What is (3-z)/(z+1)>= 1 ?

The solution to (3-z)/(z+1)>= 1 is -1<z<= 1

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(3-z)/(z+1)>= 1

\frac{3 - z}{z + 1} \geq 1

- 1 < z \leq 1

Interval Notation

(- 1, 1]

Solution steps

\frac{3 - z}{z + 1} \geq 1

Rewrite in standard form

\frac{- 2 z + 2}{z + 1} \geq 0

Factor

\frac{- 2 z + 2}{z + 1} : \frac{- 2 ( z - 1 )}{z + 1}

\frac{- 2 ( z - 1 )}{z + 1} \geq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - 2 ( z - 1 ) ) ( - 1 )}{z + 1} \leq 0 \cdot (- 1)

Simplify

\frac{2 ( z - 1 )}{z + 1} \leq 0

Divide both sides by

2

\frac{\frac{2 ( z - 1 )}{z + 1}}{2} \leq \frac{0}{2}

Simplify

\frac{z - 1}{z + 1} \leq 0

Identify the intervals

- 1 < z \leq 1

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