List the sides in order from shortest to longest. (7x+8),(5x+40), and (10x)

Answer 1

I’m assuming these are sides of a triangle. I would plug in different values for x until I got valid sides of a triangle. The sum of the two smallest sides of a triangle have to be greater than the largest side and every side length has to be positive. If we plug in 4 for x, we get a valid triangle with side lengths 36, 60, and 40. Therefore, the 7x + 8 is the shortest and the 5x + 40 is the longest.

Answer 2

10x = 7x+8

3x = 8

x = 8/3

IF x>8/3 then 10x > 7x+8

if x < 8/3 then 7x+8 > 10x

5x+40 = 10x

5x = 40

x = 8

IF x>8, then 10x> 5x+40

if x< 8, then 5x+40 > 10x

7x+8 = 5x+40

2x = 32

x = 16

if x>16 then 7x+8 > 5x+40

if x<16 then 5x+40> 7x+8

for x>16, the order from shortest to longest is: 5x+40 < 7x+8 < 10x.

5x<7x<10x

for 8<x<16, the order changes to: 7x+8<5x+40<10x

for x<8/3 the order is 10x<7x+8<5x+40

or 8/3x<x<8 the order is: 7x+8<10x<5x+40

the order depends on the value of x, whether it's 0<x<8/3, 8/3<x<8, 8<x<16 or x>16

there're 4 different orders depending on which of 4 different intervals x is in (0,8/3), (8/3,8), (8,16) or (16,inf)

IF these are sides of a box, x could be in any of the 4 intervals

IF these are sides of a triangle. the longest side cannot be longer than the sum of the other 2 sides

and the longest side cannot be shorter than the difference of the other 2 sides. L>s1-s2

L<s1+s2, if L=5x+40 then 5x+40<10x + 7x+8

40-8<12x

32<12x

8/3 <x

x>8/3

otherwise there seems to be no other constraint on x