ALGEBRA II
posted by TRACI .
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.

L=S1+S21
2*S1+11=L
L+S1+S2=51
rearranging these
S1+S2L=1
2S1 L=11
S2+S2+L=51
add the two last equations
2S1+S2=40
subtract the first from the second
S1S2=12
add these two equations
3S1=28
S1=7
can you take if from here? 
... add the two last eq.
3S1+S2=28
...
4S1=28 
Thanks. I typoed and did not catch it.
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