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

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

Solve for x

$$x\in \left(-14,-4\right)$$

Solution Steps

Denominator 4+x cannot be zero since division by zero is not defined. There are two cases.

$$4+x>0 4+x<0$$

Consider the case when 4+x is positive. Move 4 to the right hand side.

$$x>-4$$

The initial inequality does not change the direction when multiplied by 4+x for 4+x>0.

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

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

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

Combine like terms.

$$-\frac{1}{2}x>7$$

Divide both sides by -\frac{1}{2}. Since -\frac{1}{2} is negative, the inequality direction is changed.

$$x<-14$$

Consider condition x>-4 specified above.

$$x\in \emptyset$$

Now consider the case when 4+x is negative. Move 4 to the right hand side.

$$x<-4$$

The initial inequality changes the direction when multiplied by 4+x for 4+x<0.

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

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

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

Combine like terms.

$$-\frac{1}{2}x<7$$

Divide both sides by -\frac{1}{2}. Since -\frac{1}{2} is negative, the inequality direction is changed.

$$x>-14$$

Consider condition x<-4 specified above.

$$x\in \left(-14,-4\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(-14,-4\right)$$