Solve -7x-7/3x+8>-1

$$\frac { - 7 x - 7 } { 3 x + 8 } > - 1$$

Solve for x

$$x\in \left(-\frac{8}{3},\frac{1}{4}\right)$$

Solution Steps

Denominator 3x+8 cannot be zero since division by zero is not defined. There are two cases.

$$3x+8>0 3x+8<0$$

Consider the case when 3x+8 is positive. Move 8 to the right hand side.

$$3x>-8$$

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

$$x>-\frac{8}{3}$$

The initial inequality does not change the direction when multiplied by 3x+8 for 3x+8>0.

$$-7x-7>-\left(3x+8\right)$$

Multiply out the right hand side.

$$-7x-7>-3x-8$$

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

$$-7x+3x>7-8$$

Combine like terms.

$$-4x>-1$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x<\frac{1}{4}$$

Consider condition x>-\frac{8}{3} specified above.

$$x\in \left(-\frac{8}{3},\frac{1}{4}\right)$$

Now consider the case when 3x+8 is negative. Move 8 to the right hand side.

$$3x<-8$$

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

$$x<-\frac{8}{3}$$

The initial inequality changes the direction when multiplied by 3x+8 for 3x+8<0.

$$-7x-7<-\left(3x+8\right)$$

Multiply out the right hand side.

$$-7x-7<-3x-8$$

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

$$-7x+3x<7-8$$

Combine like terms.

$$-4x<-1$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x>\frac{1}{4}$$

Consider condition x<-\frac{8}{3} specified above.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\frac{8}{3},\frac{1}{4}\right)$$