Solve the System of Inequalities 25<-3x+10 or -23>=-3x+10

Subtract ${\displaystyle 10}$ from both sides of the inequality.

${\displaystyle -3x>25-10}$ or ${\displaystyle -23\ge -3x+10}$

Subtract ${\displaystyle 10}$ from ${\displaystyle 25}$.

${\displaystyle -3x>15}$ or ${\displaystyle -23\ge -3x+10}$

Divide each term in ${\displaystyle -3x>15}$ by ${\displaystyle -3}$. When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

${\displaystyle \frac{-3x}{-3}<\frac{15}{-3}}$ or ${\displaystyle -23\ge -3x+10}$

Simplify the left side.

${\displaystyle x<\frac{15}{-3}}$ or ${\displaystyle -23\ge -3x+10}$

Simplify the right side.

${\displaystyle x<-5}$ or ${\displaystyle -23\ge -3x+10}$

Subtract ${\displaystyle 10}$ from both sides of the inequality.

${\displaystyle x<-5}$ or ${\displaystyle -3x\le -23-10}$

Subtract ${\displaystyle 10}$ from ${\displaystyle -23}$.

${\displaystyle x<-5}$ or ${\displaystyle -3x\le -33}$

Divide each term in ${\displaystyle -3x\le -33}$ by ${\displaystyle -3}$. When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

${\displaystyle x<-5}$ or ${\displaystyle \frac{-3x}{-3}\ge \frac{-33}{-3}}$

Simplify the left side.

${\displaystyle x<-5}$ or ${\displaystyle x\ge \frac{-33}{-3}}$

Simplify the right side.

${\displaystyle x<-5}$ or ${\displaystyle x\ge 11}$