplying equation (1) byp and equation (2) by q and then adding the results, we get:
x(p² + q²) = p(p – q) + q(p + q)

(ii) The given equations are
ax + by = c …(1)
bx – ay = 1 + c …(2)
Multiplying equation (1) by b and equation (2) by a, we get:
abx + b²y = cb …(3)
abx + a²y = a(1+ c) …(4)
Subtracting (3) from (4), we get:

(iii) The given equations may be written as: bx – ay = 0 …(1)
ax + by = a² + b² …(2)
Multiplying equation (1) by b and equation (2) by a, we get:
b²x + aby = 0 ….(3)
a²x + aby = a(a² + b²) …..(4)
Adding equation (3) and equation (4), we get:
(a² + b²)x = a (a² + b²) a(a² + b²)