ex 3.6 class 10 ncert Q1
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NCERT Solutions for Class 10 Maths Exercise 3.6
Last Updated: July 13, 2018 by myCBSEguide

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NCERT Solutions for Class 10 Maths Exercise 3.6 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.
NCERT solutions for Maths Pair of Linear Equations in Two Variables Download as PDF

NCERT Solutions for Class 10 Maths Pair of Linear Equations in Two Variables
1. Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) 

(ii) 

(iii) + 3y = 14
 − 4y = 23
(iv) 

NCERT Solutions for Class 10 Maths Exercise 3.6
(v) 7x − 2y = 5xy
8x + 7y = 15xy
(vi) 6x + 3y − 6xy = 0
2x + 4y − 5xy = 0
(vii) 

(viii) 

Ans. (i) … (1)
 … (2)
Let = p and = q
Putting this in equation (1) and (2), we get

⇒ 3p + 2q = 12 and 6 (2p + 3q) = 13 (6)
⇒ 3p + 2q = 12 and 2p + 3q =13
⇒ 3p + 2q – 12 = 0 … (3) and 2p + 3q – 13 = 0 … (4)


⇒ 
⇒ 
⇒ 
⇒ p = 2 and q = 3
But = p and = q
Putting value of p and q in this we get
x =  and y = 
(ii) … (1)
 … (2)
Let = p and = q
Putting this in (1) and (2), we get
2p + 3q = 2 … (3)
4p − 9q = −1 … (4)
Multiplying (3) by 2 and subtracting it from (4), we get
4p − 9q + 1 – 2 (2p + 3q − 2) = 0
⇒ 4p − 9q + 1 − 4p − 6q + 4 = 0
⇒ −15q + 5 = 0
⇒ q = 
Putting value of q in (3), we get
2p + 1 = 2 ⇒ 2p = 1⇒ p = ½
Putting values of p and q in (= p and = q), we get
 and 
⇒ 
⇒ x = 4 and y = 9
NCERT Solutions for Class 10 Maths Exercise 3.6
(iii) + 3y = 14 … (1)
 − 4y = 23 … (2) and Let = p … (3)
Putting (3) in (1) and (2), we get
4p + 3y = 14 … (4)
3p − 4y = 23 … (5)
Multiplying (4) by 3 and (5) by 4, we get
3 (4p + 3y – 14 = 0) and, 4 (3p − 4y – 23 = 0)
⇒ 12p + 9y – 42 = 0 … (6) 12p − 16y – 92 = 0 … (7)
Subtracting (7) from (6), we get
9y − (−16y) – 42 − (−92) = 0
⇒ 25y + 50 = 0
⇒ y = 50 – 25 = −2
Putting value of y in (4), we get
4p + 3 (−2) = 14
⇒ 4p – 6 = 14
⇒ 4p = 20⇒ p = 5
Putting value of p in (3), we get
 = 5 ⇒ x = 
Therefore, x =  and y = −2
(iv)  … (1)
 … (2)
Let 
Putting this in (1) and (2), we get
5p + q = 2
⇒ 5p + q – 2 = 0 … (3)
And, 6p − 3q = 1
⇒ 6p − 3q – 1 = 0 … (4)
Multiplying (3) by 3 and adding it to (4), we get
3 (5p + q − 2) + 6p − 3q – 1 = 0
⇒ 15p + 3q – 6 + 6p − 3q – 1 = 0
⇒ 21p – 7 = 0
⇒ p = 
Putting this in (3), we get
5 () + q – 2 = 0
⇒ 5 + 3q = 6
⇒ 3q = 6 – 5 = 1
⇒ q = 
NCERT Solutions for Class 10 Maths Exercise 3.6
Putting values of p and q in (), we get

⇒ 3 = x − 1 and 3 = y – 2
⇒ x = 4 and y = 5
(v) 7x − 2y = 5xy … (1)
8x + 7y = 15xy … (2)
Dividing both the equations by xy, we get


Let = p and = q
Putting these in (3) and (4), we get
7q − 2p = 5 … (5)
8q + 7p = 15 … (6)
From equation (5),
2p = 7q – 5
⇒ p = 
Putting value of p in (6), we get
8q + 7 () = 15
⇒ 16q + 49q – 35 = 30
⇒ 65q = 30 + 35 = 65
⇒ q = 1
Putting value of q in (5), we get
7 (1) − 2p = 5
⇒ 2p = 2⇒ p = 1
NCERT Solutions for Class 10 Maths Exercise 3.6
Putting value of p and q in (= p and = q), we get x = 1 and y = 1
(vi) 6x + 3y − 6xy = 0 … (1)
2x + 4y − 5xy = 0 … (2)
Dividing both the equations by xy, we get


Let = p and = q
Putting these in (3) and (4), we get
6q + 3p – 6 = 0 … (5)
2q + 4p – 5 = 0 … (6)
From (5),
3p = 6 − 6q
⇒ p = 2 − 2q
Putting this in (6), we get
2q + 4 (2 − 2q) – 5 = 0
⇒ 2q + 8 − 8q – 5 = 0
⇒ −6q = −3⇒ q = ½
Putting value of q in (p = 2 – 2q), we get
p = 2 – 2 (½) = 2 – 1 = 1
NCERT Solutions for Class 10 Maths Exercise 3.6
Putting values of p and q in (= p and = q), we get x = 1 and y = 2
(vii)  … (1)
 …(2)
Let 
Putting this in (1) and (2), we get
10p + 2q = 4 … (3)
15p − 5q = −2 … (4)
From equation (3),
2q = 4 − 10p
⇒ q = 2 − 5p … (5)
Putting this in (4), we get
15p – 5 (2 − 5p) = −2
⇒ 15p – 10 + 25p = −2
⇒ 40p = 8⇒ p = 
Putting value of p in (5), we get
q = 2 – 5 () = 2 – 1 = 1
Putting values of p and q in (), we get

⇒ x + y = 5 … (6) and x – y = 1 … (7)
Adding (6) and (7), we get
2x = 6 ⇒ x = 3
Putting x = 3 in (7), we get
3 – y = 1
⇒ y = 3 – 1 = 2
Therefore, x = 3 and y = 2
NCERT Solutions for Class 10 Maths Exercise 3.6
(viii)  … (1)
 … (2)
Let 
Putting this in (1) and (2), we get
p + q =  and 
⇒ 4p + 4q = 3 … (3) and 4p − 4q = −1 … (4)
Adding (3) and (4), we get
8p = 2 ⇒ p = ¼
Putting value of p in (3), we get
4 (¼) + 4q = 3
⇒ 1 + 4q = 3
⇒ 4q = 3 – 1 = 2
⇒ q = ½
Putting value of p and q in , we get

⇒ 3x + y = 4 … (5) and 3x – y = 2 … (6)
Adding (5) and (6), we get
6x = 6 ⇒ x = 1
Putting x = 1 in (5) , we get
3 (1) + y = 4
⇒ y = 4 – 3 = 1
Therefore, x = 1 and y = 1
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