# 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.

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.

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

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 $$\frac{7+3}{9+3}=\frac{10}{12}=\frac{5}{6}$$

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.