Solve 2x+3/3x-7>4

$$\frac { 2 x + 3 } { 3 x - 7 } > 4$$

Solve for x

$$x\in \left(\frac{7}{3},\frac{31}{10}\right)$$

Solution Steps

Denominator 3x-7 cannot be zero since division by zero is not defined. There are two cases.

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

Consider the case when 3x-7 is positive. Move -7 to the right hand side.

$$3x>7$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

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

The initial inequality does not change the direction when multiplied by 3x-7 for 3x-7>0.

$$2x+3>4\left(3x-7\right)$$

Multiply out the right hand side.

$$2x+3>12x-28$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$2x-12x>-3-28$$

Combine like terms.

$$-10x>-31$$

Divide both sides by -10. Since -10 is negative, the inequality direction is changed.

$$x<\frac{31}{10}$$

Consider condition x>\frac{7}{3} specified above.

$$x\in \left(\frac{7}{3},\frac{31}{10}\right)$$

Now consider the case when 3x-7 is negative. Move -7 to the right hand side.

$$3x<7$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

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

The initial inequality changes the direction when multiplied by 3x-7 for 3x-7<0.

$$2x+3<4\left(3x-7\right)$$

Multiply out the right hand side.

$$2x+3<12x-28$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$2x-12x<-3-28$$

Combine like terms.

$$-10x<-31$$

Divide both sides by -10. Since -10 is negative, the inequality direction is changed.

$$x>\frac{31}{10}$$

Consider condition x<\frac{7}{3} specified above.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{7}{3},\frac{31}{10}\right)$$