Solve 3(r-6)+2<4(r+2)-20

$$3 ( r - 6 ) + 2 < 4 ( r + 2 ) - 20$$

Solve for r

$$r>-4$$

Solution Steps

Use the distributive property to multiply 3 by r-6.

$$3r-18+2<4\left(r+2\right)-20$$

Add -18 and 2 to get -16.

$$3r-16<4\left(r+2\right)-20$$

Use the distributive property to multiply 4 by r+2.

$$3r-16<4r+8-20$$

Subtract 20 from 8 to get -12.

$$3r-16<4r-12$$

Subtract 4r from both sides.

$$3r-16-4r<-12$$

Combine 3r and -4r to get -r.

$$-r-16<-12$$

Add 16 to both sides.

$$-r<-12+16$$

Add -12 and 16 to get 4.

$$-r<4$$

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

$$r>-4$$