Solve 3(x+3)<6(x-1)+7

$$3 ( x + 3 ) < 6 ( x - 1 ) + 7$$

Solve for x

$$x>\frac{8}{3}$$

Solution Steps

Use the distributive property to multiply 3 by x+3.

$$3x+9<6\left(x-1\right)+7$$

Use the distributive property to multiply 6 by x-1.

$$3x+9<6x-6+7$$

Add -6 and 7 to get 1.

$$3x+9<6x+1$$

Subtract 6x from both sides.

$$3x+9-6x<1$$

Combine 3x and -6x to get -3x.

$$-3x+9<1$$

Subtract 9 from both sides.

$$-3x<1-9$$

Subtract 9 from 1 to get -8.

$$-3x<-8$$

Divide both sides by -3. Since -3 is negative, the inequality direction is changed.

$$x>\frac{-8}{-3}$$

Fraction \frac{-8}{-3} can be simplified to \frac{8}{3} by removing the negative sign from both the numerator and the denominator.

$$x>\frac{8}{3}$$