Solve -7*(x-1)-2/3<3*(x+2)+1/3

$$- 7 \cdot ( x - 1 ) - \frac { 2 } { 3 } < 3 \cdot ( x + 2 ) + \frac { 1 } { 3 }$$

Solve for x

$$x>0$$

Solution Steps

Use the distributive property to multiply -7 by x-1.

$$-7x+7-\frac{2}{3}<3\left(x+2\right)+\frac{1}{3}$$

Convert 7 to fraction \frac{21}{3}.

$$-7x+\frac{21}{3}-\frac{2}{3}<3\left(x+2\right)+\frac{1}{3}$$

Since \frac{21}{3} and \frac{2}{3} have the same denominator, subtract them by subtracting their numerators.

$$-7x+\frac{21-2}{3}<3\left(x+2\right)+\frac{1}{3}$$

Subtract 2 from 21 to get 19.

$$-7x+\frac{19}{3}<3\left(x+2\right)+\frac{1}{3}$$

Use the distributive property to multiply 3 by x+2.

$$-7x+\frac{19}{3}<3x+6+\frac{1}{3}$$

Convert 6 to fraction \frac{18}{3}.

$$-7x+\frac{19}{3}<3x+\frac{18}{3}+\frac{1}{3}$$

Since \frac{18}{3} and \frac{1}{3} have the same denominator, add them by adding their numerators.

$$-7x+\frac{19}{3}<3x+\frac{18+1}{3}$$

Add 18 and 1 to get 19.

$$-7x+\frac{19}{3}<3x+\frac{19}{3}$$

Subtract 3x from both sides.

$$-7x+\frac{19}{3}-3x<\frac{19}{3}$$

Combine -7x and -3x to get -10x.

$$-10x+\frac{19}{3}<\frac{19}{3}$$

Subtract \frac{19}{3} from both sides.

$$-10x<\frac{19}{3}-\frac{19}{3}$$

Subtract \frac{19}{3} from \frac{19}{3} to get 0.

$$-10x<0$$

Product of two numbers is <0 if one is >0 and the other is <0. Since -10<0, x must be >0.

$$x>0$$