# How do you solve 3x+1=1/2?

Oct 8, 2015

$x = - \frac{1}{6}$

#### Explanation:

Subtract $1$ from both sides

$3 x + 1 - 1 = \frac{1}{2} - 1$

$3 x = - \frac{1}{2}$

And then divide both sides by $3$

$\frac{3 x}{3} = \frac{- \frac{1}{2}}{3}$

$- \frac{1}{2}$ divided by $3$ can also be written as $- \frac{1}{2}$ multiplied by $\frac{1}{3}$

$x = \left(- \frac{1}{2}\right) \cdot \frac{1}{3}$

$x = - \frac{1}{6}$

OR in simpler terms* *

Bring $1$ to the other side, making sure to change its sign

$3 x = \frac{1}{2} - 1$

$3 x = - \frac{1}{2}$

And divide both numerators by $3$

$\frac{3 x}{3} = - \frac{1}{2 \cdot 3}$

$x = - \frac{1}{6}$

Oct 8, 2015

$x = - \frac{1}{6}$

#### Explanation:

People use the term "balance the equation" to describe how to solve this. Consider 5 = 5. Which is true. Write it as (5) = (5), the brackets are only there to show what is the original 'elements' of the equation as a group. Now consider the false statement that (5) -1 = 5. To balance this we need to write (5) -1 = (5) - 1. We have applied the same process to both sides.

Solving the original equation using the same idea:
The target is to get $x$ on one side of the equals sign and everything else on the other.

First isolate all the elements with $x$ in them. That is "$3 x$"

$\left(3 x + 1\right) = \left(\frac{1}{2}\right)$

Step1. Subtract 1 from both sides giving:
$\left(3 x + 1\right) - 1 = \left(\frac{1}{2}\right) - 1$
$3 x = - \frac{1}{2}$

Step2. Now we need to separate the 3 from the $x$ and move it to the other side of the equals sign. This is done by changing 3 into $1$ as $1 \times x = x$. Dividing by three is the same as multiply by $\frac{1}{3}$.

Divide both sides by three

$\left(3 x\right) \times \frac{1}{3} = \left(- \frac{1}{2}\right) \times \frac{1}{3}$

$\frac{3}{3} \times x = \frac{- 1 \times 1}{2 \times 3}$

thus $x = - \frac{1}{6}$