Ineed the explanation to how to solve these questions like how to set up the equation or inequality.

2) 40 and 41

3) L=23 and W=11

4) W≥10 while L≥14

Step-by-step explanation:

2)

The sum of 2 consecutive integers is less than 83.

Find a pair of integers with the greatest sum.

Let n be an integer, then (n+1) is the next integer.

So we have the following inequality from the first line:

n+(n+1)<83

n+n+1<83

2n+1<83

Subtract 1 on both sides:

2n<82

Divide both sides by 2:

n=41

n+1=41+1=42.

So 41+42 is equal to 83.

We want it to be less than 83.

So the pair 40 and 41 would work

40+41 would be the next greater sum and it satisfies the restriction of 2 consecutive pair of integers having sum less than 83.

3)

The length of a rectangle is 12 meters longer than it's width.

L=12+W

The perimeter is 68. A rectangles opposite sides are congruent and there are 4 sides.  The perimeter is the sum of the side measurements:

L+L+W+W=68

Combine l ike terms:

2L+2W=68

Plug 1st equation into the perimeter equation:

2(12+W)+2W=68

Distribute:

24+2W+2W=68

Combine like terms:

24+4W=68

Subtract 24 on both sides:

4W=44

Divide both sides by 4:

W=11

L=12+W=12+11=23.

4) The length of a rectangle is 4 cm more than the width:  L=4+W.

The perimeter is at least 48 cm:  2L+2W≥48. (At least mean equals to or greater than.)

Plug the first equation into the inequality there:

2(4+W)+2W≥48.

Distribute:

8+2W+2W≥48

Combine like terms:

8+4W≥48

Subtract 8 on both sides:

4W≥40

Divide both sides by 4:

W≥10

L=4+W

Solving this for L gives us L-4=W (I subtracted 4 on both sides).

So if W≥10 and W=L-4 then:

L-4≥10

Add 4 on both sides:

L≥14.

So while the width is greater than or equal to 10, the length is 14 or greater.

(23*2)-12=x     x=34

step-by-step explanation: