# Maharashtra Board Class 9 Math Chapter 5 Linear Equations in Two Variables Solution

## Maharashtra Board Class 9 Math Solution Chapter 5 – Linear Equations in Two Variables

Balbharati Maharashtra Board Class 9 Math Solution Chapter 5: Linear Equations in Two Variables. Marathi or English Medium Students of Class 9 get here Linear Equations in Two Variables full Exercise Solution.

 Std Maharashtra Class 9 Subject Math Solution Chapter Linear Equations in Two Variables

### Practice Set 5.1

(1) By using variables x and y form any five linear equations in two variables

Solution:

x + y =6, 3x + 4y = 11, 11x – 16y = 27, 6x + 21y = 47, 2x – 3y = 6.

(2) Write five solutions of the equation x + y = 7.

Solution:

Fine solutions of x + 7 = 7 are:

(0, 7), (1, 6), (2, 5), (3, 4), (4, 3)

(3) Solve the following sets of simultaneous equations.

(i) x + y = 4, 2x – 5y = 1

Ans:

x + y = 4

=) x = 4 – y

∴ 2x – 5y = 1

=) 2 (4 – y) – 5y = 1

=) 8 – 2y – 5y = 1

=) 7y = 7

=) y = 1

∴ x = 4 – 1 = 3

(x, y) = (3, 1)

(ii) 2x + y = 5; 3x – y = 5

Ans:

2x + y = 5 —– (i)

3x – y = 5 —— (ii)

(i) + (ii) = 5x = 10

= x = 2

(i) = 2 × 2 + y = 5

=) y = 1

(x, y) = (2, 1)

(iii) 3x – 5y = 16, x – 3y = 8

Ans:

x – 3y = 8

=) x = 8 + 3y

∴ 3x – 5y = 16

=) 3 (8 + 3y) – 5y = 16

=) 24 + 9y – 5y = 16

= 4y = – 8

=) y = – 2

∴ x = 8 + 3 (-2)

= 8 – 6 = 2

(x, y) = (2, – 2)

(iv) 2y – x = 0, 10x + 15y = 105

Ans:

2y – x = 0

=) x = 2y

∴ 10x + 15y = 105

=) 10 × 2y + 15y = 105

= 35y = 105

=) y = 3

∴ x = 2×3 = 6

∴ (x, y) = (6, 3)

(v) 2x + 3y + 4 = 0, x – 5y = 11

Ans:

x – 5y = 11

=) x = 11 + 5y

∴ 2x + 3y + 4 = 0

=) 2 (11 + 5y) + 3y + 4 = 0

=) 22 + 10y + 3y + 4 = 0

=) 13y + 26 = 0

=) y = -2

∴ x = 11 + 5 (-2) = 1

(x, y) = (1, -2)

(vi) 2x – 7y = 7; 3x + y = 22

Ans:

3x + y = 22

= y = 22 – 3x

∴ 2x – 7y = 7

= 2x – 7 (22 – 3x) = 7

= 23x – 154 = 7

= x = 161/23 = 7

∴ y = 22 – 3 × 7 = 1

(x, y) = (7, 1)

#### Page 89:

(4) The population of a certain town was 50,000. In a year, male population was increased by 5% and female population was increased by 3%. Now the population became 52020. Then what was the number of males and females in the previous year?

Solution:

Let, x be number of male & y be number of females in previous year.

No. x + y = 50,000 —- (1)

And

(1 + 5/100) x + (1 + 3/100) y = 52020

= 105x + 103y = 5202000 —– (ii)

Multiplying (i) by 103,

103x + 103y = 5150000 —– (iii)

(iii) – (ii) = -2x = -52000

= x = 26000

∴ (i) = y = 50000 – 26000

= 24000

∴ Number of males = 26000

Number of females = 24000

#### Page 90:

Activity (I): Ans:

Sum of length & breadth is 36.

x + y = 36 —— (i)

My breadth is 5/7 times of the length

y = 5/7 x = 5x – 7y = 0 —– (ii)

If the breadth is subtracted from twice the length, the answer is 27.

