решите прошу пожалуйста . Created by polinkaburlakova. matematika-ru

1)y ∈ ( -∞ ; -0,5 ]2)x ∈ [ 6 ; +∞ )3)y ∈ [ ; +∞ )4)y ∈ ( -∞ ; 1 )5)y ∈ ( -∞ ; 0,5 )6)x ∈ ( -∞ ; 1 )

решите прошу пожалуйста ...

polinkaburlakova @polinkaburlakova

November 2021 1 5 Report

решите прошу пожалуйста

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ksyuwi

$5y+9\leq 3-7y\\5y+7y\leq 3-9\\12y\leq -6\\y\leq -6:12\\y\leq -0,5$

y ∈ ( -∞ ; -0,5 ]

$3x+1\leq 4x-5\\3x-4x\leq -5-1\\-x\leq -6\\x\geq -6:(-1)\\x\geq 6$

x ∈ [ 6 ; +∞ )

$\frac{1}{4}-\frac{y}{3}\geq \frac{1}{3}-y\\\\-\frac{y}{3} +y\geq \frac{1}{3}-\frac{1}{4}\\\\-\frac{4y}{12}+\frac{12y}{12}\geq \frac{4}{12}-\frac{3}{12}\\\\\frac{8y}{12}\geq \frac{1}{12} \\\\y\geq \frac{1}{12}:\frac{8}{12}\\\\y\geq \frac{1}{12}*\frac{12}{8}\\\\y\geq \frac{1}{8}$

y ∈ [ $\frac{1}{8}$ ; +∞ )

$6-5y>3y-2\\-5y-3y>-2-6\\-8y>-8\\y$

y ∈ ( -∞ ; 1 )

$3-7y>5y-3\\-7y-5y>-3-3\\-12y>-6\\y$

y ∈ ( -∞ ; 0,5 )

$\frac{x}{6} +\frac{1}{2} >x-\frac{1}{3}\\\\\frac{x}{6} -x>-\frac{1}{3} -\frac{1}{2} \\\\\frac{x}{6}-\frac{6x}{6}>-\frac{2}{6}-\frac{3}{6}\\\\-\frac{5x}{6}>-\frac{5}{6} \\\\x$

x ∈ ( -∞ ; 1 )

3 votes Thanks 2

polinkaburlakova спасибо большое

pomogite pozhalujstaebc41adb0c7055b450f8f43a94c9fc69 49151

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polinkaburlakova November 2021 | 0 Ответы

proshu pomogite pozhalujsta64ad98aebf2064ea338e538ed3dd45ac 41094

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Answers & Comments

pomogite pozhalujstaebc41adb0c7055b450f8f43a94c9fc69 49151

proshu pomogite pozhalujsta64ad98aebf2064ea338e538ed3dd45ac 41094

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