Respostas

2013-12-10T11:29:43-02:00
 Questão de rapidez usamos o método da adição:

x-3y=-15       (Termos simétricos cancelamos)
2x+3y=60  
 
 x=-15
2x=60

3x=45
x=45/3
x=15

2x+3y=60
2.15+3y=60
30+3y=60
3y=60-30
3y=30
y=30/3
y=10

b)
x+3y=35            -2
2x-5y=15

-2y-6y=-70
2x-5y=15

-6y=-70
-5y=15

-11y=-55     (-1)
11y=55
y=55/11
y=5

x+3y=35
x+3.5=35
x+15=35
x=35-15
x=20

c)
x-5y=15
2x+y=19                  5

x-5y=15
10x+5y=95

11x=110
x=110/11
x=10

x-5y=15
10-5y=15
-5y=15-10
-5y=5  (-1)
5y=-5
y=-5/5
y=-1

d)
x+y=67        2
x-2y=46

2x+2y=134
x-2y=46

3x=180
x=180/3
x=60

x-2y=46
60-2y=46
-2y=46-60
-2y=-14     (-1)
2y=14
y=14/2
y=7

2013-12-10T11:39:20-02:00
A) x-3y= -15  
   2x+3y=60
   3x      =45
x= 45/3
x= 15

B) x+3y = 35  *(-2)           -2x-6y = -70
   2x-5y= 15                     2x-5y = 15
                                           -11y= -55  *(-1)
                                            11y= 55
                                              y=5

C) x-5y = 15  *(-2)            -2x+10y= -30               
  2x+y = 19                      2x+y  =  19  
                                            11y = -11
                                               y= -1

D) x+y = 67   *(-1)          -x-y= -67
    x-2y = 46                   x-2y= 46
                                       -3y= -21  *(-1)
                                        3y=21
                                         y=7