JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4

Question 1.
Solve the following pair of linear equations by the elimination method and the substitution method:
1. x + y = 5 and 2x – 3y = 4
2. 3x + 4y = 10 and 2x – 2y = 2
3. 3x – 5y – 4 = 0 and 9x = 2y + 7
4. \(\frac{x}{2}+\frac{2 y}{3}=-1\) and x – \(\frac{y}{3}\) = 3
Solution:
1. Elimination method:
x + y = 5 ………(1)
2x – 3y = 4 ……..(2)
We multiply equation (1) by 3 and equation (2) by 1 to get following equations:
3x + 3y = 15 ……(3)
2x – 3y = 4 ……(4)
Adding equations (3) and (4), we get
(3x + 3y) + (2x – 3y) = 15 + 4
∴ 5x = 19
∴ x = \(\frac{19}{5}\)
Substituting x = \(\frac{19}{5}\) in equation (1), we get
\(\frac{19}{5}\) + y = 5
∴ y = 5 – \(\frac{19}{5}\)
∴ y = \(\frac{6}{5}\)
Thus, the solution of the given pair of linear equations is x = \(\frac{19}{5}\), y = \(\frac{6}{5}\)
Substitution method:
x + y = 5 ……(1)
2x – 3y = 4 ……(2)
From equation (1), we get y = 5 – x.
Substituting y = 5 – x in equation (2), we get
2x – 3(5 – x) = 4
∴ 2x – 15 + 3x = 4
∴ 5x = 19
∴ x = \(\frac{19}{5}\)
Substituting x = \(\frac{19}{5}\) in y = 5 – x, we get
y = 5 – \(\frac{19}{5}\)
∴ y = \(\frac{6}{5}\)
Thus, the solution of the given pair of linear equations is x = \(\frac{19}{5}\), y = \(\frac{6}{5}\)

2. Elimination method:
3x + 4y = 10 ……(1)
2x – 2y = 2 ……(2)
We multiply equation (1) by 1 and equation (2) by 2 to get following equations:
3x + 4y = 10 ……(3)
4x – 4y = 4 ……(4)
Adding equations (3) and (4), we get
(3x + 4y) + (4x – 4y) = 10 +4
∴ 7x = 14
∴ x = 2
Substituting x = 2 in equation (1), we get
3(2) + 4y = 10
∴ 4y = 10 – 6
∴ 4y = 4
∴ y = 1
Thus, the solution of the given pair of linear equations is x = 2, y = 1.

Substitution method:
3x + 4y = 10 ……(1)
2x – 2y = 2 ……(2)
From equation (2), we get 2x = 2y + 2.
i.e., x = y + 1.
Substituting x = y + 1 in equation (1), we get
3(y + 1) + 4y = 10
∴ 3y + 3 + 4y = 10
∴ 7y = 7
∴ y = 1
Substituting y = 1 in x = y + 1, we get
x = 1 + 1
x = 2
Thus, the solution of the given pair of linear equations is x = 2, y = 1.

3. Elimination method:
3x – 5y – 4 = 0 ……(1)
9x = 2y + 7 ……(2)
i.e., 3x – 5y = 4 ……(3)
9x – 2y = 7 ……(4)
We multiply equation (3) by 3 and equation (4) by 1 to get following equations:
9x – 15y = 12 ……(5)
9x – 2y = 7 ……(6)
Subtracting equation (5) from equation (6),
we get
(9x – 2y) – (9x – 15y) = 7 – 12
∴ 9x – 2y – 9x + 15y = -5
∴ 13y = -5
∴ y = –\(\frac{5}{13}\)
Substituting y = –\(\frac{5}{13}\) in equation (5), we get
9x – 15(-\(\frac{5}{13}\)) = 12
∴ 9x + \(\frac{75}{13}\) = 12
∴ 9x = 12 – \(\frac{75}{13}\)
∴ 9x = \(\frac{75}{13}\)
∴ x = \(\frac{9}{13}\)
Thus, the solution of the given pair of linear equations is x = \(\frac{9}{13}\), y = –\(\frac{5}{13}\)

Substitution method:
3x – 5y – 4 = 0 ………(1)
9x = 2y + 7 ………(2)
From equation (2), we get x = \(\frac{2 y+7}{9}\).
Substituting x = \(\frac{2 y+7}{9}\) in equation (1).
we get
3(\(\frac{2 y+7}{9}\)) – 5y – 4 = 0
∴ \(\frac{2 y+7}{9}\) – 5y – 4 = 0
∴ 2y + 7 – 15y – 12 = 0 (Multiplying by 3)
∴ -13y – 5 = 0
∴ -13y = 5
∴ y = –\(\frac{5}{13}\)
Substituting y = –\(\frac{5}{13}\) in x = \(\frac{2 y+7}{9}\)
we get,
JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 1
Thus, the solution of the given pair of linear equations is x = \(\frac{9}{13}\), y = –\(\frac{5}{13}\).

4. Elimination method:
\(\frac{x}{2}+\frac{2 y}{3}=-1\) ………(1)
\(x-\frac{y}{3}=3\) ………(2)
We multiply equation (1) by 3 and equation (2) by 6 to get following equations:
\(\frac{3}{2}\)x + 2y = -3 ………(3)
6x – 2y = 18 ………(4)
Adding equations (3) and (4), we get
(\(\frac{3}{2}\)x + 2y) + (6x – 2y) = -3 + 18
∴ \(\frac{15}{2}\)x = 15
∴ x = 2
Substituting x = 2 in equation (3), we get
\(\frac{3}{2}\)(2) + 2y = -3
∴ 3 + 2y = -3
∴ 2y = -6
∴ y = -3
Thus, the solution of the given pair of linear equations is x = 2, y = -3.

