# How do you solve and write the following in interval notation: 3x+4≥5x-8?

Jun 22, 2016

$x \in \left(- \infty , 6\right]$

#### Explanation:

Remember you can add/subtract the same amount and multiply/divide by any amount greater than zero to both sides of an inequality without invalidating the inequality.

$3 x + 4 \ge 5 x - 8$
$\textcolor{w h i t e}{\text{XXX}} 3 x + 12 \ge 5 x$

$\textcolor{w h i t e}{\text{XXX}} 12 \ge 2 x$

$\textcolor{w h i t e}{\text{XXX}} 6 \ge x$

or (reversing so $x$ is on the left, which is more common)
$\textcolor{w h i t e}{\text{XXX}} x \le 6$

Jun 22, 2016

You may add or subtract on both sides of the inequality.

#### Explanation:

Add $8$ to both sides:
$3 x + 4 + 8 \ge 5 x - \cancel{8} + \cancel{8} \to 3 x + 12 \ge 5 x$

Subtract $3 x$ from both sides:
$\cancel{3 x} - \cancel{3 x} + 12 \ge 5 x - 3 x \to 12 \ge 2 x \to$

Divide by $2$, a positive number:
$6 \ge x \to x \le 6$

$\to x \in \left[6 , \infty\right)$, where [# means 'inclusive'.

Note: you may be used to a different notation, but I think the above is clear enough to translate.