# How do you solve for x in 4-6x \le 2(2x+3)?

Dec 19, 2014

To solve an inequality of the first order (this is one because there isn't a higher exponent of x, than 1), you have to first bring every factor of x to one part, and every number to the other part.

$4 - 6 x \le 2 \left(2 x + 3\right)$
$4 - 6 x \le 4 x + 6$ (apply the distributive property)
$4 - 6 x - 6 \le 4 x + 6 - 6$ (substract $6$ from both sides)
$- 2 - 6 x \le 4 x$
$- 2 - 6 x + 6 x \le 4 x + 6 x$ (add $6 x$ to both sides)
$- 2 \le 10 x$
$\frac{- 2}{10} \le \frac{10 x}{10}$ (divide both sides by ten$\text{^"(1)}$)
$- \frac{1}{5} \le x$
This is your answer, you can flip it if that seams easier:
$x \ge - \frac{1}{5}$
The set of solutions is all the numbers that are greater than or equal to $- \frac{1}{5}$.

Hope this helped.

(1): If you were to divide by a negative number, you would have to flip the inequality sign. In this case, $\le$ would become $\ge$