Question

Sendo U = R, resolva, no caderno, as equações de 1° grau com uma incógnita e determine o conjunto solução de cada uma delas.​

Answer 1

Resposta:

$c) \: 3 - (3x - 6) = 2x + 1 \times (4 - x) \\ 3 - 3x + 6 = 2x + 4 - x \\ - 3x + 9 = x + 4 \\ - 3x + 9 - x = 4 \\ - 3x - x = 4 - 9 \\ - 4x = - 5 \times ( - 1) \\ 4x = 5 \\ \\ x = \frac{5}{4}$

$d) \: \frac{2x}{5} - 4 = \frac{x}{4} + 2 \\ \\ \frac{8x - 80 = 5x + 40}{20} \\ \\ 8x - 80 = 5x + 40 \\ 8x - 80 - 5x = 40 \\ 8x - 5x = 40 + 80 \\ 3x = 120 \div (3) \\ x = 40$

$e) \: 20x - 13 = 47 \\ 20x = 47 + 13 \\ 20x = 60 \\ \\ x = \frac{60}{20} \\ \\ x = 3$

$f) \: \frac{x}{2} - 4 = \frac{x}{3} + \frac{1}{3} \\ \\ \frac{3x - 24 = 2x + 2}{6} \\ \\ 3x - 24 = 2x + 2 \\ 3x - 24 - 2x = 2 \\ 3x - 2x = 2 + 24 \\ x = 26$

$g) \: 4 \times (2x + 1) = 5 \times (3x - 2) \\ 8x + 4 = 15x - 10 \\ 8x + 4 - 15x = - 10 \\ 8x - 15x = - 10 - 4 \\ - 7x = - 14 \times ( - 1) \\ 7x = 14 \div (7) \\ x = 2$

$h) \: 21x - 1 = 11x - 6 \\ 21x - 1 - 11x = - 6 \\ 21x - 11x = - 6 + 1 \\ 10x = - 5 \div (5) \\ 2x = - 1 \\ \\ x = - \frac{1}{2}$

$i) \: \frac{3x}{2} + 2 = 3x - 2 \\ \\ \frac{3x + 4 = 6x - 4}{2} \\ \\ 3x + 4 = 6x - 4 \\ 3x + 4 - 6x = - 4 \\ 3x - 6x = - 4 - 4 \\ - 3x = - 8 \times ( - 1) \\ 3x = 8 \\ \\ x = \frac{8}{3}$

$j) \: 8 \times (x - 3) + 6 = 2 \times (2x + 10) - 4 \\ 8x - 24 + 6 = 4x + 20 - 4 \\ 8x - 18 = 4x + 16 \\ 8x - 18 - 4x = 16 \\ 8x - 4x = 16 + 18 \\ 4x = 34 \div (2) \\ 2x = 17 \\ \\ x = \frac{17}{2}$

$k) \: 14x - 10 = 2 \times ( - 5x + 7) \\14x - 10 = - 10x + 14 \\ 14x - 10 + 10x = 14 \\ 14x + 10x = 14 + 10 \\ 24x = 24 \div (24) \\ x = 1$

$l) \: \frac{x}{5} + 24 = x \\ \\ \frac{x + 120 = 5x}{5} \\ \\ x + 120 = 5x \\ x = 5x - 120 \\ x - 5x = - 120 \\ - 4x = - 120 \times ( - 1) \\ 4x = 120 \\ \\ x = \frac{120}{4} \\ \\ x = 30$