Substitution method:
\(\frac{x}{2}+\frac{2 y}{3}=-1\) ………(1)
\(x-\frac{y}{3}=3\) ………(2)
From equation (2), we get x = \(\frac{y}{3}+3\)
Substituting x = \(\frac{y}{3}+3\) in equation (1).
we get
JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 2
Substituting y = -3 in x = \(\frac{y}{3}+3\), we get
x = \(\frac{-3}{3}+3\)
∴ x = -1 + 3
∴ x = 2
Thus, the solution of the given pair of linear equations is x = 2, y = -3.

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method:

Question 1.
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?
Solution:
Let the numerator of the required fraction be x and the denominator be y.
Then, the required fraction is \(\frac{x}{y}\)
From the first condition given, we get
\(\frac{x+1}{y-1}=1\)
∴ x + 1 = y – 1
∴ x – y = -2 …..(1)
From the second condition, we get
\(\frac{x}{y+1}=\frac{1}{2}\)
∴ 2x = y + 1
∴ 2x – y = 1 ………(2)
Now, subtracting equation (1) from equation (2), we get
(2x – y) – (x – y) = 1 – (-2)
∴ 2x – y – x + y = 3
∴ x = 3
Substituting x = 3 in equation (1), we get
3 – y = -2
∴ -y = – 2 – 3
∴ -y = -5
∴ y = 5
So, the fraction \(\frac{x}{y}=\frac{3}{5}\)
Thus, the pair of linear equations formed is x – y = -2 and 2x – y = 1; and the required fraction is \(\frac{3}{5}\)

Question 2.
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 ?
Solution:
Let the present age of Nuri be x years. and the present age of Sonu be y years.
Five years ago, the age of Nuri was (x – 5) years and the age of Sonu was (y – 5) years.
Then, from the first condition given, we get
(x – 5) = 3(y – 5)
∴ x – 5 = 3y – 15
∴ x – 3y = -10 ……….(1)
Ten years later the ages of Nuri and Sonu will be (x + 10) years and (y + 10) years respectively.
Then, from the second condition given, we get
(x + 10) = 2(y + 10)
∴ x + 10 = 2y + 20
∴ x – 2y = 10 ………..(2)
Subtracting equation (1) from equation (2).
we get
(x – 2y) – (x – 3y) = 10 – (-10)
∴ x – 2y – x + 3y = 10 + 10
∴ y = 20
Substituting y = 20 in equation (2).
we get
x – 2 (20) = 10
∴ x – 40 = 10
∴ x = 50
So, the present ages of Nuri and Sonu are 50 years and 10 years.
Thus, the pair of linear equations formed is x – 3y = -10 and x – 2y = 10 and the present ages of Nuri and Sonu are 50 years and 20 years respectively.

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

Question 3.
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.
Solution:
Let the digit at tens place be x and the digit at units place be y in the original number.
Then, the original number = 10x + y.
From the first condition given, we get x + y = 9 ………..(1)
If the digits are reversed, in the new number the digit at tens place is y and the digit at units place is x.
Then, the new number = 10y + x.
From the second condition given, we get
9(10x + y) = 2(10y + x)
∴ 90x + 9y = 20y + 2x
∴ 88x – 11y = 0
∴ 8x – y = 0 (Dividing by 11) ……….(2)
Adding equations (1) and (2), we get
(x + y) + (8x − y) = 9 + 0
∴ 9x = 9
∴ x = 1
Substituting x = 1 in equation (1), we get
1 + y = 9
∴ y = 8
So, the original number = 10x + y
= 10(1) + 8 = 18
Thus, the pair of linear equations formed is x + y = 9 and 8x – y = 0; and the original number is 18.

Question 4.
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.
Solution:
Suppose Meena received x notes of ₹ 50 and y notes of ₹ 100.
So, the total amount received by her = ₹ (50x + 100y)
From the first condition given, the total amount is ₹ 2000. So, we get
50x + 100y = 2000
∴ x + 2y = 40 (Dividing by 50) ……(1)
From the second condition given, we get
x + y = 25 ……(2)
Subtracting equation (2) from equation (1),
we get
(x + 2y) – (x + y) = 40 – 25
∴ y = 15
Substituting y = 15 in equation (2), we get
x + 15 = 25
∴ x = 10
Hence, Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.
Thus, the pair of linear equations formed is x + 2y = 40 and x + y = 25, and Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

Question 5.
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:
Let the fixed charge for first three days be ₹ x and the additional charge for each day exceeding the first three days be ₹ y. Saritha kept the book for 7 days.
So, she has to pay the fixed charge plus the additional charge for 4(7 – 3) days. Hence, we get the following equation for Saritha:
x + 4y = 27 …………(1)
Similarly, Susy has to pay the fixed charge plus the addition charge for 2 (5 – 3) days.
Hence, we get the following equation for Susy:
x + 2y = 21 ……….(2)
Subtracting equation (2) from equation (1),
we get
(x + 4y) – (x + 2y) = 27 – 21
∴ 2y = 6
∴ y = 3
Substituting y = 3 in equation (1), we get
x + 4(3) = 27
∴ x + 12 = 27
∴ x = 15
Hence, the fixed charge for first three days is ₹ 15 and the addition charge for each day thereafter is ₹ 3.
Thus, the pair of linear equations formed is x + 4y = 27 and x + 2y = 21; and the fixed charge and the additional charge per day are ₹ 15 and ₹ 3 respectively.

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

Question 1.
Solve the following pair of linear equations by the substitution method:
1. x + y = 14
x – y = 4
2. s – t = 3
\(\frac{s}{3}+\frac{t}{2}=6\)
3. 3x – y = 3
9x – 3y = 9
4. 0.2x + 0.3y = 1.3
0.4x + 0.5 y = 2.3
5. \(\sqrt{2}\)x + \(\sqrt{3}\)3y = 0
\(\sqrt{3}\)x – \(\sqrt{8}\)y = 0
6. \(\frac{3 x}{2}-\frac{5 y}{3}=-2\)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)
Solution:
1. x + y = 14 ……….(1)
x – y = 4 ……..(2)
From equation (1), we get y = 14 – x.
Substituting y = 14 – x in equation (2).
we get
x – (14 – x) = 4
∴ x – 14 + x = 4
∴ 2x = 4 + 14
∴ 2x = 18
∴ x = 9
Substituting x = 9 in equation (1), we get
9 + y = 14
∴ y = 5
Thus, the solution of the given pair of linear equations is x = 9, y = 5.
Verification: x + y = 9 + 5 = 14 and x – y = 9 – 5 = 4.
Hence, the solution is verified.

