The perimeter of a triangle is 58 centimeters. The longest side is 2 centimeters less than the sum of the other two sides. Twice the shortest is 12 centimeters less than the longest side. Find the length of each side of the triangles.**Solution**

Let x be the longest side.

Let y be the shortest side.

Let z be the last side.

The perimeter of a triangle is 58 centimeters.

The equation for the sentence above is the following:

x + y + z = 58 **equation 1**

The longest side is 2 centimeters less than the sum of the other two sides.

The equation for the sentence above is the following:

x = y + z - 2 **equation 2**

Twice the shortest is 12 centimeters less than the longest side.

The equation for the sentence above is the following:

2y = x - 12 **equation 3**

You now have a system to solve

x + y + z = 58 **equation 1**

x = y + z - 2 **equation 2**

2y = x - 12 **equation 3**

Equation 2 is the same as x - y - z = -2

The system becomes

x + y + z = 58 **equation 1**

x - y - z = - 2 **equation 2**

2y = x - 12 **equation 3**

equation 1 + equation 2 gives the following

2x = 56

x = 28

Use equation 3 to get y

2y = 28 - 12

2y = 16

y = 8

Use equation 1 to get z

28 + 8 + z = 58

36 + z = 58

z = 58 - 36

z = 22

The longest side of this triangle is 28, the shortest side is 8, and the third side is 22