r = numarul initial de bile rosii din cutie
a = numarul initial de bile albastre din cutie
g = numarul initial de bile galbene din cutie
Avem 3 necunoscute.
Trebuie sa scriem un sistem de 3 ecuatii.
Prima ecuatie o scriem din numarul total de bile
care sunt initial in cutie.
r + a + g = 164
Pentru celelalte 2 ecuatii ne vom ocupa de bilele din vitrina
deoarece problema spune:
"Ioana observa ca in vitrina va avea
acelasi numar de bile din fiecare culoare."
.
[tex]\displaystyle\bf\\Rezolvare:\\\\r + a + g = 164\\\\\frac{r}{2}-6=\frac{a}{3}-5=\frac{g}{4}+4\\\\..........\\\\r + a + g = 164\\\\\frac{r-12}{2}=\frac{a-15}{3}=\frac{g+16}{4}~~\Big|\times12\\\\..........\\\\r + a + g = 164\\\\6(r-12)=4(a-15)=3(g+16)\\\\..........\\\\r + a + g = 164\\\\6r-72=4a-60=3g+48\\\\..........\\\\r + a + g = 164\\\\6r-72=4a-60\\\\6r-72=3g+48[/tex]
.
[tex]\displaystyle\bf\\..........\\\\r + a + g = 164\\\\6r-4a=72-60\\\\6r-3g=72+48\\\\..........\\\\r + a + g = 164\\\\6r-4a=12~~\Big|:2\\\\6r-3g=120~~\Big|:3\\\\..........\\\\r + a + g = 164\\\\3r-2a=6\\\\2r-g=40[/tex]
.
[tex]\displaystyle\bf\\..........\\\\g=2r-40~~~~~~\boxed{\bf~Substitutia~1}\\\\a=\frac{3r-6}{2}~~~~~~~\boxed{\bf~Substitutia~2}\\\\..........\\\\r + \frac{3r-6}{2} + 2r-40 = 164~~~\Big|\times2\\\\2r+3r-6+4r-80=3289r=328+80+6\\\\9r=414\\\\r=\frac{414}{9}=46\\\\\boxed{\bf~r=46~de~bile~rosii~au~fost~initial~in~cutie}\\\\Ne intoarcem la substitutii:\\\\g=2r-40=2\times46-40=92-40=52\\\\\boxed{\bf~g=52~de~bile~galbene~au~fost~initial~in~cutie}[/tex]
.
[tex]\displaystyle\bf\\a=\frac{3r-6}{2}=\frac{3\times46-6}{2}=\frac{138-6}{2}=\frac{132}{2}=66\\\\\boxed{\bf~a=66~de~bile~albastre~au~fost~initial~in~cutie}[/tex]
