1) 3x-2m=0 <=> 3x = 2m <=> x = 2m/3
2) En se ramenant à un système linéaire, résoudre le système suivant dans IR
y =14x²+21y²-35=0
y= -2,4x²-3,6y²+6= 0
donc
14x²+21y²-35 = -2,4x²-3,6y²+6 <=> 16.4x² - 41 = -24.6y² <=> (16.4x² - 41)/(-24.6) = y²
<=> 5/3 - (2/3) x² = y² <=> y = racine ( 5/3 - (2/3) x² )
On veut résoudre y = 0
donc racine ( 5/3 - (2/3) x² ) = 0
donc 5/3 - (2/3) x² =0 <=> 5/3 = (2/3) x² <=> 5 = 2x² <=> x² = 2/5 <=> x = racine(2/5)
3)
y=-3x+2
y=-2/3x+2/3
donc
-3x+2 = -(2/3)x+2/3 <=> -3x + (2/3)x = 2/3 - 2 <=> (7/3) x = 4/3 <=> x = 4/7
donc y= -3x+2 = -3(4/7) + 2 = 2/7
Voilà @+
