Solve the System of Inequalities 1/8x-5<-8 or 7x-4>=38

$\frac{1}{8} x - 5 < - 8$ or $7 x - 4 \geq 38$

Simplify the first inequality.

Combine $\frac{1}{8}$ and $x$ .

$\frac{x}{8} - 5 < - 8$ or $7 x - 4 \geq 38$

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

Add $5$ to both sides of the inequality.

$\frac{x}{8} < - 8 + 5$ or $7 x - 4 \geq 38$

Add $- 8$ and $5$ .

$\frac{x}{8} < - 3$ or $7 x - 4 \geq 38$

Multiply both sides by $8$ .

$\frac{x}{8} \cdot 8 < - 3 \cdot 8$ or $7 x - 4 \geq 38$

Simplify.

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{x}{8} \cdot 8 < - 3 \cdot 8$ or $7 x - 4 \geq 38$

Rewrite the expression.

$x < - 3 \cdot 8$ or $7 x - 4 \geq 38$

Simplify the right side.

Multiply $- 3$ by $8$ .

$x < - 24$ or $7 x - 4 \geq 38$

Simplify the second inequality.

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

Add $4$ to both sides of the inequality.

$x < - 24$ or $7 x \geq 38 + 4$

Add $38$ and $4$ .

$x < - 24$ or $7 x \geq 42$

Divide each term in $7 x \geq 42$ by $7$ and simplify.

Divide each term in $7 x \geq 42$ by $7$ .

$x < - 24$ or $\frac{7 x}{7} \geq \frac{42}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$x < - 24$ or $\frac{7 x}{7} \geq \frac{42}{7}$

Divide $x$ by $1$ .

$x < - 24$ or $x \geq \frac{42}{7}$

Simplify the right side.

Divide $42$ by $7$ .

$x < - 24$ or $x \geq 6$

The result can be shown in multiple forms.

Inequality Form:

$x < - 24 or x \geq 6$

Interval Notation:

$(- \infty, - 24) \cup [6, \infty)$