Solve x^2+18x+17<x^2+6x+9

$${ x }^{ 2 } +18x+17 < { x }^{ 2 } +6x+9$$

Solve for x

$$x<-\frac{2}{3}$$

Solution Steps

Subtract x^{2} from both sides.

$$x^{2}+18x+17-x^{2}<6x+9$$

Combine x^{2} and -x^{2} to get 0.

$$18x+17<6x+9$$

Subtract 6x from both sides.

$$18x+17-6x<9$$

Combine 18x and -6x to get 12x.

$$12x+17<9$$

Subtract 17 from both sides.

$$12x<9-17$$

Subtract 17 from 9 to get -8.

$$12x<-8$$

Divide both sides by 12. Since 12 is positive, the inequality direction remains the same.

$$x<\frac{-8}{12}$$

Reduce the fraction \frac{-8}{12} to lowest terms by extracting and canceling out 4.

$$x<-\frac{2}{3}$$