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Solve the inequality
\frac{3-z}{z+1} \ge 2(z + 4).
Write your answer in interval notation.

 Feb 14, 2024
 #1
avatar+1622 
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\(\frac{3-z}{z+1} \ge 2(z + 4)\)

 

Case 1: z + 1 is greater than 0, so when you multiply both sides by z + 1, the inequality sign will NOT flip. z + 1 > 0, z > -1. (z cannot be -1, or else denominator will be 0 and thus an undefined expression on the left side of the inequality...)

\(3-z{\ge}2(z+4)(z+1)\)

\(3-z{\ge}2z^2+10z+8\)

\(2z^2+11z+5{\le}0\)

\((2z+1)(z+5){\le}0\)

z has to be [-5, -1/2] to satisfy the inequality. However, we know z > -1, so the domain is (-1, -1/2].

*(2z + 1)(z + 5) have to be one positive * one negative (or one is 0) to be less than equal to 0, so by picking [-5, -1/2], we can ensure their sign + or - is always the opposite (except at 0...)*

 

Case 2: z + 1 is less than 0, so when you multiply both sides by z + 1, the inequality sign WILL flip. z + 1 < 0, z < -1.

Follow the factoring from above...

\((2z+1)(z+5){\ge}0\)

Both factors have to be either both positive or both negative, so z has to be (-inf, -5] U [-1/2, inf) to satisfy the inequality. However, we know that z < -1, so we take the left side of the union; (-inf, -5].

 

Combining the two cases (joining their unions) would be (-inf, -5] U (-1, -1/2].

 Feb 14, 2024

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