How do you solve $(x + 5) - 2 (4 x - 1) = 0$ $(x+5) - 2(4x-1)=0$ ?

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Answer 1 · 2017-07-13T20:44:45

See a solution process below:

First, expand the terms in the right parenthesis by multiply each term within the parenthesis by the term outside the parenthesis:

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

$(x + 5) - (2 \times 4 x) + (2 \times 1) = 0$ $(x + 5) - (color(red)(2) xx 4x) + (color(red)(2) xx 1) = 0$

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

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

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

$1 x - 8 x + 5 + 2 = 0$ $1x - 8x + 5 + 2 = 0$

$(1 - 8) x + (5 + 2) = 0$ $(1 - 8)x + (5 + 2) = 0$

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

Next, subtract $7$ $color(red)(7)$ from each side of the equation to isolate the $x$ $x$ term while keeping the equation balanced:

$- 7 x + 7 - 7 = 0 - 7$ $-7x + 7 - color(red)(7) = 0 - color(red)(7)$

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

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

Now, divide each side of the equation by $- 7$ $color(red)(-7)$ to solve for $x$ $x$ while keeping the equation balanced:

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

$(color(red)(cancel(color(black)(-7)))x)/cancel(color(red)(-7)) = 1$

$x = 1$

How do you solve (x+5) - 2(4x-1)=0(x+5)−2(4x−1)=0(x+5) - 2(4x-1)=0?