Solve 3+2x/4>x

$$\frac { 3 + 2 x } { 4 } > x$$

Solve for x

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

Solution Steps

Divide each term of 3+2x by 4 to get \frac{3}{4}+\frac{1}{2}x.

$$\frac{3}{4}+\frac{1}{2}x>x$$

Subtract x from both sides.

$$\frac{3}{4}+\frac{1}{2}x-x>0$$

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

$$\frac{3}{4}-\frac{1}{2}x>0$$

Subtract \frac{3}{4} from both sides. Anything subtracted from zero gives its negation.

$$-\frac{1}{2}x>-\frac{3}{4}$$

Multiply both sides by -2, the reciprocal of -\frac{1}{2}. Since -\frac{1}{2} is negative, the inequality direction is changed.

$$x<-\frac{3}{4}\left(-2\right)$$

Express -\frac{3}{4}\left(-2\right) as a single fraction.

$$x<\frac{-3\left(-2\right)}{4}$$

Multiply -3 and -2 to get 6.

$$x<\frac{6}{4}$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

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