# ALGEBRA II

the perimeter of a triangle is 51 centimeters, the longest side is 1 centimeter less than the sum of the other two sides. twice the shortest is 11 centimeters less than the longest side. find the lengths of each side of the triangle.

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1. L=S1+S2-1
2*S1+11=L
L+S1+S2=51

rearranging these
S1+S2-L=1
2S1 -L=-11
S2+S2+L=51

2S1+S2=40
subtract the first from the second
S1-S2=-12
3S1=28
S1=7

can you take if from here?

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bobpursley
2. ... add the two last eq.
3S1+S2=28
...
4S1=28

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3. Thanks. I typoed and did not catch it.

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bobpursley

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