2. s – t = 3 ……….(1)
\(\frac{s}{3}+\frac{t}{2}=6\) ……..(2)
From equation (1), we get s = t + 3.
Substituting s = t + 3 in equation (2),
we get
\(\frac{t+3}{3}+\frac{t}{2}=6\)
∴ 2 (t + 3) + 3t = 36 (Multiplying by 6)
∴ 2t + 6 + 3t = 36
∴ 5t = 30
∴ t = 6
Substituting t = 6 in equation (1), we get s – 6 = 3
∴ s = 9
Thus, the solution of the given pair of linear equations is s = 9, t = 6.
Verification: s – t = 9 – 6 = 3 and
\(\frac{s}{3}+\frac{t}{2}=\frac{9}{3}+\frac{6}{2}=6\)
Hence, the solution is verified.

3. 3x – y = 3 ……….(1)
9x – 3y = 9 ……….(2)
From equation (1), we get y = 3x – 3.
Substituting y = 3x – 3 in equation (2).
we get
9x – 3(3x – 3) = 9
∴ 9x – 9x + 9 = 9
∴ 9 = 9
Here, we do not get the value of x, but we get a true statement 9 = 9.
Hence, the given pair of linear equations has infinitely many solutions given by y = 3x – 3, where x is any real number.

4. 0.2x + 0.3y = 1.3 ……….(1)
0.4x + 0.5y = 2.3 ……….(2)
Not mandatory, but for convenience we multiply both the equations by 10 and get equations with integer coefficients as:
2x + 3y = 13 ……(3)
4x + 5y = 23 ……..(4)
From equation (3), we get x = \(\frac{13-3 y}{2}\).
Substituting x = \(\frac{13-3 y}{2}\) in equation (4),
we get
4(\(\frac{13-3 y}{2}\)) + 5y = 23
∴ 26 – 6y + 5y = 23
∴ -y = -3
∴ y = 3
Substituting y = 3 in x = \(\frac{13-3 y}{2}\)
x = \(\frac{13-3(3)}{2}\)
∴ x = \(\frac{4}{2}\)
∴ x = 2
Thus, the solution of the given pair of linear equations is x = 2, y = 3.
Verification:
0.2x + 0.3y = (0.2) (2) + (0.3) (3) = 1.3 and
0.4x + 0.5y = (0.4) (2) = (0.5) (3) = 2.3.
Hence, the solution is verified.

5. \(\sqrt{2}\)x + \(\sqrt{3}\)y = 0 ……….(1)
\(\sqrt{3}\)x – \(\sqrt{8}\)y = 0 ……….(2)
From equation (2), we get x = \(\frac{\sqrt{8}}{\sqrt{3}}\)y.
Substituting x = \(\frac{\sqrt{8}}{\sqrt{3}}\)y in equation (1),
we get
\(\sqrt{2}\left(\frac{\sqrt{8}}{\sqrt{3}} y\right)+\sqrt{3} y=0\)
∴ \(\frac{4}{\sqrt{3}} y+\sqrt{3} y=0\)
∴ \(y\left(\frac{4}{\sqrt{3}}+\sqrt{3}\right)=0\)
∴ y = 0
Substituting y = 0 in x = \(\frac{\sqrt{8}}{\sqrt{3}}\)y, we get
x = \(\frac{\sqrt{8}}{\sqrt{3}}\)(0)
∴ x = 0
Thus, the solution of the given pair of linear equations is x = 0, y = 0.

6. \(\frac{3 x}{2}-\frac{5 y}{3}=-2\) ……….(1)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\) ……….(2)
Not mandatory, but for convenience we multiply both the equations by 6 and get
9x – 10y = – 12 ……….(3)
2x + 3y = 13 ……….(4)
From equation (3), we get x = \(\frac{10 y-12}{9}\).
Substituting x = \(\frac{10 y-12}{9}\) in equation (4).
we get
2(\(\frac{10 y-12}{9}\)) + 3y = 13
∴ 2(10y – 12) + 27y = 117 (Multiplying by 9)
∴ 20y – 24 + 27y = 117
∴ 47y = 141
∴ y = 3
Substituting y = 3 in x = \(\frac{10 y-12}{9}\), we get
x = \(\frac{10(3)-12}{9}\)
∴ x = \(\frac{18}{9}\)
∴ x = 2
Thus, the solution of the given pair of linear equations is x = 2, y = 3.

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx +3.
Solution:
2x + 3y = 11 …..(1)
2x – 4y = -24 …..(2)
From equation (2), we get x = \(\frac{4 y-24}{2}\) = 2y – 12.
Substituting x = 2y – 12 in equation (1), we get
2 (2y – 12) + 3y = 11
∴ 4y – 24 + 3y = 11
∴ 7y = 35
∴ y = 5
Substituting y = 5 in x = 2y – 12, we get
x = 2(5) – 12
∴ x = 10 – 12
∴ x = -2
Now, for x = -2 and y = 5, y = mx + 3 gives
5 = m(-2) + 3
∴ 5 = -2m + 3
∴ 2m = 3 – 5
∴ 2m = -2
∴ m = -1
Thus, the solution of the given pair of equations is x = -2, y = 5 and m = -1 satisfies y = mx +3.

Form the pair of linear equations for the following problems and find their solution by substitution method:

Question 1.
The difference between two numbers is 26 and one number is three times the other. Find them.
Solution:
Let the greater number be x and the smaller number be y
Then, from the given information, we get the following pair of linear equations:
x – y = 26 ……….(1)
x = 3y ………….(2)
Substituting x = 3y in equation (1).
we get
3y – y = 26
∴ 2y = 26
∴ y = 13
Then, x = 3y gives x = 3 × 13 = 39.
Thus, the required numbers are 39 and 13.
Verification: The difference of numbers = 39 – 13 = 26 and 39 = three times 13.

