**Group Challenge Problem Discussion**

**Question One**

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(2L)(1/10L) = 2 2/10 L

2/10L = 2 2/10

L = 22/10 x 10/2

L = 220/20

Length = 11 and Width = 11/10

**Question Two**

x(x+2) =15

x^2+2x=15

(x-3)(x+5)

x=3 or x = -5

The shorter side is 3”

**Question Three**

7 – 4b =< -3(b-1)

7-4b =< -3b+3

7-3 =< -3b+4b

b >= 4

{b| b>= 4}

**Question Four**

-1/5*2x = -2/5x

-1/5*-25=5

-2/5x+5

And

1/3*4x=4/3x

1/3*9=3

ADD

-2/5x+4/3x=14/15x

5+3=8

14/15x+8 > 1

-8 -8

14/15x > -7

DIVIDE

x > -15/2

**Question Five**

x+y+z+k=9900

y=x+1/7x

z=(x+y) + 300

k=(x+y+z) + 300

**Simplify**

7y=7x+x

7y=8x

**SIMPLFY**

-8x+7y=0

-x-y+z=300

-x-y-z+k=300

x+y+z+k=9900

-x-y-z+k=300

2k=10200

k=5100

x+y+z+5100=9900

-8x+7y=0

-x-y+z=300

x+y+z=4800

ADD

-x-y+z=300

2z=5100

z=2550

x+y+2550=4800

-8x+7y=0

Simplify

x+y = 2250

-8x+7y=0

Find y by multiplying x by 8 gives you

15y=18000

Y=1200

x=1050, y =1200, z= 2550, k =5100

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**Question**

As a group member, you will post an initial reply that includes the work and solutions to each question provided in the document by using the Equation Editor (Links to an external site.)Links to an external site. In the Rich Content Editor.

Once your initial reply has been posted, note whether or not your answers match the solutions of your fellow group members.

Come together as a group and start a dialogue.

Where are the errors in the proposed solutions (mine or classmates’)?

Which solution is correct? Why? Is there more than one way to get to the correct solution?

Have we given thorough explanations for our work?

- The length of a rectangle is a two-digit number with identical digits (aa). The width is 1/10 of the perimeter, which is 2 times the area of the rectangle. Find the length and the width.
- How long is the shorter side of a rectangle when one side is 2” longer than the other, and the area is 15 sq. in?
- Solve the following inequality for b. 7 – 4b ≤ −3(b − 1) Show the solution in set builder notation and interval
- Solve the following inequality for
*x*. –^{1}(2𝑥 − 25) +^{1}(4𝑥 + 9) > 1 5 3 - Four numbers have a sum of 9900. The second exceeds the first by 1/7 of the first. The third exceeds the sum of the first two by 300. The fourth exceeds the sum of the first three by 300. Find the four