ដោះស្រាយ 3x-3/x+5leq2

$$\frac { 3 x - 3 } { x + 5 } \leq 2$$

ដោះស្រាយសម្រាប់ x

$$x\in (-5,13]$$

ជំហានចម្លើយ

Denominator x+5 cannot be zero since division by zero is not defined. There are two cases.

$$x+5>0 x+5<0$$

Consider the case when x+5 is positive. Move 5 to the right hand side.

$$x>-5$$

The initial inequality does not change the direction when multiplied by x+5 for x+5>0.

$$3x-3\leq 2\left(x+5\right)$$

Multiply out the right hand side.

$$3x-3\leq 2x+10$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$3x-2x\leq 3+10$$

បន្សំតួដូចគ្នា។

$$x\leq 13$$

Consider condition x>-5 specified above.

$$x\in (-5,13]$$

Now consider the case when x+5 is negative. Move 5 to the right hand side.

$$x<-5$$

The initial inequality changes the direction when multiplied by x+5 for x+5<0.

$$3x-3\geq 2\left(x+5\right)$$

Multiply out the right hand side.

$$3x-3\geq 2x+10$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$3x-2x\geq 3+10$$

បន្សំតួដូចគ្នា។

$$x\geq 13$$

Consider condition x<-5 specified above.

$$x\in \emptyset$$

ចម្លើយចុងក្រោយ គឺជាប្រជុំនៃចម្លើយដែលទទួលបាន។

$$x\in (-5,13]$$