Solve 10x+10/10x+1leq1.1

$$\frac { 10 x + 10 } { 10 x + 1 } \leq 1.1$$

Solve for x

$$x\in (-\infty,-\frac{1}{10})\cup [8.9,\infty)$$

Solution Steps

Denominator 10x+1 cannot be zero since division by zero is not defined. There are two cases.

$$10x+1>0 10x+1<0$$

Consider the case when 10x+1 is positive. Move 1 to the right hand side.

$$10x>-1$$

Divide both sides by 10. Since 10 is positive, the inequality direction remains the same.

$$x>-\frac{1}{10}$$

The initial inequality does not change the direction when multiplied by 10x+1 for 10x+1>0.

$$10x+10\leq 1.1\left(10x+1\right)$$

Multiply out the right hand side.

$$10x+10\leq 11x+1.1$$

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

$$10x-11x\leq -10+1.1$$

Combine like terms.

$$-x\leq -8.9$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x\geq 8.9$$

Now consider the case when 10x+1 is negative. Move 1 to the right hand side.

$$10x<-1$$

Divide both sides by 10. Since 10 is positive, the inequality direction remains the same.

$$x<-\frac{1}{10}$$

The initial inequality changes the direction when multiplied by 10x+1 for 10x+1<0.

$$10x+10\geq 1.1\left(10x+1\right)$$

Multiply out the right hand side.

$$10x+10\geq 11x+1.1$$

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

$$10x-11x\geq -10+1.1$$

Combine like terms.

$$-x\geq -8.9$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x\leq 8.9$$

Consider condition x<-\frac{1}{10} specified above.

$$x<-\frac{1}{10}$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-\frac{1}{10})\cup [8.9,\infty)$$