SOLUTION: the perimeter of a triangle is 59 cm. the longest side is 7 cm less then the sum of the other two sides. twice the shortest side is 10 cm less than the longest side. find the perim

Question 386801: the perimeter of a triangle is 59 cm. the longest side is 7 cm less then the sum of the other two sides. twice the shortest side is 10 cm less than the longest side. find the perimeter of each side of the triangle.
Answer by gwendolyn(128) (Show Source): You can put this solution on YOUR website!
Let a be the length of the shortest side
Let b be the length of the mid-length side
Let c be the length of the longest side
These form a perimeter of 59 cm, so:
1. a + b + c = 59

The longest side is 7 cm less then the sum of the other two sides. So:
2. c = a + b - 7

Twice the shortest side is 10 cm less than the longest side. So:
3. 2*a = c - 10

First, substitute the value of c from eq 2 into eq 1:
a + b + c = 59
a + b + (a + b - 7) = 59
2*a + 2*b - 7 = 59
Add 7 to both sides:
2a + 2b - 7 + 7 = 59 + 7
4. 2a + 2b = 66

Now substitute the value of c from eq 2 into eq 3:
2*a = c - 10
2a = (a + b - 7) - 10
2a = a + b -17
Subtract a from both sides:
2a - a = a + b - 17 - a
5. a = b - 17

Now plug the expression for a in eq 5 back into eq 4:
2a + 2b = 66
2(b - 17) + 2b = 66
Distribute the 2:
2b - 17*2 + 2b = 66
2b - 34 + 2b = 66
Combine the b terms:
4b - 34 = 66
Add 34 to both sides to isolate the term with the variable (b):
4b - 34 + 34 = 66 + 34
4b = 100
Divide both sides by 4 to solve for b:
(4b)/4 = 100/4
b = 25

Now use eq 5 to solve for a:
a = b - 17
a = 25 - 17
a = 8

Now use eq 2 to solve for c:
c = a + b - 7
c = 8 + 25 - 7
c = 26
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