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a fost răspuns

Daca 3a+6b=42 si 3b+5a+c=32, unde a,b,c sunt numere naturale, sa se calculeze 7(a+b)+c

Răspuns :

3a+6b=42 /÷3 => a+2b=14 => a=14-2b

3b+5a+c=32 => c=32-5a-3b=>

=> c=32-5(14-2b)-3b

c= 32-70+10b-3b

c= 7b-38

a=14-2b

b

c=7b-38

7(a+b)+c = 7(14-2b+b)+7b-38 =

= 98-14b+7b+7b-38

=60-14b+14b

=60-0

=60

Dianageorgiana794

Răspuns:

3a+6b=42=>3(a+2b)=42=>a+2b=42:3=>a+2b=14 I·2

2a+4b=28

3b+5a+c=32

2a+4b+3b+5a+c=28+32

7a+7b+c=7(a+b)+c=60

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