इयत्ता 10वी गणित प्रकरण 1 – सर्वसंच (Ganit 1) | पूर्ण स्पष्टीकरण व सोपी उत्तरे | SSC Maharashtra Board

इयत्ता 10 गणित 1- प्रकरण 1 | दोन चलांतील एकसामयिक समीकरणे

इयत्ता 10 वी - गणित 1

प्रकरण 1 - दोन चलांतील एकसामयिक समीकरणे

1.1 सर्वसंच

प्रश्न 2.(1)
3a + 5b = 26; a + 5b = 22
SOLUTION 3a + 5b = 26 ....(i) a + 5b = 22 ....(ii) समीकरण (i) मधून समीकरण (ii) वजा करूं, 3a + 5b = 26 − (a + 5b = 22) --------------- 2a = 4 ∴ a = 4/2 = 2 a = 2 ही किंमत समीकरण (ii) मध्ये ठेवून, a + 5b = 22 2 + 5b = 22 5b = 22 − 2 5b = 20 ∴ b = 20/5 = 4 ∴ (a, b) = (2, 4) ही दिलेल्या एकसामीक समीकरणांची उत्तरे आहेत.
प्रश्न 2.(2)
x + 7y = 10; 3x - 2y = 7
SOLUTION x + 7y = 10 ....(i) 3x - 2y = 7 ....(ii) समीकरण (i) ला 3 ने गुणूं, 3(x + 7y) = 3 × 10 3x + 21y = 30 ....(iii) समीकरण (iii) मधून समीकरण (ii) वजा करूं, 3x + 21y = 30 − (3x - 2y = 7) --------------- 23y = 23 ∴ y = 23/23 = 1 y = 1 ही किंमत समीकरण (i) मध्ये ठेवून, x + 7(1) = 10 x + 7 = 10 x = 10 - 7 x = 3 ∴ (x, y) = (3, 1) ही दिलेल्या एकसामीक समीकरणांची उत्तरे आहेत.
प्रश्न 2.(3)
2x - 3y = 9; 2x + y = 13
SOLUTION 2x - 3y = 9 ....(i) 2x + y = 13 ....(ii) समीकरण (ii) मधून समीकरण (i) वजा करूं, 2x + y = 13 − (2x - 3y = 9) --------------- 4y = 4 ∴ y = 4/4 = 1 y = 1 ही किंमत समीकरण (ii) मध्ये ठेवून, 2x + 1 = 13 2x = 13 - 1 2x = 12 ∴ x = 12/2 = 6 ∴ (x, y) = (6, 1) ही दिलेल्या एकसामीक समीकरणांची उत्तरे आहेत.
प्रश्न 2.(4)
5m - 3n = 19; m - 6n = -7
SOLUTION 5m - 3n = 19 ....(i) m - 6n = -7 ....(ii) समीकरण (ii) ला 5 ने गुणूं, 5(m - 6n) = 5 × (-7) 5m - 30n = -35 ....(iii) समीकरण (iii) मधून समीकरण (i) वजा करूं, 5m - 30n = -35 − (5m - 3n = 19) --------------- -27n = -54 ∴ n = -54/-27 = 2 n = 2 ही किंमत समीकरण (ii) मध्ये ठेवून, m - 6(2) = -7 m - 12 = -7 m = -7 + 12 m = 5 ∴ (m, n) = (5, 2) ही दिलेल्या एकसामीक समीकरणांची उत्तरे आहेत.
प्रश्न 2.(5)
5x + 2y = -3; x + 5y = 4
SOLUTION 5x + 2y = -3 ....(i) x + 5y = 4 ....(ii) समीकरण (ii) ला 5 ने गुणूं, 5(x + 5y) = 5 × 4 5x + 25y = 20 ....(iii) समीकरण (iii) मधून समीकरण (i) वजा करूं, 5x + 25y = 20 − (5x + 2y = -3) --------------- 23y = 23 ∴ y = 23/23 = 1 y = 1 ही किंमत समीकरण (ii) मध्ये ठेवून, x + 5(1) = 4 x + 5 = 4 x = 4 - 5 x = -1 ∴ (x, y) = (-1, 1) ही दिलेल्या एकसामीक समीकरणांची उत्तरे आहेत.
प्रश्न 2.(6)
(1/3)x + y = 10/3; 2x + (1/4)y = 11/4
SOLUTION (1/3)x + y = 10/3 ....(i) 2x + (1/4)y = 11/4 ....(ii) समीकरण (i) ला 3 ने गुणूं, 3 × [(1/3)x + y] = 3 × (10/3) x + 3y = 10 ....(iii) समीकरण (ii) ला 4 ने गुणूं, 4 × [2x + (1/4)y] = 4 × (11/4) 8x + y = 11 ....(iv) समीकरण (iv) ला 3 ने गुणूं, 3 × (8x + y) = 3 × 11 24x + 3y = 33 ....(v) समीकरण (v) मधून समीकरण (iii) वजा करूं, 24x + 3y = 33 − (x + 3y = 10) --------------- 23x = 23 ∴ x = 23/23 = 1 x = 1 ही किंमत समीकरण (iii) मध्ये ठेवून, 1 + 3y = 10 3y = 10 - 1 3y = 9 ∴ y = 9/3 = 3 ∴ (x, y) = (1, 3) ही दिलेल्या एकसामीक समीकरणांची उत्तरे आहेत.
प्रश्न 2.(7)
99x + 101y = 499; 101x + 99y = 501
SOLUTION 99x + 101y = 499 ....(i) 101x + 99y = 501 ....(ii) समीकरण (i) आणि समीकरण (ii) ची बेरीज करूं, 99x + 101y = 499 + 101x + 99y = 501 ----------------- 200x + 200y = 1000 दोन्ही बाजूंना 200 ने भागूं, x + y = 5 ....(iii) समीकरण (ii) मधून समीकरण (i) वजा करूं, 101x + 99y = 501 − (99x + 101y = 499) ----------------- 2x - 2y = 2 दोन्ही बाजूंना 2 ने भागूं, x - y = 1 ....(iv) समीकरण (iii) आणि समीकरण (iv) ची बेरीज करूं, x + y = 5 + x - y = 1 ----------- 2x = 6 ∴ x = 6/2 = 3 x = 3 ही किंमत समीकरण (iii) मध्ये ठेवून, 3 + y = 5 y = 5 - 3 y = 2 ∴ (x, y) = (3, 2) ही दिलेल्या एकसामीक समीकरणांची उत्तरे आहेत.
प्रश्न 2.(8)
49x - 57y = 172; 57x - 49y = 252
SOLUTION 49x - 57y = 172 ....(i) 57x - 49y = 252 ....(ii) समीकरण (i) आणि समीकरण (ii) ची बेरीज करूं, 49x - 57y = 172 + 57x - 49y = 252 ----------------- 106x - 106y = 424 दोन्ही बाजूंना 106 ने भागूं, x - y = 4 ....(iii) समीकरण (ii) मधून समीकरण (i) वजा करूं, 57x - 49y = 252 − (49x - 57y = 172) ----------------- 8x + 8y = 80 दोन्ही बाजूंना 8 ने भागूं, x + y = 10 ....(iv) समीकरण (iii) आणि समीकरण (iv) ची बेरीज करूं, x - y = 4 + x + y = 10 ----------- 2x = 14 ∴ x = 14/2 = 7 x = 7 ही किंमत समीकरण (iv) मध्ये ठेवून, 7 + y = 10 y = 10 - 7 y = 3 ∴ (x, y) = (7, 3) ही दिलेल्या एकसामीक समीकरणांची उत्तरे आहेत.

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