NCERT
10th Mathematics
Exercise 3.6
Topic: 3 Pair of Linear Equations in Two Variables
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EXERCISE 3.6
Q1. Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) 1/(2x) + 1/(3y) = 2; 1/(3x) + 1/(2y) = 13/6,
(ii) (2/√x) + (3/√y) = 2; (4/√x) - (9/√y) = - 1,
(iii) 4/x + 3y = 14; 3/x - 4y = 23,
(iv) 5/(x - 1) + 1/(y - 2) = 2; 6/(x - 1) - 3/(y - 2) = 1,
(v) (7x - 2y)/xy = 5; (8x + 7y)/xy = 15,(vi) 6x + 3y = 6xy; 2x + 4y = 5xy,(vii) 10/(x + y) + 2/(x - y) = 4; 15/(x + y) - 5/(x - y) = -2(viii) 1/(3x + y) + 1/(3x - y) = 3/4; 1/[2(3x + y)] - 1/[2(3x - y)] = -1/8.
Explanation:
1) Here, let x and y be two variables.
2) Convert the variables suitably to get linear equations in two variables.
3) Then solve these equations to get the values of our variables.
Solution:
(i) 1/(2x) + 1/(3y) = 2; 1/(3x) + 1/(2y) = 13/6.
1) Given equations are
1/(2x) + 1/(3y) = 2; 1/(3x) + 1/(2y) = 13/6
2) In the equation, 1/(2x) + 1/(3y) = 2, put 1/x = p, and 1/y = q, so we have,
1/(2x) + 1/(3y) = 2
p/2 + q/3 = 2
(3p + 2q)/6 = 2
3p + 2q = 2(6)
3p + 2q = 12 ---------------equation 1
3) In the equation, 1/(3x) + 1/(2y) = 13/6, put 1/x = p, and 1/y = q, so we have,
1/(3x) + 1/(2y) = 13/6
p/3 + q/2 = 13/6
(2p + 3q)/6 = 13/6
2p + 3q = 13
2p = 13 - 3q
p = (13 - 3q)/2 ---------------equation 2
4) Substitute the value of p = (13 - 3q)/2 from equation 2 in equation 1, we get
3p + 2q = 12
3(13 - 3q)/2 + 2q = 12
(39 - 9q)/2 + 2q = 12
(39 - 9q + 4q)/2 = 12
(39 - 5q) = 12 (2)
(39 - 5q) = 24
5q = 39 - 24
q = 15/5
q = 3 ------------ equation 3
5) Put the value of q = 3 from equation 3 in equation 2, and we get
p = (13 - 3q)/2p = (13 - 3(3))/2
p = (13 - 9)/2
p = (4)/2
p = 2 ------------ equation 4
6) The value of p = 2 and the value of q = 3.
7) As, p = 1/x and p = 2, we have x = 1/2
8) As, q = 1/y and q = 3, we have y = 1/3.
(ii) (2/√x) + (3/√y) = 2; (4/√x) - (9/√y) = - 1
1) Given equations are
(2/√x) + (3/√y) = 2; (4/√x) - (9/√y) = - 1
2) In the equation, (2/√x) + (3/√y) = 2, put 1/√x = p, and 1/√y = q, so we have,
(2/√x) + (3/√y) = 2
2p + 3q = 2 ---------------equation 1
3) In the equation, (4/√x) - (9/√y) = - 1, put 1/√x = p, and 1/√y = q, so we have,
(4/√x) - (9/√y) = - 1
4p - 9q = - 1
4p = 9q - 1
p = (9q - 1)/4 ---------------equation 2
4) Substitute the value of p = (9q - 1)/4 from equation 2 in equation 1, we get,
2p + 3q = 2
2(9q - 1)/4 + 3q = 2
(9q - 1)/2 + 3q = 2
(9q - 1 + 6q)/2 = 2
15q = 4 + 1
q = 5/15
q = 1/3 ------------ equation 3
5) Put the value of q = 1/3 from equation 3 in equation 2, and we get
p = (9q - 1)/4p = (9(1/3) - 1)/4p = (3 - 1)/4
p = (2)/4
p = 1/2 ------------ equation 4
6) The value of p = 1/2 and the value of q = 1/3.
7) As, p = 1/√x and p = 1/2, we have √x = 2, so x = 4.
8) As, q = 1/√y and q = 1/3, we have √y = 3, so y = 9.
