HOME PAGE
Class ten NCERT Solutions Chapter 3 maths
Chapter 3: Pair of linear equations in two variables
TEXTBOOK’S EXERCISE 3.6
1. Solve the following pairs of equations by reducing them to a pair of linear equations :
Sol. (i) Given equations are :
3u + 2v = 12 …(3)
From equation (2), we have
Multiplying equation (3) by 2 and equation (4) by 3, and subtracting we get
Putting the value of v in equation (3), we have
3u + 2(3) = 12
⇒ 3u + 6 = 12 ⇒ 3u = 12 – 6 = 6
⇒ u = 6/3 = 2
(ii) Given equations are :
in equation (1) and (2), we have
2u + 3v = 2 …(3)
4u – 9v = –1 …(4)
Multiplying equation (3) by 2, and subtracting it from (4), we have
⇒ v = 15/ 5 = 3
Putting and in equation (1) and (2), we have
Putting the value of v in equation (3), we have
3u + 2(3) = 12
⇒ 3u + 6 = 12 ⇒ 3u = 12 – 6 = 6
⇒ u = 6/3 = 2
Now, 1/x = u and 1/y = v
⇒ x = 1/2 and y = 1/3
(ii) Given equations are :
Putting
in equation (1) and (2), we have
2u + 3v = 2 …(3)
4u – 9v = –1 …(4)
Multiplying equation (3) by 2, and subtracting it from (4), we have
v = 5/15 = 1/3
Putting and in equation (1) and (2), we have
Hence, x = 4 and y = 9
(iii) Given equations are :
Putting 1/x =v in equation (1) and equation (2), we have
4 v+ 3 y = 14 …(3)
3 v – 4 y = 23 …(4)
Multiplying equation (3) by (3) and equation (4) by 4, and subtracting we have
y = −50/ 25 = – 2
Putting the value of y in equation (3), we have
4v + 3(–2) = 14 ⇒ 4v – 6 = 14
∴ v = 20/ 4 = 5
Now, 1/x= v ⇒ x = 1/v = 1/5
Hence, x = 1/5 and y = –2.
(iv) Given equations are:
in equation (1) and (2), we have
5u + v = 2 …(3)
6u – 3v = 1 …(4)
Multiplying equation (3) by 3, we have
15u + 3v = 6 …(5)
Adding equation (4) and equation (5), we have
Putting the value of u in equation (3), we have
x – 1 = 3 and y – 2 = 3
∴ x = 3 + 1 and y = 3 + 2
or x = 4 and y = 5
Hence, x = 4 and y = 5.
(v) Given equations are :
Putting 1/x = u and 1/y = v in equation (1) and (2), we have
–2u + 7v = 5 …(3)
7u + 8v = 15 …(4)
Multiplying equation (3) by 7, equation (4) by 2, and adding, we have
–14u + 49v = 35
14u + 16v = 30
65u = 65
u = 65/ 65 = 1
Putting the value of u in equation (3), we have
–2(1) + 7v = 5 ⇒ 7v = 5 + 2
⇒ v = 7/7 = 1
But, 1/x = u and 1/y = v
∴ x = 1/u and y = 1/v
x = 1/1 and y = 1/1
Hence, x = 1 and y = 1.
(vi) Dividing each equation by xy, we get
Putting 1/x = u and 1/y = v in equation (1) and (2), we have
u + 2v = 2 …(3)
4u + 2v = 5 …(4)
Subtracting equation (3) from equation (4) we have
⇒ u = 3/3 = 1
Putting the value of u in equation (3), we have
1 + 2v = 2 ⇒ 2v = 2 – 1 = 1 ⇒ v = ½
But, 1/x = u and 1/y v = ⇒ 1 1 /x = and 1/y = 1/2
⇒ x = 1 and y = 2
(vii) Given equations are
10u + 2v = 4 or 5u + v = 2 …(3)
And, 15u – 5v = –2 …(4)
Multiplying euqation (3) by 5, and then adding to (4) we have
u = 8/ 40 = 15 m
Putting the value of u in equation (3), we have
⇒ 1 + v = 2 ⇒ v = 1
⇒ x + y = 5 …(5)
and x – y = 1 …(6)
Adding equations (5) and equation (6), we have
2x = 6 ⇒ x = 6/2 = 3
Putting the value of x in equation (5), we have
3 + y = 5 ⇒ y = 5 – 3 = 2
Hence, x = 3 and y = 2.
(viii) Given equations are :
in equation (1) and equation (2), we have
4u + 4v = 3 …(3)
and 4u – 4v = –1
Adding equation (3) and equation (4), we have
Putting the value of u in equation (3), we have
4v = 2
∴ v = 2/4= ½
3x + y = 4 …(5)
And, 3x – y = 2 …(6)
Adding equations (5) and (6), we have
6x = 6 ⇒ x = 1
Putting the value of x in equation (5), we have
3(1) + y = 4
⇒ 3 + y = 4
∴ y = 4 – 3 = 1
Hence, x = 1 and y = 1