लेखक
M.A., B.Ed., DNYS, MASSCOM
20/1/15/11/2021
सूत्र द्वारा
TYPE 06
(1) जब समीकरण
(a1x + b1)(a2x + b2) = (a3x + b3) (a4x + b4)
के रूप में हो।
तो सूत्र
[(b4) × (b3)] – [(b2) × (b1)]
x = __________________________
(b1) + (b2) – (b3) – (b4)
Exercise 01
(1). Solve the following:
निम्न को सरल करो
01. (x + 7)(x + 9) = (x + 3) (x + 21)
02. (x + 7)(x + 9) = (x + 3) (x + 22)
03. (x + 1)(x + 3) = (x + 5) (x + 7)
04. (x + 3)(x + 5) = (x + 7) (x + 9)
05. (x + 5)(x + 9) = (x + 7) (x + 12)
06. (x + 1)(x + 2) = (x – 3) (x – 4)
07. (x + 2)(x + 3) = (x – 4) (x – 5)
08. (x + 3)(x + 4) = (x – 5) (x – 6)
09. (x + 4)(x + t) = (x – 6) (x – 7)
10. (x + 5)(x + 6) = (x – 7) (x – 8)
11. (x – 2)(x – 5) = (x – 3) (x – 4)
12. (x – 2)(x – 5) = (x – 1) (x – 4)
13. (x – 7)(x – 9) = (x – 3) (x – 22)
14. (x – 2)(x – 5) = (x – 3) (x – 4)
15. (x – 3)(x – 4) = (x – 5) (x – 6)
16. (x – 6)(x + 7) = (x + 3) (x + 11)
17. (x – 7)(x + 9) = (x + 5) (x + 13)
18. (x – 3)(x + 4) = (x + 1) (x + 8)
19. (x – 2)(x + 7) = (x + 5) (x + 11)
20. (x – 1)(x + 2) = (x + 3) (x + 4)
21. (x + 7)(x + 9) = (x – 8) (x – 11)
22. (x + 2)(x + 5) = (x – 3) (x – 4)
23. (x + 1)(x + 2) = (x – 3) (x – 4)
24. (x + 2)(x + 3) = (x – 4) (x – 5)
25. (x + 4)(x + 5) = (x – 6) (x – 7)
26. (x – 6)(x + 7) = (x + 3) (x + 11)
27. (x – 1)(x + 2) = (x + 3) (x + 4)
28. (x – 2)(x + 8) = (x + 4) (x + 11)
29. (x – 3)(x + 7) = (x + 3) (x + 5)
30. (x – 5)(x + 9) = (x + 2) (x + 9)
Sol. 01.
(x + 7)(x + 9) = (x + 3) (x + 21)
[(b4) × (b3)] – [(b2) × (b1)]
x = __________________________
(b1) + (b2) – (b3)– (b4)
(3× 21) – (9 × 7)
= ______________________
(7) + (9) – (3) – (21)
(63) – (63)
= _____________
–8
0
= ____
–8
= 0
Sol. 05.
(x + 1)(x + 2) = (x – 3) (x – 4)
[(b4) × (b3)] – [(b2) × (b1)]
x = __________________________
(b1)+ (b2) – (b3) – (b4)
[(–3) × (–4)] – [(1) × (2)]
= ____________________________
(1) + (2) – (–3) – (–4)
(12) – (2)
= _____________
10
10
= ____
10
= 1
Sol. 11.
(x – 2)(x – 5) = (x – 3) (x – 4)
[(b4) × (b3)] – [(b2) × (b1)]
x = __________________________
(b1)+ (b2) – (b3) – (b4)
[(–3) × (–4)] – [(–2) × (–5)]
= ____________________________
(–2) + (–5) – (–3) – (–4)
(12) – (10)
= _____________
10
10
= ____
10
= 1
TYPE 07
(2) जब समीकरण
(a1x + b1) C1
_________ = _____
(a2x + b2) C2
के रूप में हो।
तो सूत्र
b2.C1 – b1.C2
x = _______________
a1.C2 – a2.C1
Exercise 02
(2). Solve the following:
निम्न को सरल करो
01.
(2x + 3) 4
_______ = ___
(3x + 5) 5
02.
(2x + 3) 8
_______ = ___
(x + 5) 10
03.
(3x – 4) 5
_______ = ___
(5x – 4) 9
04.
x + 2 7
_______ = ___
x – 3 3
05.
x + 7 3
_______ = ___
3x + 3 5
06.
3x + 2 3
_______ = ___
3x – 12 2
Sol. 01
(2x + 3) 4
_______ = ___
(3x + 5) 5
तो सूत्र
C1 b2 – C2 b1
x = _______________
C2 a1 – C1a2
4×5 – 5×3
x = ____________
5×2 – 4×3
20 – 15
x = _________
10 – 12
5
x = _____
– 2
Post a Comment
Post a Comment