Solve the Inequality for x x+8<=3x+1

$x + 8 \leq 3 x + 1$

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

Subtract $3 x$ from both sides of the inequality.

$x + 8 - 3 x \leq 1$

Subtract $3 x$ from $x$ .

$- 2 x + 8 \leq 1$

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

Subtract $8$ from both sides of the inequality.

$- 2 x \leq 1 - 8$

Subtract $8$ from $1$ .

$- 2 x \leq - 7$

Divide each term in $- 2 x \leq - 7$ by $- 2$ and simplify.

Divide each term in $- 2 x \leq - 7$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 2 x}{- 2} \geq \frac{- 7}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} \geq \frac{- 7}{- 2}$

Divide $x$ by $1$ .

$x \geq \frac{- 7}{- 2}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x \geq \frac{7}{2}$

The result can be shown in multiple forms.

Inequality Form:

$x \geq \frac{7}{2}$

Interval Notation:

$[\frac{7}{2}, \infty)$

Solve the Inequality for x x+8<=3x+1

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