Solve the System of Inequalities 4x<=12 or 9-x<0

$4 x \leq 12$ or $9 - x < 0$

Divide each term in $4 x \leq 12$ by $4$ and simplify.

Divide each term in $4 x \leq 12$ by $4$ .

$\frac{4 x}{4} \leq \frac{12}{4}$ or $9 - x < 0$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} \leq \frac{12}{4}$ or $9 - x < 0$

Divide $x$ by $1$ .

$x \leq \frac{12}{4}$ or $9 - x < 0$

Simplify the right side.

Divide $12$ by $4$ .

$x \leq 3$ or $9 - x < 0$

Simplify the second inequality.

Subtract $9$ from both sides of the inequality.

$x \leq 3$ or $- x < - 9$

Divide each term in $- x < - 9$ by $- 1$ and simplify.

Divide each term in $- x < - 9$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$x \leq 3$ or $\frac{- x}{- 1} > \frac{- 9}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$x \leq 3$ or $\frac{x}{1} > \frac{- 9}{- 1}$

Divide $x$ by $1$ .

$x \leq 3$ or $x > \frac{- 9}{- 1}$

Simplify the right side.

Divide $- 9$ by $- 1$ .

$x \leq 3$ or $x > 9$

The result can be shown in multiple forms.

Inequality Form:

$x \leq 3 or x > 9$

Interval Notation:

$(- \infty, 3] \cup (9, \infty)$

Solve the System of Inequalities 4x<=12 or 9-x<0

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