another perimeter question!?

Let A, B and C be the lengths of the respective sides, in centimeters:

"The perimeter of a triangle is 73 centimeters."
A + B + C = 73

"The first side is 7 centimeters longer than the second side."
A = B + 7

"The third side is 5 centimeters longer than triple the length of the first side."
C = 3A + 5

Substitute this last equation into the first equation:
A + B + C = 73
A + B + 3A + 5 = 73
4A + B + 5 = 73
4A + B = 68

Now take your second equation and substitute it into this equation:
4(B + 7) + B = 68
4B + 28 + B = 68
5B + 28 = 68
5B = 40
B = 40/5
B = 8

Then just plug this back in to figure the other sides:
A = B + 7
A = 8 + 7
A = 15

C = 3A + 5
C = 3(15) + 5
C = 45+5
C = 50

The sides are 15, 8 and 50. Unfortunately, this doesn't make a valid triangle because the longest side is greater than the sum of the other two sides.

Answer:
Algebraically the answer should be 8 cm but in reality, this is an impossible triangle.

lets length of the second side be x
the first side is x+7
the third side length is 3*(x+7)+5 = 3x +26

perimeter length is x+x+7+3x+26= 5x+33=73
so x=8 the first 8+7=15 and the third would be 50

there is an impossibility the third side would be longer than sum of the two others

side 1 = x
side 2 = y
side 3 = z

eq1: x+y+z = 73cm
eq2: x = y + 7
eq3: z = 3x + 5

substitute eq2 into eq3:
z = 5 + 3(y + 7)
z = 5 + 3y + 21
z = 3y + 26

eq4: z = 3y + 26

rearrange eq1 and substitute eq2 and eq4:
y = 73 - x - z
y = 73 - (y + 7) - (3y + 26)
y = 73 - 4y - 7 - 26
5y = 40
therefore y = 8cm

as x = y +7
x = 8 + 7
therefore x = 15cm

as z = 3y + 26
z = 3(8) + 26
z = 24 + 26
therefore z = 50cm

shortest side is y (the second side) and its 8cm long.
and my batteries melted inside my calculator so that was all in my head :(

x = length of the shortest side

[x + 7] + x + [3(x + 7) + 5] = 73
x + 7 + x + [3*x + 3*7 + 5] = 73
2x + 7 + [3x + 26] = 73
2x + 7 + 3x + 26 = 73
5x + 33 = 73
5x = 73 - 33
x = 40/5
x = 8

∴ the shortest side of the triangle is 8 centimeters.

Let the side of the triangles be a,b and c
Given
a+b+c = 73 -----(1) (perimeter)
b = a-7 -----(2)
c = 3a+5------ (3)
Sub. (2) and (3) in (1), we get
a+a-7+3a+5 = 73
5a-2 = 73
5a = 75
a = 75/5 =15
Sub. the value of a in (2) and (3), we get b and c

b =15 - 7 = 8
c = 3*15+5 = 45+5 = 50
So length of shortest side = 8 centimeters

let x= the second side.

first side is 7 cm longer than the second side..
x+7

the third side is 5 cm longer than triple the length of the first side.
5+3(x+7)

x+(x+7)+[5+3(x+7)]=73

then just solve.

Well that depends what material each side is made off.
Some materials need to be corrugated and may take more materials than a side with a different material. Some materials can be stretched out so less amount of material can cover a greater distance.

side 1------x + 7 cm
side 2-------x cm
side 3------3(x + 7) + 5

3x + 21 + 5 + x + x + 7 = 73
5x = 40
x = 8
Side 1----15 cm
Side 2-----8 cm
Side 3---- 50 cm

Shortest side is 8 cm

Say the second side of the triangle is x cm.
The first side is 7 cm longer than the second. So, the first side = 7+x cm.
The third side is 3 times and 5cm longer than the second. So, the third side = 3x+5 cm.

Perimeter = 73
x + (7+x) + (3x+5) = 73
5x+12 = 73
5x = 61
x = 12.2

Therefore, the length of the shortest side = 12.2 cm