Question 185511
Let x = length of first piece, y = length of second piece, and z = length of third piece




Since "The second piece is three times the length of the first piece", this tells us that {{{y=3x}}} and because "the third piece is 10cm longer than twice the first piece", this means that {{{z=2x+10}}}



Now the the sum of the three pieces should be equal to the total original length 190. In other words, {{{x+y+z=190}}}



{{{x+y+z=190}}} Start with the given equation.



{{{x+3x+2x+10=190}}} Plug in {{{y=3x}}} and {{{z=2x+10}}}



{{{6x+10=190}}} Combine like terms on the left side.



{{{6x=190-10}}} Subtract {{{10}}} from both sides.



{{{6x=180}}} Combine like terms on the right side.



{{{x=(180)/(6)}}} Divide both sides by {{{6}}} to isolate {{{x}}}.



{{{x=30}}} Reduce.



So the first piece is 30 cm 


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{{{y=3x}}} Go back to the first equation



{{{y=3(30)}}} Plug in {{{x=30}}} 



{{{y=90}}} Multiply



So the second piece is 90 cm


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{{{z=2x+10}}} Go back to the second equation



{{{z=2(30)+10}}} Plug in {{{x=30}}} 



{{{z=60+10}}} Multiply



{{{z=70}}} Add



So the third piece is 70 cm



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Answer:



So the lengths of the three pieces are 


1st: 30 cm
2nd: 90 cm
3rd: 70 cm