Solve 6x+2/3x-1-5leq0

$$\frac{ 6x+2 }{ 3x-1 } -5 \leq 0$$

Solve for x

$$x\in (-\infty,\frac{1}{3})\cup [\frac{7}{9},\infty)$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 5 times \frac{3x-1}{3x-1}.

$$\frac{6x+2}{3x-1}-\frac{5\left(3x-1\right)}{3x-1}\leq 0$$

Since \frac{6x+2}{3x-1} and \frac{5\left(3x-1\right)}{3x-1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{6x+2-5\left(3x-1\right)}{3x-1}\leq 0$$

Do the multiplications in 6x+2-5\left(3x-1\right).

$$\frac{6x+2-15x+5}{3x-1}\leq 0$$

Combine like terms in 6x+2-15x+5.

$$\frac{-9x+7}{3x-1}\leq 0$$

For the quotient to be ≤0, one of the values 7-9x and 3x-1 has to be ≥0, the other has to be ≤0, and 3x-1 cannot be zero. Consider the case when 7-9x\geq 0 and 3x-1 is negative.

$$7-9x\geq 0 3x-1<0$$

The solution satisfying both inequalities is x<\frac{1}{3}.

$$x<\frac{1}{3}$$

Consider the case when 7-9x\leq 0 and 3x-1 is positive.

$$7-9x\leq 0 3x-1>0$$

The solution satisfying both inequalities is x\geq \frac{7}{9}.

$$x\geq \frac{7}{9}$$

The final solution is the union of the obtained solutions.

$$x<\frac{1}{3}\text{; }x\geq \frac{7}{9}$$