Solve 3-4x/x-3-1<2x+3/x-3

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

Solve for x

$$x\in \left(-\infty,\frac{3}{7}\right)\cup \left(3,\infty\right)$$

Solution Steps

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

$$\frac{3-4x}{x-3}-\frac{x-3}{x-3}<\frac{2x+3}{x-3}$$

Since \frac{3-4x}{x-3} and \frac{x-3}{x-3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3-4x-\left(x-3\right)}{x-3}<\frac{2x+3}{x-3}$$

Do the multiplications in 3-4x-\left(x-3\right).

$$\frac{3-4x-x+3}{x-3}<\frac{2x+3}{x-3}$$

Combine like terms in 3-4x-x+3.

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

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

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

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

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

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

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

Combine like terms in 6-5x-2x-3.

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

For the quotient to be negative, 3-7x and x-3 have to be of the opposite signs. Consider the case when 3-7x is positive and x-3 is negative.

$$3-7x>0 x-3<0$$

The solution satisfying both inequalities is x<\frac{3}{7}.

$$x<\frac{3}{7}$$

Consider the case when x-3 is positive and 3-7x is negative.

$$x-3>0 3-7x<0$$

The solution satisfying both inequalities is x>3.

$$x>3$$

The final solution is the union of the obtained solutions.

$$x<\frac{3}{7}\text{; }x>3$$