Como resolver $(x + 5) - 2 (4 x - 1) = 0$ $(x + 5) -2 (4x-1) = 0$ ?

$x = 1$ $x=1$

$(x + 5) - 2 (4 x - 1) = 0$ $(x+5) - 2 ( 4x - 1) = 0$

$(x + 5) - 2 \cdot (4 x - 1) = 0$ $(x+5) - 2 * ( 4x - 1) = 0$

$x + 5 + (- 2) \cdot (4 x) + (- 2) \cdot (- 1) = 0$ $x+5 + (-2) * (4x) + ( -2) * (- 1) = 0$

$x + 5 - 8 x + 2 = 0$ $x+5 -8x +2 = 0$

$x - 8 x + 2 + 5 = 0$ $x-8x +2 + 5= 0$

$- 7 x + 7 = 0$ $-7x + 7= 0$

$- 7 x = - 7$ $-7x = - 7$

$x = \frac{- 7}{- 7}$ $x = (-7)/(-7)$

$x = 1$ $x=1$

Como resolver (x + 5) -2 (4x-1) = 0 (x+5)−2(4x−1)=0 (x + 5) -2 (4x-1) = 0 ?