(iii) 4/x + 3y = 14; 3/x - 4y = 23
1) Given equations are
4/x + 3y = 14; 3/x - 4y = 23
2) In the equation, 4/x + 3y = 14, put 1/x = p, so we have,
4/x + 3y = 14
4p + 3y = 14 ---------------equation 1
3) In the equation, 1/(3x) + 1/(2y) = 13/6, put 1/x = p, and 1/y = q, so we have,
3/x - 4y = 23
3p - 4y = 23
3p = 23 + 4y
p = (23 + 4y)/3 ---------------equation 2
4) Substitute the value of p = (23 + 4y)/3 from equation 2 in equation 1, we get,
4p + 3y = 14
(4(23 + 4y)/3) + 3y = 14
(92 + 16y)/3 + 3y = 14
(92 + 16y + 9y)/3 = 14
(92 + 25y)/3 = 14
(92 + 25y) = 14 (3)
(92 + 25y) = 42
(92 + 25y) = 42
25y = 42 - 92
25y = - 50
y = - 50/25
y = - 2 ------------ equation 3
5) Put the value of y = - 2 from equation 3 in equation 2, and we get
p = (23 + 4y)/3
p = (23 + 4(-2))/3
p = (23 - 8)/36) The value of p = 5 and the value of y = - 2.
p = (15)/3
p = 5 ------------ equation 4
7) As, p = 1/x and p = 5, we have x = 1/5 and y = - 2.
(iv) 5/(x - 1) + 1/(y - 2) = 2; 6/(x - 1) - 3/(y - 2) = 1
1) Given equations are
5/(x - 1) + 1/(y - 2) = 2; 6/(x - 1) - 3/(y - 2) = 1
2) In the equation, 5/(x - 1) + 1/(y - 2) = 2, put 1/(x - 1) = p, and 1/(y - 2) = q,
so we have,
5/(x - 1) + 1/(y - 2) = 2
5p + q = 2
q = 2 - 5p ---------------equation 1
3) In the equation, 6/(x - 1) - 3/(y - 2) = 1, put 1/(x - 1) = p, and 1/(y - 2) = q,
so we have,
6/(x - 1) - 3/(y - 2) = 1
6p - 3q = 1 ---------------equation 2
4) Substitute the value of q = (2 - 5p) from equation 1 in equation 2, we get,
6p - 3q = 1
6p - 3(2 - 5p) = 1
6p - 6 + 15p = 1
6p + 15p = 1 + 6
21p = 7
p = 7/21
p = 1/3 ---------------equation 3
5) Put the value of p = 1/3 from equation 3 in equation 1, and we get
q = 2 - 5p
q = 2 - 5(1/3)
q = (6 - 5)/3
q = 1/3 ------------ equation 4
6) The value of p = 1/3 and the value of q = 1/3.
7) As, p = 1/(x - 1) and p = 1/3, we have (x - 1) = 3, so x = 4.
8) As, q = 1/(y - 2) and q = 1/3, we have (y - 2) = 3, so y = 5.
(v) (7x - 2y)/xy = 5; (8x + 7y)/xy = 15
1) First we will simplify the first equation (7x - 2y)/xy = 5:
(7x - 2y)/xy = 5
(7x)/xy - (2y)/xy = 5
7/y - 2/x = 5
- 2/x + 7/y = 5, here, put 1/x = p and 1/y = q, we get,
- 2p + 7q = 5
2p - 7q = - 5
2p = 7q - 5
p = (7q - 5)/2 ---------------equation 1
2) First we will simplify the second equation (8x + 7y)/xy = 15:
(8x + 7y)/xy = 15
(8x)/xy + (7y)/xy = 15
8/y + 7/x = 15
7/x + 8/y = 15, here, put 1/x = p and 1/y = q, we get,
7p + 8q = 15 ---------------equation 2
3) Substitute the value of p = (7q - 5)/2 from equation 1 in equation 2, we get,
7p + 8q = 15
7(7q - 5)/2 + 8q = 15
(49q - 35)/2 + 8q = 15
(49q - 35 + 16q)/2 = 15
(49q + 16q - 35) = 15 (2)
(65q - 35) = 30
65q = 35 + 30
65q = 65
q = 1 ---------------equation 3
4) Put the value of q = 1 from equation 3 in equation 1, and we get
p = (7q - 5)/2
p = (7(1) - 5)/2
p = (7 - 5)/2
p = 2/2
p = 1 ------------ equation 4
5) The value of p = 1 and the value of q = 1.
6) As, p = 1/x and p = 1, we have x = 1.
7) As, q = 1/y and q = 1, we have y = 1.
