Răspuns:
a) {y=6-3x
2x+3y=11
2x+3(6-3x) =11
x=1
y=6-3
y=3
(x, y) =(1,3)
b) {-x+y=-3
x=-3-2y
-(-3-2y)+y=-3
y=-2
x=-3-2×(-2)
x=1
(x,y) = (1, - 2)
c) {y=-11-5x
2x+3y=-7
2x+3(-11-5x) =-7
x=-2
y=-11-5×(-2)
y=-1
(x, y) = (-2, -1)
d){x=1/7+3/7 y
2x+5y=12
2(1/7+3/7y)+5y= 12
y=2
x=1/7+3/7×2
x=1
(x, y) =(1, 2)
e){y=3-2x
-x+4y=21
-x+4(3-2x)=21
x=-1
y=3-2×(-1)
y= 5
(x, y) = (-1,5)
f) {x=-1/2+3/2y
3x+4y=-1
3(-1/2+3/2y)-4y=-1
y=1
x=-1/2+3/2×1
x= 1
(x, y)=(1, 1)
g) {x=-5+2y
3x+y=6
3(-5+2y) +y=6
y=3
x=-5+2×3
x=1
(x,y)=(1,3)
h) {7x-6y=19
x=5+2y
7(5+2y)-6y=19
y=-2
x=5+2×(-2)
x=1
(x, y) =) (1, -2)
Explicație pas cu pas:
la primele doua operații, pui o acolada mare ca sa le cuprindă pe ambele.