2x – y = 27 —- (iii)

The difference between my length and breadth is 6 units

x – y = 6 —– (iv)

(a) Taking pairs (i) and (ii)

7 × (i) + (ii) = 7x + 5x = 252

=) x = 21

y = 5/7 × 21 = 15

∴ (x, y) = (21, 15).

Similarly taking (i) & (iii), (i) & (iii), (ii) & (iii), (ii) & (iv), (iii) ——

We get x = 21 & y = 15

Solution of each pair is (21, 15)

6 pairs can be formed from the above equation.

We can conclude that length is 21 units and breadth is 15 units.

### Practice Set 5.2

(1) In an envelope there are some 5 rupee notes and some 10 rupee notes. Total amount of these notes together is 350 rupees. Number of 5 rupee notes are less by 10 than number of 10 rupee notes. Then find the number of 5 rupee and 10 rupee notes.

Solution:

Let, x & y be number of 5 & 10 rupees notes.

5x + 10y = 350 —- (i)

(i) => 5y + 50 + 10y = 350

=> 15y = 300

=> y = 20

(ii) => x = 20 + 10 = 30

∴ Number of 50 rupees note = 30

Number of 10 rupees note = 20

(2) The denominator of a fraction is 1 less than twice its numerator. If 1 is added to numerator and denominator respectively, the ratio of numerator to denominator is 3:5. Find the fraction.

Ans:

Let, x be numerator & y be denominator.

y = 2x – 1 => 2x – y = 1 —– (i)

x+1/y+1 = 3/5

= 5x + 5 = 3y + 3

= 5x – 3y = -2 = 3y – 5x = 2 —- (ii)

(i) × 3 + (ii) = x = 5

(i) => y = 2 × 5 – 11

=> 5x + 5 = 3y + 3

=> 5x – 3y = – 2

=> 3y – 5x = 2 —- (ii)

(i) × 3 + (ii) => x = 5

(i) => y = 2 × 5 – 1 = 9

∴ The fraction is 5/9

(3) The sum of ages of Priyanka and Deepika is 34 years. Priyanka is elder to Deepika by 6 years. Then find their today’s ages.

Ans:

Let, x & y be Priyanka & Deepika age respectively.

According to Question, x + y = 34. —— (i)

x = y + 6 => x – y = 6 —– (ii)

(i) + (ii) => 2x = 40 => x = 20

(i) = y = 34 – 20 = 14

∴ Priyanka is 20 years old & Deepika is 14 years old.

(4) The total number of lions and peacocks in a certain zoo is 50. The total number of their legs is140. Then find the number of lions and peacocks in the zoo.

Ans:

Let, x & y be number of lions & peacocks respectively.

x + y = 50 —— (i)

4x + 2y = 140

=> 2x + y = 70 —- (ii)

(ii) – (i) = x = 20

(i) => y = 50 – 20 = 30

∴ Number of lines = 20

Number of peacocks = 30

(5) Sanjay gets fixed monthly income. Every year there is a certain increment in his salary. After 4 years, his monthly salary was Rs. 4500 and after 10 years his monthly salary became 5400 rupees, then find his original salary and yearly increment.

Ans:

Let, x and y be the original salary & yearly increment respectively.

x + 4y = 4500 —– (i)

x + 10y = 5400 —– (ii)

(ii) – (i) => 6y = 900

=> y = 150

∴ (ii) => x = 5400 – 10y

= 5400 – 1500

= 3900

∴ His original salary is ₹3900 and yearly increment is ₹150.

(6) The price of 3 chairs and 2 tables is 4500 rupees and price of 5 chairs and 3 tables is 7000 rupees, then find the total price of 2 chairs and 2 tables.

Ans:

Let, price of a chair and table be x and y respectively

3 + 2y = 4500 —– (i)

5x + 3y = 7000 —- (ii)

2 × (ii) – 3 × (i) => 10x – 9x = 14000 – 13500

=> x = 500

(i) => 3 × 500 + 2y = 4500

=> 2y = 4500 – 1500 = 3000

=> y = 1500

∴ Price of 2 chair = 2 × 500 = ₹1000

Price of 2 tables = 2 × 1500 = ₹3000

∴ Total prices = ₹(3000 + 1000) = ₹4000

(7) The sum of the digits in a two-digits number is 9. The number obtained by interchanging the digits exceeds the original number by 27. Find the two-digit number.

