1. For what value of k will the following system of linear equations have no solution?
3x + y = 1, (2k - 1)x + (k - 1)y = 2k +1
cx + 3y = 3; 12x + cy = 6
x + 2y = 5; 3x + ky + 15 = 0
kx + 2y = 5; 3x - 4y = 10
2x - 3y = 7
(a + b)x - (a + b - 3)y = 4a + b.)
2x + 3y = 7
(k - 1) x + (k +2)y = 3k
8. x + y = 7, 5x + 12y = 7
9. 3x - 4y = 1, 4x - 3y = 6
10. x + y/2 = 4, x/3 + 2y = 5
11. x/3 + y/4 = 4, 5x/6 - y/8 = 4
12. 2/x + 2/3y = 1/6, 3/x + 2/y = 0
13. 2u + 15v = 17uv, 5u + 5v = 36uv
14. 3(a + 3b) = 11ab, 3(2a +b) = 7ab
15. 2x/a + y/b = 2, x/a - y/b = 4
16. x/a - y/b = 0, ax + by = a2 + b2
17. (a - b)x + (a +b)y = a2 - 2ab - b2, (a +b) (x + y) = a2 + b2
18. x/a + y/b = a +b, x/a2 + y/b2 = 2
19. ax + by = c, bx + ay = 1 + c
Answers | ||||
(1) 2 | (2) 6 | (3) (i) (ii) | ||
(4) (i) (ii) | (5) a = -5, b = -1 | (6) 7 | (7) -2, -1 | |
(8) 11, -4 (13) 5, 1/7 | (9) 3, 2 (14) 1, 3/2 | (10) 3, 2 (15) 2a, -2b | (11) 6, 8 (16) a, b | (12) 6, -4 (17) a +b, -2ab/(a + b) |
(18) a2, b2 | (19) (bc - ac + b)/(b2 - a2), (bc - ac - a)/(b2 - a2 ) |
Test papers are the key of success and in my opinion linear equation is a type of algebraic equation in which each term can be a constant or can be product of the constant and its the first order equation.
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