Question

resolva os sistemas de equacoes

a) x + y = -5

[3x - 2y = 0

b) [x + y = 2
y= 3

c) [x - 2y = - 4/12
x= y + 1/12

Answer 1

a) x+y= -5
3x-2y=0

y= -5-x
3x-2.(-5-x)
3x+10+2x=0
5x=-10
x=-10/5
x=-2

3x-2y=0
3.(-2)-2y=0
-6-2y=0
-2y=6 .(-1)
2y=-6
y=-6/2
y=-3

b) x+y=2
y=3

x+3=2
x=2-3
x=-1

c) x-2y=-4/12
x=y+1/12

y+1-2y=-4/12
12y+1-24y=-4
-12y=-4-1
-12y=-5 .(-1)
12y=5
y=5/12

x=5/12+1/12
12x=5+1
12x=6
x=6/12 :2
x=3/6 :3
x=1/2

Answer 2

a) { x+y = -5 → x = -5-y
{ 3x-2y = 0

3.(-5-y)-2y = 0
-15 -3y -2y = 0
-3y -2y = 0 +15
-5y = 15
-y = 15/5
-y = 3 (-1 )
y = -3
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b) {x + y = 2
{y = 3

x + 3 = 2
x = 2 - 3
x = -1
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c) {x - 2y = -4/12
{x = y + 1/12

y + 1/12 - 2y = -4/12
12y + 1 - 24y = - 4
12y - 24y = - 4 - 1
-12y = - 5
y = - 5/-12 . (-1 )
y = 5/12

x = 5/12 + 1/12
12x = 5 + 1
12x = 6
x = 6/12 : 2
x = 3/6 : 3
x = 1/2