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## 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

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