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Question 2.
The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution:
Let the measure (in degrees) of the greater angle be x and that of the smaller angle be y.
Then, from the given data, we get the following pair of linear equations:
x + y = 180 ……….(1)
x – y = 18 ………….(2)
From equation (2), we get x = y + 18.
Substituting x = y + 18 in equation (1).
we get
y + 18 + y = 180
∴ 2y = 162
∴ y = 81
Substituting y = 81 in equation (2), we get
x – 81 = 18
∴ x = 99
Thus, the measures (in degrees) of the given angles are 99 and 81.
Verification: Larger angle-Smaller angle = 99° – 81° = 18° and Larger angle + Smaller angle = 99° + 81° = 180°, i.e., the angles are supplementary angles.

Question 3.
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.
Solution:
Let the cost of each bat be ₹ x and the cost of each ball be ₹ y.
Then, from the given data, we get the following pair of linear equations:
7x + 6y = 3800 ……….(1)
3x + 5y = 1750 ……….(2)
From equation (2), we get x = \(\frac{1750-5 y}{3}\)
Substituting x = \(\frac{1750-5 y}{3}\) in equation (1) we get
7(\(\frac{1750-5 y}{3}\)) + 6y = 3800
∴ 7(1750 – 5y) + 18y = 11400 (Multiplying by 3)
∴ 12250 – 35y + 18y = 11400
∴ -17y = 11400 – 12250
∴ -17y = -850
∴ 17y = 850
∴ y = 50
Substituting y = 50 in x = \(\frac{1750-5 y}{3}\), we get
x = \(\frac{1750-5(50)}{3}\)
∴ x = \(\frac{1500}{3}\)
∴ x = 500
Thus, the cost of each bat is ₹ 500 and the cost of each ball is ₹ 50.
Verification:
Cost of 7 bats and 6 balls = 7 × 500 + 6 × 50 = ₹ 3800
Cost of 3 bats and 5 balls = 3 × 500 + 5 × 50 = ₹ 1750

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Question 4.
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?
Solution:
Let the fixed charge be ₹ x and the charge for the distance covered be ₹ y per km.
Then, from the given data, we get the following pair of linear equations:
x + 10y = 105 ……….(1)
x + 15y = 155 ……….(2)
From equation (1), we get x = 105 – 10y.
Substituting x = 105 – 10y in equation (2), we get
(105 – 10y) + 15y = 155
∴ 105 + 5y = 155
∴ 5y = 50
∴ y = 10
Substituting y = 10 in x = 105 – 10y.
we get
x = 105 – 10(10)
∴ x = 5
Thus, the fixed charge is ₹ 5 and the charge for the distance covered is ₹ 10 per km.
So, the total charge a person has to pay for travelling d km, is given by
Total charge = ₹(5 + 10d)
∴ Total charge to be paid for travelling 25 km = ₹ (5 + 10 × 25) = ₹ 255.

Question 5.
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.
Solution:
Let the numerator of the fraction be x and the denominator of the fraction be y.
Then, the required fraction is \(\frac{x}{y}\)
Then, from the given data, we get the following pair of equations:
\(\frac{x+2}{y+2}=\frac{9}{11}\)
∴ 11(x + 2) = 9(y + 2)
∴ 11x + 22 = 9y + 18
∴ 11x – 9y = -4 is the first linear equation derived from the data.
Similarly, \(\frac{x+3}{y+3}=\frac{5}{6}\)
∴ 6x + 18 = 5y + 15
∴ 6x – 5y = -3 is the second linear equation derived from the data.
Hence, required pair of linear equations is as follows:
11x – 9y = -4 ……….(1)
6x – 5y = -3 ……….(2)
From equation (2), we get x = \(\frac{5 y-3}{6}\)
Substituting x = \(\frac{5 y-3}{6}\) in equation (1),
we get
11(\(\frac{5 y-3}{6}\)) – 9y = -4
∴ 11 (5y – 3) – 54y = -24 (Multiplying by 6)
∴ 55y – 33 – 54y = -24
∴ y = 9
Substituting y = 9 in x = \(\frac{5 y-3}{6}\), we get
x = \(\frac{5(9)-3}{6}\)
∴ x = \(\frac{42}{6}\)
∴ x = 7
Thus, the required fraction is \(\frac{7}{9}\)
Verification:
\(\frac{7+2}{9+2}=\frac{9}{11}\) and \(\frac{7+3}{9+3}=\frac{10}{12}=\frac{5}{6}\)

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Question 6.
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:
Let the present age of Jacob be x years and the present age of his son be y years.
∴ Five years hence, the age of Jacob will be (x + 5) years and the age of his son will be (y + 5) years.
Then, from the given data,
(x + 5) = 3(y + 5)
∴ x + 5 = 3y + 15
∴ x – 3y = 10
Again, five years ago, the age of Jacob was (x – 5) years and the age of his son was (y – 5) years.
Then, from the given data,
(x – 5) = 7(y – 5)
∴ x – 5 = 7y – 35
∴ x – 7y = -30
Hence, the required pair of linear equations is as follows:
x – 3y = 10 ……….(1)
x – 7y = -30 ……….(2)
From equation (2), we get x = 7y – 30.
Substituting x = 7y – 30 in equation (1).
we get
7y – 30 – 3y = 10
∴ 4y = 40
∴ y = 10
Substituting y = 10 in x = 7y – 30, we get
x = 7(10) – 30
∴ x = 40
Thus, the present ages of Jacob and his son are 40 years and 10 years respectively.

JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3

Jharkhand Board JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 6 Triangles Exercise 6.3

Question 1.
State which pairs of triangles in the given figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
Answer:
1.
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 1
In ΔABC and ΔPQR,
∠A = ∠P = 60°, ∠B = ∠Q = 80° and
∠C = ∠R = 40°
∴ By AAA criterion, ΔABC – ΔPQR.

