Question 156223
Let {{{x}}}= shortest piece of cut= 1st piece
{{{x+2)}}}= 2 centimeters longer than the shortest = 2nd piece
{{{y}}} = longest piece = 3rd piece
Take note: 2nd piece is also 2 centimeters shorter than the longest piece, {{{y-2}}} = 2nd piece
We can equate the two 2nd piece since they are the same:
{{{(x+2)=(y-2)}}}
{{{x=y-2-2=y-4}}} -------------------------------> eqn 1
Now, adding the 1st, 2nd & 3rd piece equals 60 cm:
{{{x+x+2+y=60}}} --------------------------------> eqn 2
{{{2x+y=60-2}}}
{{{2x+y=58}}}, substitute he value of {{{x}}} in eqn 1:
{{{2(y-4)+y=58}}}
{{{2y-8+y=58}}}
{{{3y=58+8}}} --------------> {{{3y=66}}}
{{{cross(3)y/cross(3)=cross(66)22/cross(3)}}}
{{{y=22cm}}} ----------------------------> 3rd piece, longest cut
go back eqn 1: {{{x=y-4=22-4=18cm}}} -----> 1st piece, shortest cut
{{{x+2=18+2=20cm}}} ----------------------> 2nd piece, 2cm longer than the shortest
In doubt? go back eqn 2,
{{{18+18+2+22=60}}}
{{{60cm=60cm}}} , total ribbon
Thank you,
Jojo