In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.3 Pdf, These solutions are solved subject experts from the latest edition books.

## MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Question 1.

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

(i) x + y = 14; x – y = 4

(ii) s – f = 3; s3+t2 = 6

(iii) 3x – y = 3; 9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3

(v) 2–√x+3–√y = 0; 3–√x−8–√y = 0

(vi) 3x/2−5y/3 = -2, x/3+y/2=13/6

**Solutions:**

(i) x+y=14⇒y=14−xSubstituting this value in the second equation, we getx−(14−x)=42x=18⇒x=9Substituting this value of x in the first equation, we get9+y=14⇒y=5

(ii) s−t=3⇒s=t+3Substituting in 2nd equation3t+3+2t=6

62t+6+3t=6

5t+6=36⇒t=6Substituting value of t in 1st equations=t+3=9

(iii) ∵a2a1=b2b1=c2c1Hence all the points lying on the line y=3x−3 like x=2,y=3; are a solution.

(iv) 0.2x+0.3y=1.3⇒x=0.21.3−0.3ySubstituting this value in the second equation0.4×0.21.3−0.3y+0.5y=2.32.6−0.6y+0.5y=2.32.6−2.3=0.6y−0.5y0.1y=0.3⇒y=3Substituting this value of y,x=0.21.3−0.3y=0.21.3−0.3×3=0.20.4x=2

(v) 2x+3y=0⇒y=−32xSubstituting this in 2nd equation 3x−8×(−32x)=03x+4x=0⇒x=0Substituting value of x,y=−32×0=0

(vi) 23x−35y=−235y=23x+2⇒y=109x+12

Substituting this in 2nd equation3x+209x+12=613

6047x+36=613

47x+36=13047x=94⇒x=2Substituting this value of x in 1st equationy=109x+12=109×2+12=3

10th maths exercise 3.3 Solution:

MP Board Class 10th Maths exercise 3.3 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 13800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

(v) A fraction becomes 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 5/6. Find the fraction.

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

Solution:

(i) Let the two numbers be X and y such that x > y

It is given that

Difference between two numbers = 26

∴ x – y = 26 … (1)

Also one number = 3 [the other number]

⇒ x = 3y … (2)

Substituting x = 3y in (1) , we get 3y – y = 26 ⇒ 2y = 26

Now, substituting y = 13 in (2) , we have

x = 3(13) ⇒ x = 39

Thus, two numbers are 39 and 13.

(ii) Let the two angles be x and y such that x > y

∵ The larger angle exceeds the smaller by 18° (Given)

∴ x = y + 18°…. (1)

Also, sum of two supplementary angles = 180°

∴ x + y = 180° … (2)

Substituting the value of x from (1) in (2) , we get,

(18° + y) + y = 180°

Substituting, y = 81° in (1) , we get

x = 18° + 81° = 99°

Thus, x = 99° and y = 81°

(iii) Let the cost of a bat = ₹ x

And the cost of a ball = ₹ y

∵ [cost of 7 bats] + [cost of 6 balls] = ₹ 3800

⇒ 7x + 6y = 3800 … (1)

Also, [cost of 3 bats] + [cost of 5 balls] = ₹ 1750

3x + 5y = 1750 …. (2)

Substituting this value of y in (1) , we have

(iv) Let fixed charges = ₹ x

and charges per km = ₹ y

∵ Charges for the journey of 10 km = ₹ 105 (Given)

∴ x + 10y = 105 … (1)

and charges for the journey of 15 km = ₹ 155

∴ x + 15y = 155 … (2)

From (1) , we have, x = 105 – 10y …. (3)

Putting the value of x in (2) , we get

(105 – 10y) + 15y = 155

⇒ 5y = 155 – 105 = 50 ⇒ y = 10

Substituting y = 10 in (3) , we get

x = 105 – 10(10) ⇒ x = 105 – 100 = 5

Thus, x = 5 and y = 10

⇒ Fixed charges = ₹ 5

and charges per km = ₹ 10

Now, charges for 25 km = x + 25y = 5 + 25(10) = 5 + 250 = ₹ 255

∴ The charges for 25 km journey = ₹ 255

MP Board 10th maths exercise 3.3 questions solutions

(v) Let the numerator = x

and the denominator = y

∴ Fraction = xy

According to the given condition,

⇒ 11(x + 2) = 9(y + 2)

⇒ 11x + 22 = 9y + 18

⇒ 11x – 9y + 4 = 0 … (1)

Substituting this value of x in (1) , we have

(vi) Let the present age of Jacob = x years

and the present age of his son = y years

∴ 5 years hence: Age of Jacob = (x + 5) years

Age of his son = (y + 5) years

According to given condition,

[Age of Jacob] = 3[Age of his son]

x + 5 = 3(y + 5) ⇒ x + 5 = 3y + 15

⇒ x – 3y -10 = 0 … (1)

5years ago : Age of Jacob = (x – 5) years,

Age of his son = (y – 5) years

According to given condition,

[Age of Jacob] = 7[Age of his son]

∴ (x – 5) = 7(y – 5) ⇒ x – 5 = 7y – 35

⇒ x – 7y + 30 = 0 … (2)

From (1) , x = [10 + 3y] … (3)

Substituting this value of x in (2) , we get

(10 + 3y) – 7y + 30 = 0

⇒ -4y = -40 ⇒ y = 10

Now, substituting y = 10 in (3) ,

we get x = 10 + 3(10)

⇒ x = 10 + 30 = 40

Thus, x = 40 and y = 10

⇒ Present age of Jacob = 40 years and present age of his son = 10 years

**More MP Board 10th Maths Solutions:**

- MP Board Class 10th Maths Solutions Chapter 5 Exercise 5.3 Arithmetic Progressions
- MP Board Class 10th Maths Solutions Chapter 5 Exercise 5.2 Arithmetic Progressions
- MP Board Class 10th Maths Solutions Chapter 5 Exercise 5.1 Arithmetic Progressions
- MP Board Class 10th Maths Solutions Chapter 4 Exercise 4.4 Quadratic Equations
- MP Board Class 10th Maths Solutions Chapter 4 Exercise 4.3 Quadratic Equations