2.
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 2
In ΔABC and ΔQRP,
\(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{2}{4}=\frac{1}{2}\), \(\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{2.5}{5}=\frac{1}{2}\) and \(\frac{\mathrm{CA}}{\mathrm{PQ}}=\frac{3}{6}=\frac{1}{2}\)
Thus, \(\frac{AB}{QR}=\frac{BC}{RP}=\frac{CA}{PQ}\)
∴ By SSS criterion, ΔABC ~ ΔQRP.

3.
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 3
No, the given triangles are not similar as
\(\frac{MP}{DE}=\frac{1}{2}\), \(\frac{LP}{DF}=\frac{1}{2}\), but \(\frac{LM}{EF}\) = \(\frac{2.7}{5}≠\frac{1}{2}\)

4.
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 4
In ΔMNL and ΔQPR,
\(\frac{\mathrm{MN}}{\mathrm{QP}}=\frac{2.5}{5}=\frac{1}{2}\), \(\frac{\mathrm{ML}}{\mathrm{QR}}=\frac{5}{10}=\frac{1}{2}\)
∠M = ∠Q = 70°
∴ By SAS criterion, ΔMNL ~ ΔQPR.

5.
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 5
No, the given triangles are not similar as two given corresponding sides are proportionate but the corresponding included angles are not equal.

6.
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 6
In ΔDEF, ∠D = 70°, ∠E = 80°
∠F = 180° – 70° – 80° = 30°
In ΔPQR, ∠Q = 80°, ∠R = 30°
∴ ∠P = 180° – 80° – 30° = 70°
Thus, in ΔDEF and ΔPQR.
∠D = ∠P, ∠E = ∠Q and ∠F = ∠R
∴ By AAA criterion, ΔDEF ~ ΔPQR.

JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3

Question 2.
In the given figure ΔODC – ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Answer:
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 7
In ΔDOC, ∠COB is an exterior angle.
∴ ∠COB + ∠DOC = 180°
∴ 125° + ∠DOC = 180°
∴ ∠DOC = 55°
Again, ∠COB = ∠ODC + ∠DCO
∴ 125° = 70° + ∠DCO
∴ ∠DCO = 55°
Now, ΔODC ~ ΔOBA
∴ ∠OAB = ∠OCD
∴ ∠OAB = 55°
Thus, ∠DOC = 55°, ∠DCO = 55° and
∴ ∠OAB = 55°.

Question 3.
Diagonal AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \(\frac{OA}{OC}=\frac{OB}{OD}\)
Answer:
Given:
In trapezium ABCD, AB || DC and diagonals AC and BD intersect at O.
To prove : \(\frac{OA}{OC}=\frac{OB}{OD}\)
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 8
Proof: In trapezium ABCD, AB || CD.
∴ ∠CAB = ∠ACD and ∠DBA = ∠BDC
(Alternate angles) ……(1)
Then, in ΔOAB and ΔOCD.
∠OAB = ∠OCD and ∠OBA = ∠ODC [By (1)]
∴ By AA criterion, ΔOAB ~ ΔOCD.
∴ \(\frac{OA}{OC}=\frac{OB}{OD}\)

Question 4.
In the given figure \(\frac{QR}{QS}=\frac{QT}{PR}\) and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
Answer:
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 9
In ΔPQR, ∠1 = ∠2, i.e., ∠PQR = ∠PRQ
∴ PR = QP
Now, \(\frac{QR}{QS}=\frac{QT}{PR}\)
∴ \(\frac{QR}{QS}=\frac{QT}{QP}\)
In ΔTQR, P and S are points on QT and QR respectively and \(\frac{QR}{QS}=\frac{QT}{QP}\)
∴ By theorem 6.2, SP || RT.
∴ ∠QPS = ∠QTR and ∠QSP = ∠QRT (Corresponding angles)
Now, in ΔPQS and ΔTQR.
∠QPS = ∠QTR,
∠QSP = ∠QRT and
∠PQS = ∠TQR (Same angle)
∴ By AAA criterion, ΔPQS ~ ΔTQR.

JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3

Question 5.
S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS
Answer:
Given:
S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS.
To prove ΔRPQ ~ ΔRTS
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 10
Proof : ∠P = ∠RTS
∴∠RPQ = ∠RTS.
In ΔRPQ and ΔRTS,
∠RPQ = ∠RTS and
∠PRQ = ∠TRS (Same angle)
∴ By AA criterion, ΔRPQ ~ ΔRTS.

Question 6.
In the given figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
Answer:
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 11
ΔABE ≅ ΔACD (Given)
∴ AB = AC and AE = AD (CPCT)
∴ \(\frac{AE}{AC}=\frac{AD}{AB}\)
Now, in ΔADE and ΔABC,
\(\frac{AE}{AC}=\frac{AD}{AB}\)
and ∠DAE = ∠BAC (Same angle)
∴ By SAS criterion, ΔADE ~ ΔABC

Question 7.
In the given figure, altitudes AD and CE of ΔABC intersect each other at the point P.
Show that:
1. ΔAEP ~ ΔCDP
2. ΔABD ~ ΔCBE
3. ΔAEP ~ ΔADB
4. ΔPDC ~ ΔBEC
Αnswer:
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 12
1. In ΔAEP and ΔCDP,
∠AEP = ∠CDP (Right angles)
∠EPA = ∠DPC (Vertically opposite angles)
∴ By AA criterion, ΔAEP ~ ΔCDP.