Ans:

Let, x and y be the tens and units place digits of the two-digit member.

x + y = 9 —- (i)

10y + x = 10x + y + 27

=) 9 (x – y) = -27

=) y – x = 3 —– (ii)

(i) + (ii) => 2y = 12 => y = 6

(ii) => x = y – 3 = 6 – 3 = 3

∴ The two-digit number is 36.

(8) In ABC, the measure of angle A is equal to the sum of the measures of B and C. Also the ratio of measures of B and C is 4:5. Then find the measures of angles of the triangle.

Ans:

According to Question, ∠A = ∠B + ∠C —– (i)

∠B/∠C = 4/5 => 5∠B – 4∠C = 0 —- (ii)

We know, sum of ∠’s of a △ = 180°

= ∠A + ∠B + ∠C = 180°

From (i), ∠A + ∠A = 180° [∵ ∠A = ∠B + ∠C

=> ∠A = 90°

∴ ∠B + ∠C = 90°

=> 4/5 ∠C + ∠C = 90° [By (ii)]

=> 9/5 ∠C = 90° = ∠C = 50°

(ii) => ∠B = 4/5 ∠C = 4/5 × 50

= 40°

∴ ∠A = 90°, ∠B = 40°, ∠C = 50°

(9) Divide a rope of length 560 cm into 2 parts such that twice the length of the smaller part is equal to 1/3 of the larger part. Then find the length of the larger part.

Ans:

Let, x be length of larger part & y be length of smaller part.

x + y = 560 —- (i)

2y = 1/3 x

=> y = 1/6 x

∴ Substituting value of y in (i),

x + 1/6 x = 560

=> 7x = 3360

=> x = 480

∴ Length of larger part is 480 cm

(10) In a competitive examination, there were 60 questions. The correct answer would carry 2 marks, and for incorrect answer 1 mark would be subtracted. Yashwant had attempted all the questions and he got total 90 marks. Then how many questions he got wrong?

Ans:

Let, x and y be the number of right and wrong questions. Yahswant got respectively.

According to Question, x + y = 60 —— (i)

2x + (-1) y = 90

=> 2x – y = 90 —- (ii)

(i) + (ii) = 3x = 150

=> x = 50

(i) => y = 60 – x = 60 – 50 = 10

∴ Yashwant got 10 questions wrong.

### Problem set – 5

(1) Choose the correct alternative answers for the following questions.

(i) If 3x + 5y = 9 and 5x + 3y = 7 then What is the value of x + y?

(A) 2

(B) 16

(C) 9

(D) 7

Ans: (A) 2.

(ii) ‘When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes 26.’ What is the mathematical form of the statement?

(A) x – y = 8

(B) x + y = 8

(C) x + y = 23

(D) 2x + y = 21

Ans:

According to question,

2 (l – 5 + b – 5) = 26

= 2 (l + b) – 20 = 26

= l + b = 46/2 = 23

(c) x + y = 23

(iii) Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay’s age?

(A) 20

(B) 15

(C) 10

(D) 5

Ans:

According to question,

x = y – 5 => y – x = 5 —– (i)

x + y = 25 —– (ii)

(ii) – (i) => 2x = 30 => x = 15

(B) 15.

(2) Solve the following simultaneous equations.

(i) 2x + y = 5 ; 3x – y = 5

Ans:

2x + y = 5 —— (i) & 3x – y = 5 —— (ii)

(i) + (ii) = 5x = 10 = x = 2

(i) => y = 5 – 2x = 5 – 4 = 1

∴ (x, y) = (2, 1).

