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# Solve The Inequality \frac{3-z}{z+1} \geq 1

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Solve The Inequality

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

Jun 15, 2022

#1
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Multiply both sides by z + 1: 3 - z >= z + 1

Then 2z <= 2, so z <= 1.

Solution: (-inf,1].

Jun 15, 2022
#3
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You're very new to math, in this case, if you include -1 in the denominator, the answer is undefined, so you're WRONG. The answer is $(-\infty, -1)U(-1, 1)$

Guest Jun 15, 2022
#4
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You're very new to math too, as if the answer is negative and lower than -1 then it will be less than 1. so the answer then, is obviously (-1, 1]

Guest Jun 15, 2022
#5
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You're all new to math, because the answer is actually (-1,2].

Guest Jun 15, 2022
#6
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You're all all new to math, the answer is (-1, inf) because the higher the number, the more greater is to 1. Simple logic.

Guest Jun 15, 2022
#7
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Guest Jun 15, 2022
#8
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thats wrong?

Guest Jun 15, 2022
#9
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That answer is wrong because you need to include the (-1 (it's undefined) to 1]

Guest Jun 16, 2022
#2
+282
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$\displaystyle \frac{3-z}{z+1} \geq{1}$

Jun 15, 2022
#10
+9458
+1

Ignoring the toxic "discussion" above and actually solving the problem:

$$\dfrac{3 - z}{z + 1} \geq 1\\ \dfrac{3 - z}{z + 1} (z + 1)^2 \geq (z + 1)^2\text{ and }z +1 \neq 0\\ (3 - z)(z + 1) \geq (z + 1)^2\text{ and }z +1 \neq 0\\ -z^2 + 2z + 3\geq z^2 + 2z + 1 \text{ and }z +1 \neq 0\\ 2z^2 - 2 \leq 0 \text{ and }z +1 \neq 0\\ z^2 - 1 \leq 0 \text{ and }z +1 \neq 0\\ -1 \leq z \leq 1 \text{ and }z \neq -1\\ -1 < z \leq 1$$

Jun 16, 2022
#11
+117737
+1

Sometimes I just gotta laugh.

$$\frac{3-z}{z+1} \geq 1 \\ \text{ Assuming that z is a real number}\\ z\ne-1\\ \frac{3-z}{z+1} \geq 1 \\~\\ If\;\;z>-1\;\;then\\ 3-z \geq z+1 \\ -2z\ge -2\\ z\le1\\ \therefore -1 or \(If\;\;z<-1\;\;then\\ 3-z \leq z+1 \\ -2z\le -2\\ z\ge1\\ \therefore no\;\;soln$$

Hence

\(-1

Jun 16, 2022
edited by Melody  Jun 16, 2022
#12
+117737
+1

Sorry, the Latex is playing up.

its (-1,1]

Melody  Jun 16, 2022
#15
+1

I WAS RIGHT >:(

Guest Jun 16, 2022
#13
+117737
0