2. In ΔABD and ΔCBE,
∠ABD = ∠CBE (Same angle)
∠ADB = ∠CEB (Right angles)
∴ By AA criterion, ΔABD ~ ΔCBE.

3. In ΔAEP and ΔADB,
∠AEP = ∠ADB (Right angles)
∠EAP = ∠DAB (Same angle)
∴ By AA criterion, ΔAEP ~ ΔADB.

4. In ΔPDC and ΔBEC,
∠PDC = ∠BEC (Right angles)
∠PCD = ∠BCE (Same angle)
∴ By AA criterion, ΔPDC ~ ΔBEC.

JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
Answer:
Given: E is a point on the side AD produced of parallelogram ABCD and BE intersects CD at F
To prove : ΔABE ~ ΔCFB
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 12
Proof: In parallelogram ABCD,
∠A = ∠C (Opposite angles)
∴ ∠BAE = ∠FCB ………..(1)
E lies on AD extended in parallelogram ABCD.
∴ AE || BC
∴ ∠AEB = ∠CBE (Alternate angles)
∴ ∠AEB = ∠CBF …………..(2)
Now, in ΔABE and ΔCFB,
∠BAE = ∠FCB [By (1)]
∠AEB = ∠CBF [By (2)]
∴ By AA criterion, ΔABE ~ ΔCFB.

Question 9.
In the given figure, ABC and AMP are two right triangles, right-angled at B and M respectively. Prove that :
1. ΔABC ~ ΔAMP
2. \(\frac{CA}{PA}=\frac{BC}{MP}\)
Answer:
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 14
In ΔABC and ΔAMP
∠ABC = ∠AMP (Right angles)
∠BAC = ∠MAP (Same angle)
∴ By AA criterion, ΔABC ~ ΔAMP [Result (1)]
Since ΔABC ~ ΔAMP, \(\frac{CA}{PA}=\frac{BC}{MP}\) [Result (2)]

Question 10.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC – ΔFEG, show that:
1. \(\frac{CD}{GH}=\frac{AC}{FG}\)
2. ΔDCB ~ ΔHGE
3. ΔDCA ~ ΔHGF
Answer:
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 15
ΔABC ~ ΔFEG
∴ ∠A = ∠F, ∠B = ∠E and ∠ACB = ∠FGE …………….(1)
CD is the bisector of ∠ACB and GH is the bisector of ∠FGE.
∴ ∠ACD = ∠BCD = \(\frac{1}{2}\)∠ACB ……(2)
and ∠FGH = ∠EGH = \(\frac{1}{2}\)∠FGE ………..(3)
So, from (1), (2) and (3).
∠ACD = ∠FGH and ∠BCD = ∠EGH ……(4)
Now, in ΔDCB and ΔHGE,
∠B = ∠E [By (1)]
∠BCD = ∠EGH [By (4)]
Hence, by AA criterion,
ΔDCB ~ ΔHGE [Result (2)]
Again, in ΔDCA and ΔHGF
∠A = ∠F [By (1)]
∠ACD = ∠FGH [By (4)]
Hence, by AA criterion,
ΔDCA – ΔHGF [Result (3)]
Now, ΔDCA – ΔHGF
∴ \(\frac{CD}{GH}=\frac{AC}{FG}\) [Result (1)]

JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3

Question 11.
In the given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.
Answer:
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 16
In ΔABC, AB = AC
∴ ∠ABC = ∠ACB
∴ ∠ABD = ∠ECF (∵ E lies on CB extended and F lies on AC.)
AD ⊥ BC
∴ ∠ADB = 90°
EF ⊥ AC
∴ ∠EFC = 90°
Now, in ΔABD and ΔECF
∠ABD = ∠ECF
∠ADB = ∠EFC (Both right angles)
∴ By AA criterion, ΔABD ~ ΔECF

Question 12.
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see the given figure). Show that ΔABC ~ ΔPQR.
Answer:
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 17
Hence, by SSS criterion, ΔABD ~ ΔPQM.
∴ ∠ABD = ∠PQM
∴ ∠ABC = ∠PQR
Now, in ΔABC and ΔPQR,
\(\frac{AB}{PQ}=\frac{BC}{QR}\) and ∠ABC = ∠PQR
Hence, by SAS criterion, ΔABC ~ ΔPQR.

Question 13.
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB. CD.
Answer:
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 18
In ΔCDA and ΔCAB.
∠ADC = ∠BAC (Given)
∠ACD = ∠BCA (Same angle)
∴ By AA criterion, ΔCDA ~ ΔCAB
∴ \(\frac{CD}{CA}=\frac{CA}{CB}\)
∴ CB . CD = CA . CA
∴ CA² = CB.CD

JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3

Question 14.
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides Pg and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.
Answer:
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 19
In ΔABC, AD is a median.
∴ BD = DC
Take point E on AD extended such that AD = DE and draw BE and CE.
∴ AE = 2AD.
In quadrilateral ABEC, diagonals AE and BC bisect each other.
∴ ABEC is a parallelogram.
∴ BE = AC (Opposite sides) ……(1)
Similarly, in ΔPQR, PM is a median.
∴ QM = MR
Take point N on PM extended such that PM MN and draw QN and RN.
∴ PN = 2PM
In quadrilateral PQNR, diagonals PN and QR bisect each other.
∴ PQNR is a parallelogram.
∴ QN = PR (Opposite sides) …………..(2)
Now,
∴\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\) (Given)
∴ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BE}}{\mathrm{QN}}=\frac{2 \mathrm{AD}}{2 \mathrm{PM}}\) [By (1) and (2)]
∴\(\frac{AB}{PQ}=\frac{\mathrm{BE}}{\mathrm{QN}}=\frac{\mathrm{AE}}{\mathrm{PN}}\)
∴ By SSS criterion ΔABE ~ ΔPQN.
∴ ∠BAE = ∠QPN
∴ ∠BAD = ∠QPM …………..(3)
In the same manner, it can be proved that ΔACE ~ ΔPRN
∴ ∠CAE = ∠RPN
∴ ∠CAD = ∠RPM …………..(4)
Adding (3) and (4).
∠BAD + ∠CAD = ∠QPM + ∠RPM
∴ ∠BAC = ∠QPR
Now, in ΔABC and ΔPQR,
\(\frac{AB}{PQ}=\frac{AC}{PR}\) and ∠BAC = ∠QPR
Hence, by SAS criterion, ΔABC ~ ΔPQR.

