Question

Match each inequality to the number line that represents its solution

Answer 1

Part 1) shaded area at left of x=8 (close circle) --->
`-(x)/(10)+(1)/(5) \geq-(33)/(55)`

Part 2) shaded area at left of x=-5 (close circle) --->
`-(50x)/(3)-(11)/(6) \geq (163)/(2)`

Part 3) shaded area at left of x=-6 (close circle) --->
`(3x)/(2)+105 \leq 96`

Part 4) shaded area at left of x=7 (close circle) --->
`-(13x)/(18)+(5)/(9) \geq -(81)/(18)`

see the attached figure

Explanation:

Part 1) we have

`-(x)/(10)+(1)/(5) \geq-(33)/(55)`

Multiply by -10 both sides

`x-2 \leq 6`

Adds 2 both sides

`x \leq 6+2`

`x \leq 8`

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

`-(50x)/(3)-(11)/(6) \geq (163)/(2)`

Multiply by -6 both sides

`100x+11 \leq -489`

Subtract 11 both sides

`100x \leq -489-11`

`100x \leq -500`

Divide by 100 both sides

`x \leq -5`

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

`(3x)/(2)+105 \leq 96`

Multiply by 2 both sides

`3x+210 \leq 192`

Subtract 210 both sides

`3x \leq 192-210`

`3x \leq -18`

Divide by 3 both sides

`x \leq -18/3`

`x \leq -6`

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

`-(13x)/(18)+(5)/(9) \geq -(81)/(18)`

Multiply by -18 both sides

`13x-10 \leq 81`

Adds 10 both sides

`13x \leq 91`

Divide by 13 both sides

`x \leq 91/13`

`x \leq 7`

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)