Question

6. Daca a,b,c sunt numere naturale astfel inacat a+b+c=32 si 2*a+b+2*c=57, calculati (a+c)*(3*a+2-b+3*c).
VA ROG FRUMOS SA MA AJUTATI

Answer 1

    
[tex]Se~ da: \\ a+b+c = 32 \\ 2a+b+2c=57 \\ \\ Se ~cere: \\ (a+c)(3a+2-b+3c) = ? \\ \\ Rezolvare: \\ \text{Rezolvam sistemul de ecuatii:} \\ a+b+c = 32 \\ 2a+b+2c=57 \\ --------- \\ (a+c) + b = 32 ~~~~~|\times (-1) \\ 2(a+c) + b = 57\\--------------\\-(a+c) - b = -32\\ 2(a+c) + b = 57\\\text{--------------------------adunam}[/tex]
[tex] \\ (a+c) = \boxed{25} \\ b= 32 - (a+b) = 32 - 25 = \boxed{7} \\ \\ (a+c)(3a+2-b+3c)= \\ = (a+c)[3(a+c) +2-b] = \\ = 25 \times [3 \times 25 +2-7] = \\ = 25 \times [75 +2-7] = \\ = 25 \times [77-7] = 25 \times 70 = \boxed{\boxed{1750}}[/tex]