Solve the Inequality for x 4-3x<=-(1+8x)

$4 - 3 x \leq - (1 + 8 x)$

Step 1

Simplify $- (1 + 8 x)$ .

Rewrite.

$4 - 3 x \leq 0 + 0 - (1 + 8 x)$

Simplify by adding zeros.

$4 - 3 x \leq - (1 + 8 x)$

Apply the distributive property.

$4 - 3 x \leq - 1 \cdot 1 - (8 x)$

Multiply.

Multiply $- 1$ by $1$ .

$4 - 3 x \leq - 1 - (8 x)$

Multiply $8$ by $- 1$ .

$4 - 3 x \leq - 1 - 8 x$

Step 2

Move all terms containing $x$ to the left side of the inequality.

Add $8 x$ to both sides of the inequality.

$4 - 3 x + 8 x \leq - 1$

Add $- 3 x$ and $8 x$ .

$4 + 5 x \leq - 1$

Step 3

Move all terms not containing $x$ to the right side of the inequality.

Subtract $4$ from both sides of the inequality.

$5 x \leq - 1 - 4$

Subtract $4$ from $- 1$ .

$5 x \leq - 5$

Step 4

Divide each term in $5 x \leq - 5$ by $5$ and simplify.

Divide each term in $5 x \leq - 5$ by $5$ .

$\frac{5 x}{5} \leq \frac{- 5}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} \leq \frac{- 5}{5}$

Divide $x$ by $1$ .

$x \leq \frac{- 5}{5}$

Simplify the right side.

Divide $- 5$ by $5$ .

$x \leq - 1$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 1$

Interval Notation:

$(- \infty, - 1]$