Din inversa proportionalitate⇒6(x+y)=8(x+z)=4(y+z) impartim relatia prin 2 si obtinem relatia 3(x+y)=4(x+z)=2(y+z)⇒3x+3y=4x+4z=2y+2z
Din relatia x+y+z+78 il scoatem pe z in functie de x si y si inlocuim in relatia de mai sus: z=78-x-y, 3x+3y=4x+4(78-x-y)=2y+2(78-x-y)⇒3x+3y=4x+312-4x-4y=2y+156-2x-2y reducem termenii asemenea 3x+3y=312-4y=156-2x
Formam un sistem de 2 ecuatii cu 2 necunoscute:
[tex] \left \{ {{3x+3y+4y=312} \atop {2x-4y=-156}} \right. [/tex]
Impartim a doua ecuatie prin 2 ⇒[tex] \left \{ {{3x+7y=312} \atop {x-2y=-78}} \right. [/tex]⇔[tex] \left \{ {{3x+7y=312} \atop {-3x+6y=234}} \right. [/tex]⇒13y=546⇒y=42
Inlocuim y obtinut in a doua relatie a sistemului⇒x-84=-78⇒x=6
6+42+z=78⇒z=30
