# How do you solve - 3= 3- 4( x + 1)?

May 24, 2018

$x = 0.5$

#### Explanation:

$- 3 = 3 - 4 \left(x + 1\right)$

$- 3 - 3 = - 4 \left(x + 1\right)$

$- 6 = - 4 \left(x + 1\right)$

$\frac{- 6}{-} 4 = \frac{- 4 \left(x + 1\right)}{-} 4$

$1.5 = x + 1$

So

$1.5 - 1 = x$

$x = 0.5$

May 24, 2018

$x = \frac{1}{2}$

#### Explanation:

$- 3 = 3 - \setminus \textcolor{red}{4 \left(x + 1\right)}$

Let's solve the red part. Do the Distributive Property:
$- 4 \left(x + 1\right) = - 4 \left(x\right) + \left(- 4\right) \left(1\right) = - 4 x - 4$

So now we have
$- 3 = 3 + \left[\setminus \textcolor{red}{- 4 x - 4}\right]$
$\setminus \Rightarrow - 3 = 3 - 4 x - 4$

Add anything without an $x$ and move to the left side.
$- 3 = 3 - 4 - 4 x$
$\setminus \Rightarrow - 3 = - 1 - 4 x$
$\setminus \Rightarrow - 2 = - 4 x$

Divide to find $x$
$- 2 \setminus \div \left(- 4\right) = x$
$\frac{1}{2} = x$

-3\stackrel(?)(=)3-4(\color(seagreen)(1/2)+1)
-3\stackrel(?)(=)3-4(1/2)-4(1)
-3\stackrel(?)(=)3-2-4
-3\stackrel(?)(=)1-4
$- 3 = - 3 \setminus \textcolor{l i g h t g r e e n}{\setminus \sqrt{}}$