(ii) x – 2y = -1 ; 2x- y = 7

Ans:

x – 2y = -1 —– (i), 2x – y = 7 —– (ii)

2 × (i) – (ii) = -3y = -9

= y = 3

(i) => x – 2y – 1 = 6 – 1 = 5

(x, y) = (5, 3)

(iii) x + y = 11 ; 2x – 3y = 7

Ans:

x + y = 11 —— (i), 2x – 3y = 7 —– (ii)

3 × (i) + (ii) = 5x = 40

= x = 8

(i) => y = 11 – 8 = 3

(x, y) = (8, 3)

(iv) 2x + y = -2 ; 3x – y = 7

Ans:

2x + y = – 2 —– (i), 3x – y = 7 —- (ii)

(i) + (ii) => 5x = 5 => x = 1

(i) => y = -2 – 2 = -4

∴ (x, y) = (1, -4)

(v) 2x – y = 5 ; 3x + 2y = 11

Ans:

2x – y = 5 —- (i), 3x + 2y = 11 —– (ii)

2 × (i) + (ii) => 7x = 21 => x = 3

(i) => y = 2x – 5 = 1

(x, y) = (3, 1)

(vi) x – 2y = -2, x + 2y = 10

Ans:

x – 2y = -2 —- (i), x + 2y = 10 —– (ii)

(i) + (ii) => 2x = 8 => x = 4

(i) => 2y = x + 2 => y = 6/4 = 3

(x, y) = (4, 3)

(3) By equating coefficients of variables, solve the following equations.

(i) 3x – 4y = 7; 5x + 2y = 3

Ans:

3x – 4y = 7 —— (i), 5x + 2y = 3 —– (ii)

(i) + 2 × (ii) => 13x = 13 => x = 1

(i) => 4y = 7 – 3x => y = 7-3/4 = 1

(x, y) = (1, 1)

(ii) 5x + 7y=17; 3x – 2y = 4

Ans:

5x + 7y = 17 —– (i), 3x – 2y = 4 —– (ii)

2 × (i) + 7 × (ii) => 31x = 34 + 28

=> x = 2

(ii) 2y = 3x – 4 => y = 6-4/2 = 1

(x, y) = (2, 1)

(iii) x – 2y = -10; 3x – 5y = -12

Ans:

x – 2y = -10 —– (i), 3x – 5y = -12 —– (ii)

3 × (i) – (ii) => -y = -18 => y = 18

(i) => x = 2y – 10 = 36 – 10 = 26

(x, y) = (26, 18)

(iv) 4x + y = 34; x + 4y = 16

Ans:

4x + y = 34 —– (i), x + 4y = 16 —– (ii)

(i) – 4 × (ii) = -15y = -30 => y = 2

(ii) => x = 16 – 4 × 2 = 8

(x, y) = (8, 2)

(4) Solve the following simultaneous equations.

(i) x/y + y/4 = 4; x/2 – y/4 = 1

Ans:

x/3 + y/4 = 4 = 4x + 3y = 48 —– (i)

x/2 – y/4 = 1 = 2x – y = 4 —– (ii)

(i) + 3 × (ii) => 10x = 60 => x = 6

(ii) => 2x – y = 4 = y = 2x + 4

= 16

∴ (x, y) = (6, 16)

(ii) x/3 + 5y = 13; 2x + y/2 = 19

Ans:

x/3 + 5y = 13 => x + 15y = 39 —– (i)

2x + y/2 = 19 => 4x + y = 38 —- (ii)

(i) – 15 × (ii) => -59x = -531

= x = 9

∴ (ii) => y = 38 – 4 × 9 = 38 – 36

= 2

∴ (x, y) = (9, 2)

(iii) 2/x + 3/y = 13; 5/x – 4/y = -2

Ans:

2/x + 3/y = 13 —– (i)

5/x – 4/y = -2 —– (ii)

4 × (i) + 3 × (ii) = 8/x + 15/x = 46

=> x = 23/46 = ½

∴ (i) = 2/(1/2) + 3/y = 13

=> 4 + 3/y = 13 => 3/y = 9 => y = 3/9 = 1/3

∴ (x, y) = (1/2, 1/3)

(5) A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to this number, the sum is equal to the number obtained by interchanging the digits. Find the number.

Ans:

Let, x & y be the tens and units digit of the number.

According to Question, 10x + y = 4(x + y) + 3

= 6x – 3y = 3

= 2x – y = 1

And 10x + y + 18 = 10y + x

=> 9x – 9y + 18 = 0

=> y – x = 2 —- (ii)

(i) + (ii) = x = 2

(ii) => y = 2 + x = 4

∴ The digit is 24.

