Algebra word problems.
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Algebra word problems.

[From: ] [author: ] [Date: 12-05-13] [Hit: ]
x = 30.perimeter = x + (x-3)+(x-7)=101 ,......
the second side of a triangle is 4 inches longer than the first side. the third side of the triangle is 3 inches longer than the second side. if the perimeter of the triangle is 101 inches, find the length of the third side of the triangle.

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Let the length of the first side be x

the length of the second side will be x + 4

the length of the third side will be x + 4 + 3 = x + 7, so:

x + x + 4 + x + 7 = 101, so

3x + 11 = 101, so:

3x = 90, so

x = 30.

The length of the third side = 37

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1st Side - x
2nd side - x+4
3rd Side - x+7
3x + 11 = 101 (-11)
3x = 90 (/3)
x=30
first side = 30 inches
second side = 34 inches
third side = 37 inches

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the third side =x
===>
2nd side = x-3
1st side = x-7

perimeter = x + (x-3)+(x-7)=101 , now solve for x
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