Question

resuelve los siguientes ejercicios de Ecuaciones 1 grado con productos indicados. x + 3(x-1) = 6 – 4(2x+3)

5(x-1) + 16(2x+3) = 3(2x-7) – x

2(3x+3) – 4(5x-3) = x(x-3) – x(x+5)

184 – 7(2x+5) = 301 + 6(x-1) – 6

7(18-x) – 6(3-5x) = 2(7x+9) – 3(2x+5) – 12

3(x-3) + 5(x+7) – x(x+1) – 2(x+7) + 4 = 0

-3(2x+7) + (-5x+6) – 8(1-2x) – (x-3) = 0

Answer 1

Respuesta:

5(x-1) + 16(2x+3) = 3(2x-7) – x

5x-5 + 32x+48 = 6x-21 - x

37x + 43 = 5x - 21

37x - 5x = -43 - 21

32x = -64

x = -64/32

x = -2

2(3x+3) – 4(5x-3) = x(x-3) – x(x+5)

6x + 6 - 20x + 12 = $x^{2}$ - 3x - $x^{2}$ - 5x

- 14x + 18 = - 8x

- 14x + 8x = -18

- 6x = -18

x = -18 / -6

x = 3

184 – 7(2x+5) = 301 + 6(x-1) – 6

184 - 14x - 35 = 301 + 6x - 6 -6

- 14x - 149 = 289 + 6x

- 14x - 6x = 289 + 149

- 20x = 140

x = 140 / -20

x = - 7

7(18-x) – 6(3-5x) = 2(7x+9) – 3(2x+5) – 12

126 - 7x - 18 + 30x = 14x + 18 - 6x - 15 -12

23x + 108 = 8x - 9

23x -8x = -108 -9

15x = - 117

x = -117 / 15

x = - 7.8

3(x-3) + 5(x+7) – x(x+1) – 2(x+7) + 4 = 0

3x - 9 + 5x + 35 - $x^{2}$ - x - 2x - 14 + 4 = 0

-$x^{2}$ + 5x + 16 = 0

-$x^{2}$ + 5x = - 16

-$x^{2}$ + x = -16 / 5

-$x^{2}$ + x = - 3.2

x + x = $\sqrt{3.2}$

2x = 1.78

x = 1.78 / 2

x = .89

-3(2x+7) + (-5x+6) – 8(1-2x) – (x-3) = 0

- 6x - 21 - 5x + 6 - 8 + 16x - x + 3 = 0

4x - 20 = 0

4x = 20

x = 20 / 4

x = 5

Answer 2

Respuesta:

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