NCERT SOLUTIONS FOR CLASS 10 MATHS CHAPTER 3 EXERCISE 3.6 QUESTION 1 ALL IN HINDI AND ENGLISH.
दोस्तों अब हम क्लास 10 गणित प्रश्नावली 3. 6 (NCERT SOLUTIONS FOR CLASS 10 MATHS CHAPTER 3 EXERCISE 3.6 QUESTION 1 ALL IN HINDI AND ENGLISH ) के सभी क्वेश्चन हल करेंगे।
प्रश्न -( 1 ) निम्न समीकरणों के युग्मो को रेखिक समीकरणों के युग्म में बदल करके हल कीजिये ?
(i) 1 / 2x + 1 / 3y =2 ; 1 / 3x + 1 / 2y = 13 / 6
हल : माना 1 /x = s तथा 1 / y = t हो , तो
s / 2 + t / 3 = 2 => 3s + 2 t = 12 ----> (1 )
s / 3 + t / 2 = 13 / 6 => 2s + 3t = 13 -----> (2 )
समीकरण (1 ) से , 3s + 2 t = 12
3s = 12 - 2t
s = (12 -2t ) / 3 ------> (3)
s का यह मान समीकरण (2) में रखने पर ,
2s + 3t = 13
2 (12 - 2t ) / 3 + 3t = 13
3 से भाग देने पर ,
2 * 3 (12 -2t ) / 3 + 3t * 3 = 13 *3
2 (12 -2t ) + 9t = 39
24 - 4t + 9t = 39
24 + 5t = 39
5t = 39 - 24
5t = 15
t = 15 / 5
t = 3
1 / y = 3 /1
y = 1 / 3 ( व्युत्क्रम लेने पर )
t का मान समीकरण (3 ) में रखने पर ,
s = (12 -2t ) / 3 ------> (3)
s =( 12 - 2 * 3 )/ 3
s = (12 - 6 )/ 3
s = 6 / 3
s = 2
1 /x = 2 / 1
x = 1 / 2
X = 1 / 2 , Y = 1 / 3 : Answer
(ii )
Question -( 1 ) Solve the following pair of equations by converting them to a pair of linear equations?
(i) 1/2x + 1/3y =2 ; 1/3x + 1/2y = 13 / 6
Solution : Let 1 /x = s and 1 / y = t, then
s / 2 + t / 3 = 2 => 3s + 2 t = 12 ----> (1 )
s / 3 + t / 2 = 13 / 6 => 2s + 3t = 13 -----> (2)
From Equation (1) , 3s + 2 t = 12
3s = 12 - 2t
s = (12 -2t ) / 3 ------> (3)
Substituting this value of s in equation (2),
2s + 3t = 13
2 (12 - 2t ) / 3 + 3t = 13
On dividing by 3,
2 * 3 (12 -2t ) / 3 + 3t * 3 = 13 *3
2 (12 -2t ) + 9t = 39
24 - 4t + 9t = 39
24 + 5t = 39
5t = 39 - 24
5t = 15
t = 15 / 5
t = 3
1 / y = 3 /1
y = 1 / 3 (taking the inverse)
Substituting the value of t in equation (3),
s = (12 -2t ) / 3 ------> (3)
s =( 12 - 2 * 3 )/3
s = (12 - 6)/3
s = 6 / 3
s = 2
1 /x = 2 / 1
x = 1/2
X = 1/2 , Y = 1 / 3 : Answer
(ii )
In equation (3), multiplying by 3,
s = 1/2
Substituting the value of s in equation (1),
(iii) 4 / x + 3y = 14 ------->(A )
3 / x - 4y = 23 -------->(B )
Solution : Let 1 / x = z Then,
4z + 3y = 14 -------> (1)
3z -4y = 23 ---------> (2)
On multiplying 4 in equation (1) and 3 in equation (2),
16Z + 12Y = 56
9Z - 12Y = 69
On adding equation (5) and equation (6),
25z = 125
z = 125 / 25
z = 5
1 / x = 5 / 1
On taking the inverse,
x = 1 / 5
Substituting the value of x in Equation (A),
( 4 * 5 ) / 1 + 3Y = 14
=> 20 +3Y = 14
=> 3Y = 14 - 20
=> 3Y = -6
=> Y = -6/3
=> Y = -2
Answer: X = 1 / 5, Y = -2
(iv)
21 s = 1 + 6
x = 4 , y = 5 => Answer