Solve the System of Inequalities r/3<-6 OR 3r+1>100

$\frac{r}{3} < - 6$ OR $3 r + 1 > 100$

Solve $\frac{r}{3} < - 6$ for $r$ .

Multiply both sides by $3$ .

$\frac{r}{3} \cdot 3 < - 6 \cdot 3$

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{r}{3} \cdot 3 < - 6 \cdot 3$

Rewrite the expression.

$r < - 6 \cdot 3$

Simplify the right side.

Multiply $- 6$ by $3$ .

$r < - 18$

Solve $3 r + 1 > 100$ for $r$ .

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

Subtract $1$ from both sides of the inequality.

$3 r > 100 - 1$

Subtract $1$ from $100$ .

$3 r > 99$

Divide each term in $3 r > 99$ by $3$ and simplify.

Divide each term in $3 r > 99$ by $3$ .

$\frac{3 r}{3} > \frac{99}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 r}{3} > \frac{99}{3}$

Divide $r$ by $1$ .

$r > \frac{99}{3}$

Simplify the right side.

Divide $99$ by $3$ .

$r > 33$

Find the intersection of $r < - 18$ and $r > 33$ .

No solution