# How do you solve |7 - 6x| <= -4?

Aug 30, 2016

$x \ge + \frac{11}{6} \mathmr{and} x \le \frac{1}{2}$

Set the +value $\le - 4 \mathmr{and} - \text{value} \le - 4$ and solve.

#### Explanation:

The absolute value can be either a negative or a positive value. It is the distance from the value to zero. So the answer that comes out is the same regardless if the answer inside is negative or positive.

Set $+ 1 \times \left(7 - 6 x\right) \le - 4$ and solve for the positive value.

This gives

$7 - 6 x \le - 4 \text{ }$ subtract -7 from both sides.

$7 - 7 - 6 x \le - 4 - 7 \text{ }$ This gives

$- 6 x \le - 11 \text{ }$ divide everything by -1 . The sign will change.

$\frac{- x}{-} 1 \ge \frac{- 11}{6 \times - 1} \text{ }$ This gives

$x \ge + \frac{11}{6}$

$\frac{- 1}{-} 1 = + 1$ the opposite of -a - is +a (Dividing by a negative always gives you the opposite of what you start with)

When you divide an inequality by a negative number the inequality sign in the middle changes around.

$\frac{- 11}{- 1 \times 6} = + \frac{11}{6}$ (Dividing by a negative always gives you the opposite sign of what you start with)

Then set $- 1 \left(7 - 6 x\right) \le - 4 \text{ }$ and solve for x .This gives

$- 7 + 6 x \le - 4 \text{ }$ add seven to both sides

$- 7 + 7 + 6 x \le - 4 + 7 \text{ }$ this gives

$6 x \le + 3 \text{ }$ Divide both sides by 6

$\frac{6 x}{6} \le + \frac{3}{6} \text{ }$ this gives

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

Note that as the x is positive, dividing by +1 would not change any of the values, so is not necessary.