Se dau a; d; c a caror suma este 95.stiind ca inyre multimea {a; d+2; c+3} si {5; 12; 8} se stabileste o proportie directa , sa se afle cele trei numere

a+b+c=95a/5=(b+2)/12=(c+3)/18
a/5=(b+2)/1212a=5(b+2)12a=5b+10a=(5b+10)/12
(b+2)/12=(c+3)/1818 (b+2)=12 (c+3)18b+36=12c+3618b=12cc=18b/12
=> (5b+10)/12+b+18b/12=9523b+10+12b=114035b=1130b=1130/35b=226/7
a=(5*226/7+10)/12a=[(1130+70)/7]/12a=1200/7*1/12a=100/7
c=(18*226/7)/12c=4068/7*1/12c=339/7
verificare :100/7+226/7+339/7=665/7=95