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