How do you solve $4 (x - 1) - 2 (3 x + 5) = - 3 x - 1$ $4(x-1)-2(3x+5)=-3x-1$ ?

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How do you solve $4 (x - 1) - 2 (3 x + 5) = - 3 x - 1$ $4(x-1)-2(3x+5)=-3x-1$ ?

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Answer 1 · 2017-02-21T22:20:12

See the entire solution process below:

Explanation:

First, expand the two terms within parenthesis on the left side of the equation:

$4 (x - 1) - 2 (3 x + 5) = - 3 x - 1$ $color(red)(4)(x - 1) - color(blue)(2)(3x + 5) = -3x - 1$

$(4 \times x) - (4 \times 1) - (2 \times 3 x) - (2 \times 5) = - 3 x - 1$ $(color(red)(4) xx x) - (color(red)(4) xx 1) - (color(blue)(2) xx 3x) - (color(blue)(2) xx 5) = -3x - 1$

$4 x - 4 - 6 x - 10 = - 3 x - 1$ $4x - 4 - 6x - 10 = -3x - 1$

Next, combine like terms on the left side of the equation:

$4 x - 6 x - 4 - 10 = - 3 x - 1$ $4x - 6x - 4 - 10 = -3x - 1$

$(4 - 6) x - 14 = - 3 x - 1$ $(4 - 6)x - 14 = -3x - 1$

$- 2 x - 14 = - 3 x - 1$ $-2x - 14 = -3x - 1$

Now, add $3 x$ $color(red)(3x)$ and $14$ $color(blue)(14)$ to each side of the equation to solve for $x$ $x$ while keeping the equation balanced:

$- 2 x - 14 + 3 x + 14 = - 3 x - 1 + 3 x + 14$ $-2x - 14 + color(red)(3x) + color(blue)(14) = -3x - 1 + color(red)(3x) + color(blue)(14)$

$- 2 x + 3 x - 14 + 14 = - 3 x + 3 x - 1 + 14$ $-2x + color(red)(3x) - 14 + color(blue)(14) = -3x + color(red)(3x) - 1 + color(blue)(14)$

$(- 2 + 3) x - 0 = 0 + 13$ $(-2 + 3)x - 0 = 0 + 13$

$1 x = 13$ $1x = 13$

$x = 13$ $x = 13$

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How do you solve $4 (x - 1) - 2 (3 x + 5) = - 3 x - 1$ $4(x-1)-2(3x+5)=-3x-1$ ?

1 Answer

Explanation:

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How do you solve 4(x−1)−2(3x+5)=−3x−14(x-1)-2(3x+5)=-3x-1?

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How do you solve $4 (x - 1) - 2 (3 x + 5) = - 3 x - 1$ $4(x-1)-2(3x+5)=-3x-1$ ?