Solve -1/4+1/3x>3x-1/2

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

Solve for x

$$x<\frac{3}{32}$$

Solution Steps

Subtract 3x from both sides.

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

Combine \frac{1}{3}x and -3x to get -\frac{8}{3}x.

$$-\frac{1}{4}-\frac{8}{3}x>-\frac{1}{2}$$

Add \frac{1}{4} to both sides.

$$-\frac{8}{3}x>-\frac{1}{2}+\frac{1}{4}$$

Least common multiple of 2 and 4 is 4. Convert -\frac{1}{2} and \frac{1}{4} to fractions with denominator 4.

$$-\frac{8}{3}x>-\frac{2}{4}+\frac{1}{4}$$

Since -\frac{2}{4} and \frac{1}{4} have the same denominator, add them by adding their numerators.

$$-\frac{8}{3}x>\frac{-2+1}{4}$$

Add -2 and 1 to get -1.

$$-\frac{8}{3}x>-\frac{1}{4}$$

Multiply both sides by -\frac{3}{8}, the reciprocal of -\frac{8}{3}. Since -\frac{8}{3} is negative, the inequality direction is changed.

$$x<-\frac{1}{4}\left(-\frac{3}{8}\right)$$

Multiply -\frac{1}{4} times -\frac{3}{8} by multiplying numerator times numerator and denominator times denominator.

$$x<\frac{-\left(-3\right)}{4\times 8}$$

Do the multiplications in the fraction \frac{-\left(-3\right)}{4\times 8}.

$$x<\frac{3}{32}$$