x/2=y/6
y/5=z/4 (am tinut cont ca dac sunt i.p cu1/5 si1/4sunt d.p.cu 5 sicu4)
x/4+y/12-z/15=5
3 ec , 3 nec,., decirezolvabil
in sistemede acest gen se fol met.substitutiei. aica incercam sa exprima 2 din nec functiede a treai, in locuin d in ecuatia incare apar toate
x=y/3 deci y=3x
z=4y/5=4*3x/5=12x/5
atunci
x/4+y/12-z/15=5 devine
x/4+3x/12-12x/(5*15)=5
x/4+x/4-4x/25=5
x/2-4x/25=5
25x/50-8x/50=250/50
17x=250
x=250/17
y=3x=750/17
z=12x/5=12*50/17=600/17
verificare
x=y/3 =250/17 adevarat
y/5=150/17 z/4=150/17 adevarat
x/4+y/12-z/15= 1/17(125/2+125/2-40)=1/17(125-40)=85/17=5 adevarat
adevarate toate , bine rezolvat
(la datele pecare mi le-ai dat)
