Solve the System of Inequalities (m-2)/3<-2 OR 4m+3>15

$\frac{m - 2}{3} < - 2$ OR $4 m + 3 > 15$

Step 1

Solve $\frac{m - 2}{3} < - 2$ for $m$ .

Multiply both sides by $3$ .

$\frac{m - 2}{3} \cdot 3 < - 2 \cdot 3$

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{m - 2}{3} \cdot 3 < - 2 \cdot 3$

Rewrite the expression.

$m - 2 < - 2 \cdot 3$

Simplify the right side.

Multiply $- 2$ by $3$ .

$m - 2 < - 6$

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

Add $2$ to both sides of the inequality.

$m < - 6 + 2$

Add $- 6$ and $2$ .

$m < - 4$

Step 2

Solve $4 m + 3 > 15$ for $m$ .

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

Subtract $3$ from both sides of the inequality.

$4 m > 15 - 3$

Subtract $3$ from $15$ .

$4 m > 12$

Divide each term in $4 m > 12$ by $4$ and simplify.

Divide each term in $4 m > 12$ by $4$ .

$\frac{4 m}{4} > \frac{12}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 m}{4} > \frac{12}{4}$

Divide $m$ by $1$ .

$m > \frac{12}{4}$

Simplify the right side.

Divide $12$ by $4$ .

$m > 3$

Step 3

Find the intersection of $m < - 4$ and $m > 3$ .

No solution