(vi) 6x + 3y = 6xy; 2x + 4y = 5xy
1) First we will simplify the first equation 6x + 3y = 6xy:
6x + 3y = 6xy
(6x)/xy + (3y)/xy = 6
6/y + 3/x = 6
3/x + 6/y = 6, here, put 1/x = p and 1/y = q, we get,
3p + 6q = 6
p + 2q = 2
p = 2 - 2q ---------------equation 1
2) First we will simplify the second equation 2x + 4y = 5xy:
2x + 4y = 5xy
(2x)/xy + (4y)/xy = 5
2/y + 4/x = 5
4/x + 2/y = 5, here, put 1/x = p and 1/y = q, we get,
4p + 2q = 5 ---------------equation 2
3) Substitute the value of p = 2 - 2q from equation 1 in equation 2, we get,
4p + 2q = 5
4(2 - 2q) + 2q = 5
8 - 8q + 2q = 5
8 - 6q = 5
- 6q = 5 - 8
- 6q = - 3
q = (- 3)/(- 6)
q = 1/2 ---------------equation 3
4) Put the value of q = 1/2 from equation 3 in equation 1, and we get
p = 2 - 2q
p = 2 - 2(1/2)
p = 2 - 1
p = 1 ------------ equation 4
5) The value of p = 1 and the value of q = 1/2.
6) As, p = 1/x and p = 1, we have x = 1.
7) As, q = 1/y and q = 1/2, we have y = 2.
(vii) 10/(x + y) + 2/(x - y) = 4; 15/(x + y) - 5/(x - y) = - 2
1) Given equations are
10/(x + y) + 2/(x - y) = 4; 15/(x + y) - 5/(x - y) = - 2
2) In the equation, 10/(x + y) + 2/(x - y) = 4, put 1/(x + y) = p, and 1/(x - y) = q,
so we have,
10/(x + y) + 2/(x - y) = 4
10p + 2q = 4
5p + q = 2
q = 2 - 5p ---------------equation 1
3) In the equation, 15/(x + y) - 5/(x - y) = - 2, put 1/(x + y) = p, and 1/(x - y) = q,
so we have,
15/(x + y) - 5/(x - y) = - 2
15p - 5q = - 2 ---------------equation 2
4) Substitute the value of q = (2 - 5p) from equation 1 in equation 2, we get,
15p - 5q = - 2
15p - 5(2 - 5p) = - 2
15p - 10 + 25p = - 2
15p + 25p = - 2 + 10
40p = 8
p = 8/40
p = 1/5 ---------------equation 3
5) Put the value of p = 1/5 from equation 3 in equation 1, and we get
q = 2 - 5p
q = 2 - 5(1/5)
q = 2 - 1
q = 1 ------------ equation 4
6) The value of p = 1/5 and the value of q = 1.
7) As, p = 1/(x + y) and p = 1/5, we have
(x + y) = 5 ------------ equation 5
8) As, q = 1/(x - y) and q = 1, we have
(x - y) = 1 ------------ equation 6
9) Adding equation 5 and equation 6, we get,
x + y + x - y = 5 + 1
10) put x = 3 from equation 7 in equation 5, and we get,2x = 6, so x = 3 ------------ equation 7
(x + y) = 5
(3 + y) = 5
y = 5 - 3
y = 2 ------------ equation 8.
11) So, x = 3, and y = 2.
(viii) 1/(3x + y) + 1/(3x - y) = 3/4; 1/[2(3x + y)] - 1/[2(3x - y)] = -1/8
1) Given equations are
1/(3x + y) + 1/(3x - y) = 3/4; 1/[2(3x + y)] - 1/[2(3x - y)] = -1/8
2) In the equation, 1/(3x + y) + 1/(3x - y) = 3/4, put 1/(3x + y) = p, and 1/(3x - y) = q,
so we have,
1/(3x + y) + 1/(3x - y) = 3/4
p + q = 3/4 ---------------equation 1
3) In the equation, 1/[2(3x + y)] - 1/[2(3x - y)] = -1/8, put 1/(3x + y) = p, and
1/(3x - y) = q, so we have,
1/[2(3x + y)] - 1/[2(3x - y)] = -1/8
p/2 - q/2 = -1/8
p - q = -1/4 ---------------equation 2
4) Adding equation 1, and equation 2, we get,
p + q = 3/4
p - q = -1/4
-----------------------
2p = 2/4
p = 2/8
p = 1/4 ---------------equation 3
5) Put the value of p = 1/4 from equation 3 in equation 1, and we get
p + q = 3/4
1/4 + q = 3/4
q = 3/4 - 1/4
q = (3 - 1)/4
q = 2/4
q = 1/2 ------------ equation 4
6) The value of p = 1/4 and the value of q = 1/2.
7) As, p = 1/(3x + y) and p = 1/4, we have
(3x + y) = 4 ------------ equation 5
8) As, q = 1/(3x - y) and q = 1/2, we have
(3x - y) = 2 ------------ equation 6
9) Adding equation 5 and equation 6, we get,
-----------------------3x + y = 43x - y = 2
6x = 6x = 1 ---------------equation 7
10) put x = 1 from equation 7 in equation 5, and we get,
(3x + y) = 4
(3 + y) = 4
y = 4 - 3
y = 1 ------------ equation 8.
11) So, x = 1, and y = 1.
Q2. Formulate the following problems as a pair of equations, and hence find their solutions:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Explanation:
1) Here, let x and y be two variables.
