Question

1) 5×a+7×b+c , daca 4×a+7×b=840 si a+c=120
2) Daca 9×a+6×b+12×c=690 si b+2×c=40 , calculați a si a×b+2×a×c
3) Daca a, b , ca sunt nr. naturale astfel încât a+b+c=25 si 2×a+b+2×c=57 , calculați (a+c)×(3×a+2×b+3×c)
4) Calculați 2×a+9×b+3×c , daca a+3×b=120 si b+c=40
Vreau rezolvarea cu explicatii . Multumesc

Answer 1

1)5a+7b+c=4a+7b+a+c=840+120=960
2)9a+6b+12c=9a+6(b+2c)=9a+6*40=9a+240
9a+240=690 
9a=450
a=50

a*b+2*a*c=a( b+2c)=50*40=2000

3)2a+b+2c=a+b+c+a+c=a+c+25 => 57=a+c+25 =>a+c=57-25=32
(a+c)(3a+2b+3c)=32*(2a+b+2c+a+b+c)=32*(25+57)=32*82=2624

4)2a+9b+3c=2a+6b+3b+3c=2(a+3b)+2(b+c)= 2*120+3*40=240+120=360