Solve 2x-3/x+1-1/1geq0

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

Solve for x

$$x\in (-\infty,-1)\cup [4,\infty)$$

Solution Steps

Anything divided by one gives itself.

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

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

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

Since \frac{2x-3}{x+1} and \frac{x+1}{x+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2x-3-\left(x+1\right)}{x+1}\geq 0$$

Do the multiplications in 2x-3-\left(x+1\right).

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

Combine like terms in 2x-3-x-1.

$$\frac{x-4}{x+1}\geq 0$$

For the quotient to be ≥0, x-4 and x+1 have to be both ≤0 or both ≥0, and x+1 cannot be zero. Consider the case when x-4\leq 0 and x+1 is negative.

$$x-4\leq 0 x+1<0$$

The solution satisfying both inequalities is x<-1.

$$x<-1$$

Consider the case when x-4\geq 0 and x+1 is positive.

$$x-4\geq 0 x+1>0$$

The solution satisfying both inequalities is x\geq 4.

$$x\geq 4$$

The final solution is the union of the obtained solutions.

$$x<-1\text{; }x\geq 4$$