### Inequalities - Word Problem Exercise 2 Q2

A rectangular box is (2x - 1) cm long and (x + 3) cm wide.
If the perimeter of the frame is not less 38 cm and not greater than 80 cm, find the range of possible values of x.

1. (2x-1)+(2x+1)+(x+3)+(x+3)
=2x+2x+x+x+3+3-1-1
=6x+4
38≤6x+4≤80
38-4≤6x+4-4≤80-4
34/6≤6x/6≤76/6
5/2/3≤x≤12/2/3
The possible values of x is between the range of 5/2/3 to 12/2/3

2. perimeter is 4x-2 + 2x + 6 = 6x + 4
38≤perimeter≤80
perimeter - 4 = 6x
34≤perimeter - 4≤76 = 34≤6x≤74
= 5.6<x<12.4

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4. (2x-1)+(2x-1)+(x+3)+(x+3)=6x+4
6x+4≤80
6x<80-4
6x<76
x<12 2/3
x<13(greatest possible)
6x+4>38
6x>38-4=34
x>5 2/3
x>6(smallest possible)

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4. 6 < x < 13

5. (2x-1)+(2x+1)+(x+3)+(x+3)
=2x+2x+x+x+3+3-1-1
=6x+4
38≤6x+4≤80
38-4≤6x+4-4≤80-4
34/6≤6x/6≤76/6
5/2/3≤x≤12/2/3
x= 5/2/3 to 12/2/3

6. (2x-1) x (x+3) = 2x^2 + 6x - x - 3
= 2x^2 + 5x - 3

7. Let y be the perimeter of frame.
Perimeter of frame: 38cm > y > 80cm
Highest possible perimeter: 79 cm

Total perimeter: (2x - 1)+(2x - 1)+(x + 3)+(x + 3)
= (6x + 4)
= y cm

79cm - 4cm = 75cm
75cm ÷ 6 = 12.5cm
x = 12.5cm

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9. (2x-1)+(2x+1)+(x+3)+(x+3)
= 2x-1 + 2x-1 + x+3 + x+3
= 6x + 4

38≤perimeter ≤80
6x+4 = perimeter
6x= perimeter-4

34≤6x≤ 74
= 6 ≤ x ≤ 12

10. 38cm < 80cm

perimeter in terms of x --> 2(2x-1) + 2(x+3)
=4x - 2 + 2x + 6
=6x + 4

38 < 6x+4 < 80
= 34 < 6x < 76
= 5 2/3 < x < 12 2/3
The possible values of x is from 5 2/3 to 12 2/3

11. (2x-1)+(2x-1)+(x+3)+(x+3)=6x+4
6x+4≤80
6x<80-4
6x<76
x<12 2/3
x<13(greatest possible)
6x+4>38
6x>38-4=34
x>5 2/3
x>6
cuboid