Solve x+7/3x-2-1<0

$$\frac { x + 7 } { 3 x - 2 } - 1 < 0$$

Solve for x

$$x\in \left(-\infty,\frac{2}{3}\right)\cup \left(\frac{9}{2},\infty\right)$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3x-2}{3x-2}.

$$\frac{x+7}{3x-2}-\frac{3x-2}{3x-2}<0$$

Since \frac{x+7}{3x-2} and \frac{3x-2}{3x-2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x+7-\left(3x-2\right)}{3x-2}<0$$

Do the multiplications in x+7-\left(3x-2\right).

$$\frac{x+7-3x+2}{3x-2}<0$$

Combine like terms in x+7-3x+2.

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

For the quotient to be negative, 9-2x and 3x-2 have to be of the opposite signs. Consider the case when 9-2x is positive and 3x-2 is negative.

$$9-2x>0 3x-2<0$$

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

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

Consider the case when 3x-2 is positive and 9-2x is negative.

$$3x-2>0 9-2x<0$$

The solution satisfying both inequalities is x>\frac{9}{2}.

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

The final solution is the union of the obtained solutions.

$$x<\frac{2}{3}\text{; }x>\frac{9}{2}$$