Solve 3x-1/3leq3/2x

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

Solve for x

$$x\leq \frac{2}{9}$$

Solution Steps

Subtract \frac{3}{2}x from both sides.

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

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

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

Add \frac{1}{3} to both sides. Anything plus zero gives itself.

$$\frac{3}{2}x\leq \frac{1}{3}$$

Multiply both sides by \frac{2}{3}, the reciprocal of \frac{3}{2}. Since \frac{3}{2} is positive, the inequality direction remains the same.

$$x\leq \frac{1}{3}\times \frac{2}{3}$$

Multiply \frac{1}{3} times \frac{2}{3} by multiplying numerator times numerator and denominator times denominator.

$$x\leq \frac{1\times 2}{3\times 3}$$

Do the multiplications in the fraction \frac{1\times 2}{3\times 3}.

$$x\leq \frac{2}{9}$$