Question 15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same. time a tower casts a shadow 28 m long. Find the height of the tower.
Answer:
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 20
Here, AB is the vertical pole and AC is its shadow while PQ is the tower and QR is its shadow.
As both the shadows are measured at the same time, ∠C and ∠R both represent the elevation of the sun.
∴ ∠C = ∠R
In ΔABC and ΔPQR,
∠C = ∠R
∠B = ∠Q (Right angles)
∴ By AA criterion, ΔABC ~ ΔPQR.
∴ \(\frac{AB}{PQ}=\frac{BC}{QR}\)
∴ \(\frac{6}{PQ}=\frac{4}{28}\)
∴ PQ = \(\frac{6 \times 28}{4}\)
∴ PQ = 42 m
Thus, the height of the tower is 42 m.

JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3

Question 16.
If AD and PM are medians of triangles ABC and PQR respectively; where ΔABC ~ ΔPQR prove that \(\frac{AB}{PQ}=\frac{AD}{PM}\)
Answer:
JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.3 - 21
(∵ AD and PM are medians of ΔABC and ΔPQR.)
Also, ∠ABC = ∠PQR
∴ ∠ABD = ∠PQM
Now, in ΔABD and ΔPQM,
\(\frac{AB}{PQ}=\frac{BD}{QM}\) and ∠ABD = ∠PQM
∴ By SAS criterion, ΔABD ~ ΔPQM.
∴ \(\frac{AB}{PQ}=\frac{AD}{PM}\)

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2

Question 1.
Form the pair of linear equations in the following problems, and find their solutions graphically:
1. 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.
2. 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:
1. Let the number of boys be x and the number of girls be y.
Then, the equations formed as follows:
x + y = 10 ……. (1)
and y = x + 4,
i.e., y – x = 4 …… (2)
To draw the graphs of these equations,
we find two solutions for each equation.
For equation (1), x + y = 10 gives y = 10 – x.

x 0 5
y 10 5

For equation (2), y – x = 4 gives y = x + 4.

x 0 2
y 4 6

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 1
Two lines intersect at point (3, 7). Hence, x = 3 and y = 7 is the required solution of the pair of linear equations.
Thus, 3 boys and 7 girls took part in the quiz.
Verification : x = 3 and y = 7 satisfy both the equations x + y = 10 and y – x = 4.

2. Let the cost of each pencil be ₹ x and the cost of each pen be ₹ y.
Then, from the given information, we receive the following equations:
5x + 7y = 50 ……… (1)
7x + 5y = 46 ……….. (2)
To draw the graphs of these equations, we find two solutions for each equation.
For equation (1), 5x + 7y = 50 gives
y = \(\frac{50-5 x}{7}\)

x 3 10
y 5 0

For equation (2), 7x + 5y = 46 gives
y = \(\frac{46-7 x}{5}\)

x 3 8
y 5 -2

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 2
Two lines intersect at point (3, 5). Hence, x = 3 and y = 5 is the required solution of the pair of linear equations.
Thus, the cost of each pencil is ₹ 3 and the cost of each pen is ₹ 5.
Verification : x = 3 and y = 5 satisfy both the equations 5x + 7y = 50 and 7x + 5y = 46.

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

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:
1. 5x – 4y + 8 = 0; 7x + 6y – 9 = 0
2. 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
3. 6x – 3y + 10 = 0; 2x – y + 9 = 0
Solution:
5x – 4y + 8 = 0; 7x + 6y – 9 = 0
For the given pair of linear equations,
a1 = 5, b1 = -4, c1 = 8, a2 = 7, b2 = 6 and c2 = -9
Now, \(\frac{a_1}{a_2}=\frac{5}{7}, \quad \frac{b_1}{b_2}=\frac{-4}{6}=-\frac{2}{3}\)
and \(\frac{c_1}{c_2}=\frac{8}{-9}=-\frac{8}{9}\)
Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Hence, the lines representing the given pair of linear equations intersect at a point.

2. 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
For the given pair of linear equations, a1 = 9, b1 = 3, c1 = 12, a2 = 18, b2 = 16 and c2 = 24.
Now, \(\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the lines representing the given pair of linear equations are coincident lines.

3. 6x – 3y + 10 = 0; 2x – y + 9 = 0
For the given pair of linear equations, a1 = 6, b1 = -3, c1 = 10, a2 = 2, b2 = -1 and c2 = 9.
Now, \(\frac{a_1}{a_2}=\frac{6}{2}=3, \quad \frac{b_1}{b_2}=\frac{-3}{-1}=3\)
and \(\frac{c_1}{c_2}=\frac{10}{9}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the lines representing the given pair of linear equations are parallel lines.

Question 3.
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:
1. 3x + 2y = 5; 2x – 3y = 7
2. 2x – 3y = 8; 4x – 6y = 9
3. \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7; 9x – 10y = 14
4. 5x – 3y = 11; -10x + 6y = -22
5. \(\frac{4}{3}\)x + 2y = 8; 2x + 3y = 12
Solution:
1. 3x + 2y = 5; 2x – 3y = 7
For the given pair of linear equations, a1 = 3, b1 = 2, c1 = -5, a2 = 2, b2 = -3 and c2 = -7.
Now, \(\frac{a_1}{a_2}=\frac{3}{2}, \frac{b_1}{b_2}=\frac{2}{-3}=-\frac{2}{3}\)
and \(\frac{c_1}{c_2}=\frac{-5}{-7}=\frac{5}{7}\)
Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Hence, the given pair of linear equations is consistent.

2. 2x – 3y = 8: 4x – 6y = 9
For the given pair of linear equations,
a1 = 2, b1 = -3, c1 = -8, a2 = 4, b2 =-6 and c2 = -9.
Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-8}{-9}=\frac{8}{9}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the given pair of linear equations is inconsistent.

