Question

1) Resolva As Equações

a) 5(1-x) - 2x + 1 = - 3 (+ x )

b) 4( x - 3 ) = 2x - 5

c) 2 + 3[ x - (3x + 1 )] = 5(x - (2x - 1)]

d) 1- 2x = x/3 - x/2

e) 3(x - 1) - 2x /5 = 5(x - 3 ) /6

f) 2x + 5 / 3x = 1/4

g) x/3 + x/ 2 = 15

h) x/2 + x/4 = 2 ​

Answer 1

a)

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

b)

$fazendo \: direto \\ 4x - 12 = 2x - 5 \\ 4x - 2x = - 5 + 12 \\ 2x = 7 \\ x = \frac{7}{2} = 3.5$

c)

$2 + 3(x - (3x + 1)) = 5(x - (2x - 1)) \\ 2 + 3(x - 3x - 1)) = 5(x - 2x + 1) \\ 2 + 3x - 9x - 3 = 5x - 10x + 5 \\ 3x - 9x - 5x + 10x = - 5 + 3 - 2 \\ 9x = - 4 \\ x = - \frac{4}{9}$

d)

$1 - 2x = x \div 3 - \times \div 2 \\ 1 = mmc = 6 \\ 1 = 2x - 3x + 12x \\ 1 = 11x$

e)

$3x - 3 - 2x \div 5 = 5x - 15 \div 6 \\ 3x - 5x - 2x \div 5 = - 15 \div 6 + 3 \\ mmc = 5 = mmc = 6 \\ 15x - 25x - 10x = - 15 + 18 \\ - 20x = + 3 \: \: \: \: ( - 1) \\ 20x = - 3$

f)

$2x + 5 \div 3x = \frac{1}{4} \\ mmc = 3x \\ 6x^{2} + 15 = 1 \\ 6 {x}^{2} = - 14 \\ x = - \frac{14}{6} = \frac{7}{3}$

g)

$\frac{x}{3}+ \frac{x}{2} = 15 \\ mmc = 6 \\ 2x + 3x = 15 \\ 5 \times = 15 \\ x = 3$

h)

$\frac{x}{2} + \frac{x}{4} = 2 \\ mmc = 4 \\ 2x + x = 2$

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