. Resolver las siguientes ecuaciones : a) 3(2x− 5) + 5(3x− 1) = 7(x− 3) + 15 b) 7x + 3(5x− 3) − (5x + 1) = 7(2x + 2) c) 5x− 4(3x− 1) − (6x + 1) = 5(3x + 12) − 1.
ax² + bx + c = 0
A)
3(2x− 5) + 5(3x− 1) = 7(x− 3) + 15
6x - 15 + 15x - 5 = 7x - 21 + 15 - 20 + 6 = 7x - 21x - 14 = - 14x
14x = 14
x = 14 / 14
x = 1
b)
7x + 3(5x− 3) − (5x + 1) = 7(2x + 2)
7x + 15x - 9 - 5x - 1 = 14x + 14
17x - 14x = 14 + 10
3x = 24
x = 24 / 3
x = 8
c)
5x− 4(3x− 1) − (6x + 1) = 5(3x + 12) − 1
5x - 12x + 4 - 6x - 1 = 15x + 60 - 1 - 13x - 15x = 59 - 3 - 56 = 28x - 56 / 28 = x - 2 = x
Espero te sirva ; ).
. Espero que te sirva, salu2!
7x / 10 = 28 7x = 280 x = 40 (16x + 40 - 120) / 8 = x + 2 16x - 80 = 8x + 16 8x = 96 x = 12 (2x + 2 + 14x) / 14 = x - 2 2x + 2 + 14x = 14x - 2 2x + 2 = - 2 2x = - 4 x = - 2 (16x + 24 - 25x + 10) / 40 = (31x + 27) / 6…