Respuesta :

Answer:

Part 1)  shaded area at left of x=8 (close circle) ---> [tex]-\frac{x}{10}+\frac{1}{5} \geq-\frac{33}{55}[/tex]  

Part 2) shaded area at left of x=-5 (close circle) ---> [tex]-\frac{50x}{3}-\frac{11}{6} \geq \frac{163}{2}[/tex]

Part 3) shaded area at left of x=-6 (close circle) ---> [tex]\frac{3x}{2}+105 \leq 96[/tex]

Part 4) shaded area at left of x=7 (close circle) ---> [tex]-\frac{13x}{18}+\frac{5}{9} \geq -\frac{81}{18}[/tex]

see the attached figure

Step-by-step explanation:

Part 1) we have

[tex]-\frac{x}{10}+\frac{1}{5} \geq-\frac{33}{55}[/tex]  

Multiply by -10 both sides

[tex]x-2 \leq 6[/tex]

Adds 2 both sides

[tex]x \leq 6+2[/tex]

[tex]x \leq 8[/tex]

The solution is the interval -----> (-∞,8]

All real numbers less than or equal to 8

In a number line the solution is the shaded area at left of x=8 (close circle)

Part 2) we have

[tex]-\frac{50x}{3}-\frac{11}{6} \geq \frac{163}{2}[/tex]

Multiply by -6 both sides

[tex]100x+11 \leq -489[/tex]

Subtract 11 both sides

[tex]100x \leq -489-11[/tex]

[tex]100x \leq -500[/tex]

Divide by 100 both sides

[tex]x \leq -5[/tex]

The solution is the interval -----> (-∞,-5]

All real numbers less than or equal to -5

In a number line the solution is the shaded area at left of x=-5 (close circle)

Part 3) we have

[tex]\frac{3x}{2}+105 \leq 96[/tex]

Multiply by 2 both sides

[tex]3x+210 \leq 192[/tex]

Subtract 210 both sides

[tex]3x \leq 192-210[/tex]

[tex]3x \leq -18[/tex]

Divide by 3 both sides

[tex]x \leq -18/3[/tex]

[tex]x \leq -6[/tex]

The solution is the interval -----> (-∞,-6]

All real numbers less than or equal to -6

In a number line the solution is the shaded area at left of x=-6 (close circle)

Part 4) we have

[tex]-\frac{13x}{18}+\frac{5}{9} \geq -\frac{81}{18}[/tex]

Multiply by -18 both sides

[tex]13x-10 \leq 81[/tex]

Adds 10 both sides

[tex]13x \leq 91[/tex]

Divide by 13 both sides

[tex]x \leq 91/13[/tex]

[tex]x \leq 7[/tex]

The solution is the interval -----> (-∞,7]

All real numbers less than or equal to 7

In a number line the solution is the shaded area at left of x=7 (close circle)

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