2) Apply the given conditions and frame the equations.
3) We will get two equations from the above two conditions, then solve these
equations to get the values of x and y.
Solution:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
1) Let Ritu's speed in still water be x km/h.
2) Let the speed of the current be y km/h.
3) Ritu's speed of rowing downstream (x + y) km/h.
4) Ritu's speed of rowing upstream (x - y) km/h.
5) According to the first condition, as Ritu can row downstream 20 km in 2 hrs,
2(x + y) = 20
(x + y) = 10 -------------------------- equation 1.
6) According to the first condition, as Ritu can row upstream 4 km in 2 hrs,
2(x - y) = 4
(x - y) = 2-------------------------- equation 2.
7) Adding equation 1 and equation 2, we get,
(x + y) + (x - y) = 10 + 2
2x = 12
x = 12/2
x = 6 -------------------------- equation 3.
8) Put the value of x = 6 from equation 3 in equation 1, we get,
(x + y) = 10
(6 + y) = 10
y = 10 - 6
y = 4 -------------------------- equation 4.
9) Ritu's speed of rowing in still water is 6 km/h and the speed of the current is 4
km/h.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
1) Let the number of days taken by one woman to finish the work = x. So work done
by a woman in 1 day will be 1/x
2) Let the number of days taken by one man to finish the work = y. So work done
by a man in 1 day will be 1/y
3) As 2 women and 5 men finish the work in 4 days, so we have,
(2/x + 5/y) = 1/4, put 1/x = p and 1/y = q, we have,
2p + 5q = 1/4
4(2p + 5q) = 1
8p + 20q = 1 -------------------------- equation 1.
4) As 3 women and 6 men finish the work in 3 days, so we have,
(3/x + 6/y) = 1/3, put 1/x = p and 1/y = q, we have,
3p + 6q = 1/3
3(p + 2q) = 1/3
(p + 2q) = 1/9
p = 1/9 - 2q -------------------------- equation 2.
5) Put the value of p = (1/9 - 2q) from equation 2 in equation 1, we get,
8p + 20q = 1
8(1/9 - 2q) + 20q = 1
8/9 - 16q + 20q = 1
8/9 + 4q = 1
4q = 1 - 8/9
4q = (9 - 8)/9
q = 1/36 -------------------------- equation 3.
6) Put the value of q = 1/36 from equation 3 in equation 2, and we get,
p = 1/9 - 2q
p = 1/9 - 2(1/36)
p = 1/9 - 1/18
p = 2/18 - 1/18
p = (2 - 1)/18
p = 1/18 -------------------------- equation 4.
7) The value of p = 1/18 and the value of q = 1/36.
8) As, p = 1/x and p = 1/18, we have x = 18.
9) As, q = 1/y and q = 1/36, we have y = 36.
10) Number of days taken by a woman to finish the work is 18 and by a man is 36.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
1) Let the speed of the train be x km/h.
2) Let the speed of the bus be y km/h.
3) Total distance traveled by Roohi is 300 km.
4) According to the first condition, she traveled 60 km by train and 240 km by bus in
4 hrs, so, we have,
60/x + 240/y = 4, put 1/x = p and 1/y = q, we have,
60p + 240q = 4
4(15p + 60q) = 415p + 60q = 1
15(p + 4q) = 1
p + 4q = 1/15
p = ((1/15) - 4q) -------------------------- equation 1.
5) According to the second condition, she traveled 100 km by train and 200 km by
bus in 4 hrs and 10 minutes i.e. 4 + (10/60) = 4 + (1/6) = 25/6 hrs, so, we have,
100/x + 200/y = 25/6, put 1/x = p and 1/y = q, we have,
100p + 200q = 25/6
25(4p + 8q) = 25/6
(4p + 8q) = 1/6 -------------------------- equation 2.
6) Put the value of p = ((1/15) - 4q) from equation 1 in equation 2, we get,
(4p + 8q) = 1/6
(4((1/15) - 4q) + 8q) = 1/6
(4/15 - 16q + 8q) = 1/6
(4/15 - 8q) = 1/6
8q = 4/15 - 1/6
8q = (8/30) - (5/30)
8q = 3/30
8q = 1/10
q = 1/(10(8))q = 1/80 -------------------------- equation 3.
7) Put the value of q = 1/80 from equation 3 in equation 1, we get,
p = ((1/15) - 4q)
p = ((1/15) - 4(1/80))
p = ((1/15) - 1/20))
p = ((4/60) - 3/60))
p = (4 - 3)/60
p = 1/60 -------------------------- equation 4.
8) The value of p = 1/60 and the value of q = 1/80.
9) As, p = 1/x and p = 1/60, we have x = 60.
10) As, q = 1/y and q = 1/80, we have y = 80.
11) The speed of the train is 60 km/h.
12) The speed of the bus is 80 km/h.
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