Respuesta :
[tex]\begin{cases} a = shortest\\ b = medium\\ c = longest \end{cases} \begin{array}{llll} \stackrel{\textit{perimeter is 57}~\hfill }{a + b + c = 57}~\\\\\stackrel{\textit{twice longest minus shortest}}{2c-a=22~\hfill }\\\\ \stackrel{\textit{longest plus twice others}}{c + 2(a+b) = 94~\hfill } \end{array} \\\\[-0.35em] ~\dotfill\\\\ 2c-a=22\implies 2c=a+22\implies \boxed{2c-22=a} \\\\\\ \stackrel{\textit{we know that}}{c+2(a+b)=94}\implies c+2a+2b=94\implies c+2(2c-22)+b=94[/tex]
[tex]c+4c-44+2b=94\implies 5c-44+2b=94\implies 5c+2b=138 \\\\\\ 2b=138-5c\implies \boxed{b = \cfrac{138-5c}{2}} \\\\\\ \stackrel{\textit{we know the perimeter is}}{57=a + b + c}\implies 57 = \stackrel{a}{(2c-22)}+\stackrel{b}{\cfrac{138-5c}{2}}+c \\\\\\ 57=2c-22+\cfrac{138}{2}-\cfrac{5c}{2}+c\implies 57=3c-22+69-\cfrac{5c}{2} \\\\\\ 57=3c-47+\cfrac{5c}{2}\implies 10=3c-\cfrac{5c}{2}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{20=6c-5c}[/tex]
[tex]\blacktriangleright 20=c \blacktriangleleft \\\\\\ \boxed{2c-22=a}\implies 40-22=a\implies \blacktriangleright 18=a \blacktriangleleft \\\\\\ \boxed{b = \cfrac{138-5c}{2}}\implies b=\cfrac{138-5(20)}{2}\implies b=\cfrac{38}{2}\implies \blacktriangleright 19 \blacktriangleleft[/tex]