3. \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7; 9x – 10y = 14
Multiplying the first equation by 6 and expressing both the equations in the standard form, we get following equations:
9x + 10y – 42 = 0; 9x – 10y – 14 = 0
For the given pair of linear equations, a1 = 9, b1 = 10, c1 = -42, a2 = 9, b2 = -10 and c2 = -14.
Now, \(\frac{a_1}{a_2}=\frac{9}{9}=1, \frac{b_1}{b_2}=\frac{10}{-10}=-1\)
and \(\frac{c_1}{c_2}=\frac{-42}{-14}=3\)
Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Hence, the given pair of linear equations is consistent.

4. 5x – 3y = 11; -10x + 6y = -22
For the given pair of linear equations, a1 = 5, b1 =-3, c1 = -11, a2 = -10, b2 = 6 and c2 = 22.
Now, \(\frac{a_1}{a_2}=\frac{5}{-10}=-\frac{1}{2}, \frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-11}{22}=-\frac{1}{2}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the given pair of linear equations is consistent and dependent.

5. \(\frac{4}{3}\)x + 2y = 8; 2x + 3y = 12
For the given pair of linear equations, a1 = \(\frac{4}{3}\), b1 = 2, c1 = -8, a2 = 2, b2 = 3 and c2 = -12.
Now, \(\frac{a_1}{a_2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}, \quad \frac{b_1}{b_2}=\frac{2}{3}\)
and \(\frac{c_1}{c_2}=\frac{-8}{-12}=\frac{2}{3}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the given pair of linear equations is consistent and dependent.

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

Question 4.
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
1. x + y = 5; 2x + 2y = 10
2. x – y = 8; 3x – 3y = 16
3. 2x + y – 6 = 0; 4x – 2y – 4 = 0
4. 2x – 2y – 2 = 0; 4x – 4y – 5 = 0
Solution:
1. x + y = 5; 2x + 2y = 10
For the given pair of linear equations,
a1 = 1, b1 = 1, c1 = -5, a2 = 2, b2 = 2 and c2 = -10.
Now, \(\frac{a_1}{a_2}=\frac{1}{2}, \quad \frac{b_1}{b_2}=\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-5}{-10}=\frac{1}{2}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the given pair of linear equations is consistent and dependent.
Now, we draw the graphs of both the equations.
x + y = 5 gives y = 5 – x.

x 0 5
y 5 0

2x + 2y = 10 gives y = \(\frac{10-2 x}{2}\)

x 1 3
y 4 2

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 3
Here, lines representing both the equations. are coincident. Hence, any point on the line gives a solution. In general, y = 5 – x, where x is any real number is a solution of the given pair of linear equations.

2. x – y = 8; 3x – 3y = 16
For the given pair of linear equations, a1 = 1, b1 = -1, c1 = -8, a2 = 3, b2 = -3 and c2 = -16.
Now, \(\frac{a_1}{a_2}=\frac{1}{3}, \quad \frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3}\)
and \(\frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the given pair of linear equations is inconsistent.

3. 2x + y – 6 = 0; 4x – 2y – 4 = 0
For the given pair of linear equations, a1 = 2, b1 = 1, c1 = -6, a2 = 4, b2 = -2 and c2 = -4.
Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \quad \frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-6}{-4}=\frac{3}{2}\)
Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Hence, the given pair of linear equations is consistent.
Now, we draw the graphs of both the equations.
2x + y – 6 = 0 gives y = 6 – 2x.

x 0 3
y 6 0

4x – 2y – 4 = 0 gives y = \(\frac{4 x-4}{2}\) = 2x – 2.

x 0 1
y -2 0

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 4
Here, the lines intersect at point (2, 2). Hence, x = 2 and y = 2 is the unique solution of the given pair of linear equations.

4. 2x – 2y – 2 = 0; 4x – 4y – 5 = 0
For the given pair of linear equations, a1 = 2, b1 = -2, c1 = -2, a2 = 4, b2 = -4 and c2 = -5.
Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \quad \frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-2}{-5}=\frac{2}{5}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the given pair of linear equations is inconsistent.

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

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.
Solution:
Let the length and breadth of the rectangular garden be x m and y m respectively.
Then, from the given data, x = y + 4 and 36 = \(\frac{1}{2}\)[2(x + y)] i.e., x + y = 36 as the perimeter of a rectangle = 2 (length + breadth).
To draw the graphs, we find two solutions of each equation.
x = y + 4 gives y = x – 4

x 8 24
y 4 20

x + y = 36 gives y = 36 – x

x 12 24
y 24 12

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 5
Here, the lines intersect at point (20, 16). Hence, x = 20 and y = 16 is the unique solution of the pair of linear equations.
Thus, for the rectangular garden, length = 20 m and breadth = 16 m.

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:
1. intersecting lines
2. parallel lines
3. coincident lines
Solution:
1. For intersecting lines, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\). The given equation is 2x + 3y – 8 = 0. We can give another equation as 3x + 4y – 24 = 0. Here, \(\frac{a_1}{a_2}=\frac{2}{3}\) and \(\frac{b_1}{b_2}=\frac{3}{4}\) satisfying \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

2. For parallel lines, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\). The given equation is 2x + 3y – 8 = 0. We can give another equation as 6x + 9y – 10 = 0.
Here, \(\frac{a_1}{a_2}=\frac{1}{3}, \quad \frac{b_1}{b_2}=\frac{1}{3}\) and \(\frac{c_1}{c_2}=\frac{4}{5}\) satisfying \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).

3. For coincident lines, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). The given equation is 2x + 3y – 8 = 0.
We can give another equation as 10x + 15y – 40 = 0. Here, \(\frac{a_1}{a_2}=\frac{1}{5}\), \(\frac{b_1}{b_2}=\frac{1}{5}\) and \(\frac{c_1}{c_2}=\frac{1}{5}\) satisfying \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

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:
x – y + 1 = 0 gives y = x + 1

x -1 2
y 0 3

3x + 2y – 12 = 0 gives y = \(\frac{12-3 x}{2}\)

x 0 4
y 6 0

JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 6
The vertices of the triangle formed by the given lines and the x-axis are (-1, 0), (4, 0) and (2, 3).