A 102-inch length of ribbon is to be cut into three pieces. The longest piece is to be 38 inches longer than the shortest piece, and the third piece is to be half the length of the longest. Find the length of each ribbon.

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ok... let's say, the pieces are "a", "b", and "c". "a" being the shortest and "c" being the longest we know the longest is 38 more than the shortest, so c = a + 38 and the third piece, b, is half the longest, or c/2 we know the three pieces come from the 102in ribbon, thus a + b + c = 102 [latex]\bf a+b+c=102\quad \begin{cases} c=a+38\\ b=\frac{c}{2}\\ \quad =\frac{a+38}{2} \end{cases}\implies a+(a+38)+\left( \frac{a+38}{2} \right)=102[/latex] solve for "a". c = a + 38, and b = c/2

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