Question

решите прошу пожалуйста ​

Answer 1

1)

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

y ∈ ( -∞ ; -0,5 ]

2)

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

x ∈ [ 6 ; +∞ )

3)

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

y ∈ [ $\frac{1}{8}$ ; +∞ )

4)

$6-5y>3y-2\\-5y-3y>-2-6\\-8y>-8\\y<-8:(-8)\\y<1$

y ∈ ( -∞ ; 1 )

5)

$3-7y>5y-3\\-7y-5y>-3-3\\-12y>-6\\y<-6:(-12)\\y<0,5$

y ∈ ( -∞ ; 0,5 )

6)

$\frac{x}{6} +\frac{1}{2} >x-\frac{1}{3}\\\\\frac{x}{6} -x>-\frac{1}{3} -\frac{1}{2} \\\\\frac{x}{6}-\frac{6x}{6}>-\frac{2}{6}-\frac{3}{6}\\\\-\frac{5x}{6}>-\frac{5}{6} \\\\x<-\frac{5}{6} :(-\frac{5}{6} )\\\\x<-\frac{5}{6} *(-\frac{6}{5} )\\\\x<1$

x ∈ ( -∞ ; 1 )