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

    add the two last equations
    2S1+S2=40
    subtract the first from the second
    S1-S2=-12
    add these two equations
    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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