(1 ) NCERT SOLUTIONS FOR CLASS 10 MATHS EXERCISE 3.5 CHAPTER 3 QUESTION 1 ALL
(2 ) CLASS 10 CHAPTER 3 EXERCISE 3.5 SOLUTION
(3 ) EX 3.5 CLASS 10
प्रश्न - (1) निम्न रेखिक समीकरणों के युग्मो में से किसका अद्वितीय हल हे , किसका कोई हल नहीं हे या किसके अपरिमित रूप से अनेक हल हे। अद्वितीय हल की स्थिति में , उसे वज्रगुणन विधि से ज्ञात कीजिये ?
(i) x -3y -3 = 0 ; 3x - 9y -2 =0 (ii ) 2x + y = 5 ; 3x +2y = 8 (iii ) 3x - 5y = 20 ; 6x -10y =40
(iv) x -3y -7 =0 ; 3x -3y -15 = 0
हल : ( i ) दिया गया रेखिक समीकरण ,
x -3y -3 = 0 -----> (1)
3x - 9y -2 =0 ------> (2)
इन् समीकरणों की तुलना a 1 x + b 1 y + c 1 = 0 तथा a 2 x + b 2 y + c 2 = 0 से करने पर ,
a 1 = 1 , b 1 = -3 , c 1 = -3
a 2 =3 , b2 = -9 , c 2 = -2
a 1 / a 2 = 1 / 3 , b 1 / b 2 = -3 / -9 = 3 / 9 , c 1 / c 2 = -3 / -2
स्पस्ट हे , a 1 / a 2 बराबर हे , b 1 / b 2 के लेकिन c 1 / c 2 के बराबर नहीं हे।
अतः रेखिक समीकरण युग्म का कोई हल नहीं हे। =
(ii ) 2x + y = 5 ; 3x +2y = 8
हल : दिया गया रेखिक समीकरण ,
2x + y = 5 => 2x +y -5 = 0 -------> (1)
3x +2y = 8 => 3x +2y -8 = 0 ------> (2)
इन् समीकरणों की तुलना a 1 x + b 1 y + c 1 = 0 तथा a 2 x + b 2 y + c 2 = 0 से करने पर ,
a 1 = 2 , b 1 = 1 , c 1 = -5
a 2 = 3 , b 2 = 2 , c 2 = -8
स्पस्ट हे , यहाँ पर a 1 / a 2 , b 1 / b 2 और c 1 / c 2 तीनो ही आपस बराबर नहीं हे। इस स्थिति में
(i) x -3y -3 = 0 ; 3x - 9y -2 =0 (ii) 2x + y = 5 ; 3x +2y = 8 (iii) 3x - 5y = 20 ; 6x -10y =40
(iv) x -3y -7 =0 ; 3x -3y -15 = 0
Solution: (i) Given linear equation,
x -3y -3 = 0 -----> (1)
3x - 9y -2 =0 ------> (2)
Comparing these equations with a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0,
a 1 = 1 , b 1 = -3 , c 1 = -3
a 2 =3 , b2 = -9 , c 2 = -2
a 1 / a 2 = 1 / 3 , b 1 / b 2 = -3 / -9 = 3 / 9 , c 1 / c 2 = -3 / -2
Obviously, a 1 / a 2 is equal to b 1 / b 2 but not equal to c 1 / c 2.
Hence, the pair of linear equations has no solution.
(ii) 2x + y = 5 ; 3x +2y = 8
Solution : Given linear equation,
2x + y = 5 => 2x +y -5 = 0 -------> (1)
3x +2y = 8 => 3x +2y -8 = 0 ------> (2)
Comparing these equations with a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0,
a 1 = 2 , b 1 = 1 , c 1 = -5
a 2 = 3 , b 2 = 2 , c 2 = -8
Obviously, here a 1 / a 2 , b 1 / b 2 and c 1 / c 2 are not equal. In this situation
The system of equations will have a unique solution. And in this situation we will need to work out these solutions.
x / 2 = 1 /1 and y / 1 = 1 / 1
(iii) 3x – 5y = 20; 6x -10y =40
Solution : 3x - 5y = 20 => 3x - 5y - 20 = 0 -------> (1)
6x -10y =40 => 6x - 10 y - 40 = 0 -------> (2)
Here a 1 = 3 , b 1 = -5 , c 1 = -20
a 2 = 6 , b 2 = -10 , c 2 = -40
a 1 / a 2 = 3 / 6 = 1/2 , b 1 / b 2 = -5 / -10 = 1/2 , c 1 / c 2 = -20 / -40 = 1/2
Clearly, a 1 / a 2 = b 1 / b 2 = c 1 / c 2
Hence, the given linear equation will have infinitely many solutions.
(iv) x -3y -7 =0 ; 3x -3y -15 = 0
Solution : x -3y -7 =0 ------> (1)
3x -3y -15 = 0 ------> (2)
Here a 1 = 1 , b 1 = -3 , c 1 = -7
a 2 = 3 , b 2 = -3 , c 2 = -15
a 1 / a 2 = 1 / 3 , b 1 / b 2 = -3 / -3 = 1 /1 , c 1 / c 2 = -7 / -15 = 7 / 15
Therefore, the system of equations will have a unique solution or will have a solution or be consistent.
Answer: x = 4 and y = -1