Notam:
x+2y=a
2x-y=b
a,b≠0
[tex]\left \{ {{\frac{7}{a} +\frac{2}{b}=16 } \atop {\frac{5}{a}+\frac{3}{b}=13 }} \right.[/tex]
- Aducem la acelasi numitor comun "ab" si eliminam numitorul
7b+2a=16ab
5b+3a=13ab
7b+a(2-16b)=0
7b=a(16b-2)
- [tex]a=\frac{7b}{16b-2}[/tex] si inlocuim in a doua relatie si vom avea:
[tex]5b+\frac{21b}{16b-2} =13b\times\frac{7b}{16b-2}[/tex]
Aducem la acelasi numitor comun 16b-2, amplificand pe 5b cu 16b-2 si obtinem:
5b(16b-2)+21b=91b²
91b²-80b²+10b-21b=0
11b²-11b=0
11b(b-1)=0
b=0 si b=1, dar b nu poate fi 0, deci b=1
Pentru b=1⇒
[tex]a=\frac{7}{16-2} =\frac{1}{2}[/tex]
Dar, a=x+2y si b=2x-y
Vom avea:
[tex]\left \{ {{x+2y=\frac{1}{2} } \atop {2x-y=1}} \right. \\\\\\\left \{ {{2x+4y=1} \atop {2x-y=1}} \right.[/tex]
Le scadem si obtinem:
5y=0
y=0
[tex]x+0=\frac{1}{2} \\\\\\x=\frac{1}{2}[/tex]