(6) The total cost of 8 books and 5 pens is 420 rupees and the total cost of 5 books and 8 pens is 321 rupees. Find the cost of 1 book and 2 pens.

Ans:

Let, cost of 1 book & 1 pen be x and y respectively

According to Question,

8x + 5y = 420 —- (i)

5x + 8y = 321 —– (ii)

5 × (i) – 8 × (ii) = 25y – 64y = 2100 – 2568

=> 39y = 468

=> y = 12

(ii) 5x = 321 – 8 × 12 = 321 – 96

=> x = 45

∴ cost of 1 book = ₹45

Cost of 2 book = ₹24

(7) The ratio of incomes of two persons is 9:7. The ratio of their expenses is 4:3. Every person saves rupees 200, find the income of each.

Ans:

Let, x and y be the incomes of the two persons respectively

According to Question,

x/y = 9/7 => 7x – 9y = 0 —— (i)

Savings of two persons is ₹200

∴ x-200/y-200 = 4/3 => 3(x – 200) = 4 (y – 200)

3x – 4y = -200

=> 4x – 3x = 200 —- (ii)

(i) × 3 + 7 × (ii) => -27y + 28y = 1400

=> y = 1400

(i) => 7x = 9 × 1400

=> x = 1800

∴ The income of two persons is 1800 and 1400.

(8) If the length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 8 square units. If length is reduced by 3 units and breadth is increased by 2 units, then the area of rectangle will increase by 67 square units. Then find the length and breadth of the rectangle.

Ans:

Let, l and b be the length and breadth of the rectangle

According to Question,

(l – 5) (b + 3) = lb – 8

=> lb – 5b + 3l – 15 = lb – 8

= 3l – 5b = 7 —– (i)

(l – 3) (b + 2) = lb + 67

= lb – 3b + 2l – 6 = lb + 67

= 2l – 3b = 61 —– (ii)

2 × (i) – 3 × (ii) = – 10b + 9b = 14 – 183

= b = 169

(ii) = 2l = 61 + 3 × 169

= l = 284

∴ Length = 284 units and breadth = 169 units

(9) The distance between two places A and B on road is 70 kilometers. A car starts from A and the other from B. If they travel in the same direction, they will meet after 7 hours. If they travel towards each other they will meet after 1 hour, then find their speeds.

Ans:

Let, x and y be the speeds of the car at A & B respectively

According to Question, x – y = 70/7

=> x – y = 10 —– (i)

And x + y = 70/1 = x + y = 70 —– (ii)

(i) + (ii) = 2x = 80

= x = 40

(ii) => y = 70 – x = 70 – 40 = 30

∴ The speed of the cars are 40km/h & 30km/h.

(10) The sum of a two digit number and the number obtained by interchanging its digits is 99. Find the number.

Ans:

Let, x and y be the tens and units place digit of the number

According to Question, 10x + y + 10y + x = 99

=> 11x + 11y = 99

=> x + y = 9

∴ The possible numbers are: 18, 27, 36, 45, 54, 63, 72, 81

#### Page 92:

Activity:

Ans:

Given, 11x – 6y + 18 = 0 —- (i)

x + y + 8 = 0 —– (ii)

(i) – 11 × (ii) => 5y – 70 = 0 => y = 14

(ii) => x = y – 8 = 14 – 8 = 6

∴ Given fraction = 6/14 = ¾

Verification: (i) => 11x – 6y + 18 = 11 × 6 – 6 × 14 + 18

= 66 + 18 – 84 = 0

(ii) => x – y + 8 = 6 – 14 + 8 = 14 – 4 = 0

∴ The values satisfy the given equations.

#### Page 97:

Activity I:

Amita invested some part of 35000 rupees at 4% and the rest at 5% interest for one year. Altogether her gain was Rs. 1530. Find out the amounts she had invested at the two different rates. Write your answer in words.

Ans:

Let, x and y be the investments.

According to Question,

x + y = 35000 —– (i)

4/100 x + 5/100 y = 1530

=> 4x + 5y = 153000 —– (ii)

4 × (i) – (ii) => -y = -13000

=> y = 13000

∴ (i) = x = 35000 – 13000

= 22000

Updated: September 17, 2021 — 6:34 pm