Solve 3-x/2<x

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

Solve for x

$$x>1$$

Solution Steps

Divide each term of 3-x by 2 to get \frac{3}{2}-\frac{1}{2}x.

$$\frac{3}{2}-\frac{1}{2}x<x$$

Subtract x from both sides.

$$\frac{3}{2}-\frac{1}{2}x-x<0$$

Combine -\frac{1}{2}x and -x to get -\frac{3}{2}x.

$$\frac{3}{2}-\frac{3}{2}x<0$$

Subtract \frac{3}{2} from both sides. Anything subtracted from zero gives its negation.

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

Multiply both sides by -\frac{2}{3}, the reciprocal of -\frac{3}{2}. Since -\frac{3}{2} is negative, the inequality direction is changed.

$$x>-\frac{3}{2}\left(-\frac{2}{3}\right)$$

Multiply -\frac{3}{2} times -\frac{2}{3} by multiplying numerator times numerator and denominator times denominator.

$$x>\frac{-3\left(-2\right)}{2\times 3}$$

Do the multiplications in the fraction \frac{-3\left(-2\right)}{2\times 3}.

$$x>\frac{6}{6}$$

Divide 6 by 6 to get 1.

$$x>1$$