Solve 3x-2/3leq1/2x+4

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

Solve for x

$$x\leq \frac{28}{15}$$

Solution Steps

Subtract \frac{1}{2}x from both sides.

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

Combine 3x and -\frac{1}{2}x to get \frac{5}{2}x.

$$\frac{5}{2}x-\frac{2}{3}\leq 4$$

Add \frac{2}{3} to both sides.

$$\frac{5}{2}x\leq 4+\frac{2}{3}$$

Convert 4 to fraction \frac{12}{3}.

$$\frac{5}{2}x\leq \frac{12}{3}+\frac{2}{3}$$

Since \frac{12}{3} and \frac{2}{3} have the same denominator, add them by adding their numerators.

$$\frac{5}{2}x\leq \frac{12+2}{3}$$

Add 12 and 2 to get 14.

$$\frac{5}{2}x\leq \frac{14}{3}$$

Multiply both sides by \frac{2}{5}, the reciprocal of \frac{5}{2}. Since \frac{5}{2} is positive, the inequality direction remains the same.

$$x\leq \frac{14}{3}\times \frac{2}{5}$$

Multiply \frac{14}{3} times \frac{2}{5} by multiplying numerator times numerator and denominator times denominator.

$$x\leq \frac{14\times 2}{3\times 5}$$

Do the multiplications in the fraction \frac{14\times 2}{3\times 5}.

$$x\leq \frac{28}{15}$$