Solve 6x+2/3x-2-5leq0

$$\frac { 6 x + 2 } { 3 x - 2 } - 5 \leq 0$$

Solve for x

$$x\in (-\infty,\frac{2}{3})\cup [\frac{4}{3},\infty)$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 5 times \frac{3x-2}{3x-2}.

$$\frac{6x+2}{3x-2}-\frac{5\left(3x-2\right)}{3x-2}\leq 0$$

Since \frac{6x+2}{3x-2} and \frac{5\left(3x-2\right)}{3x-2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{6x+2-5\left(3x-2\right)}{3x-2}\leq 0$$

Do the multiplications in 6x+2-5\left(3x-2\right).

$$\frac{6x+2-15x+10}{3x-2}\leq 0$$

Combine like terms in 6x+2-15x+10.

$$\frac{-9x+12}{3x-2}\leq 0$$

For the quotient to be ≤0, one of the values 12-9x and 3x-2 has to be ≥0, the other has to be ≤0, and 3x-2 cannot be zero. Consider the case when 12-9x\geq 0 and 3x-2 is negative.

$$12-9x\geq 0 3x-2<0$$

The solution satisfying both inequalities is x<\frac{2}{3}.

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

Consider the case when 12-9x\leq 0 and 3x-2 is positive.

$$12-9x\leq 0 3x-2>0$$

The solution satisfying both inequalities is x\geq \frac{4}{3}.

$$x\geq \frac{4}{3}$$

The final solution is the union of the obtained solutions.

$$x<\frac{2}{3}\text{; }x\geq \frac{4}{3}$$