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Fie A,B.C , nr . naturale pentru care A+B=120 si B+C=230
Calculati;
A+2B+C
5A+8B+3C

Răspuns :

Cesienus
A+B=120
B+C=230 le adunam si obtinem A+2B+C=350

A+B=120,inmultim relatia cu 5 si obtinem 5A+5B=600
B+C=230,inmultim relatia cu 3 si obtinem 3B+3C=690
Adunam cele doua relatii si obtinem 5A+8B+3C=1290
A+B=120⇒A=120-B
B+C=230⇒B=230-C
1)A+2B+C=120-(230-C)+2(230-C)+C=120-230+C+460-2C+C=350
2)5A+8B+3C=5[120-(230-C)]+8(230-C)+3C=5(120-230+C)+1840-8C+3C=600-1150+5C+1840-8C+3C=1290

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