Solve 1-3(x+1)<4(x-1)+3

$$1 - 3 ( x + 1 ) < 4 ( x - 1 ) + 3$$

Solve for x

$$x>-\frac{1}{7}$$

Solution Steps

Use the distributive property to multiply -3 by x+1.

$$1-3x-3<4\left(x-1\right)+3$$

Subtract 3 from 1 to get -2.

$$-2-3x<4\left(x-1\right)+3$$

Use the distributive property to multiply 4 by x-1.

$$-2-3x<4x-4+3$$

Add -4 and 3 to get -1.

$$-2-3x<4x-1$$

Subtract 4x from both sides.

$$-2-3x-4x<-1$$

Combine -3x and -4x to get -7x.

$$-2-7x<-1$$

Add 2 to both sides.

$$-7x<-1+2$$

Add -1 and 2 to get 1.

$$-7x<1$$

Divide both sides by -7. Since -7 is negative, the inequality direction is changed.

$$x>-\frac{1}{7}$$