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10 (y-x) = 11y - 12x ec. 2
2x+10 = 4y - 16x
10y -10x = 11y - 12x
2x+16x-4y = -10
12x-10x+10y-11y = 0
18x-4y = -10 ec. 3
2x -y = 0 ec. 4
si multiplicamos por -4 la segunda ecuación:
18x -4y = -10
-8x + 4y = 0
al sumarlas da:
10x = -10
x =-10/10
x = -1
al sustituir valores en al ec. 4:
2x-y = 0
2*-1-y = 0
-2-y = 0
-y = 2
y = -2
Ahora comprobémoslo:
En la ec. 1
2(-1+5) = 4(-2-(4*-1))
2*4 = 4(-2+4)
8 = 4*2
8≡8
en la ec. 2:
10(-2-(-1)) = 11*-2 - 12*-1
10*-1 = -22 - (-12)
-10 ≡ -10
Ahora El método de sustitución:
3x-4y-2(2x-7) = 0 ec. 1
5(x-1)-(2y-1) = 0 ec. 2
3x-4y-4x+14 = 0
5x-5-2y+1 = 0
3x-4x-4y= -14
5x-2y -4 = 0
-x-4y = -14 ec. 3
5x-2y = 4 ec. 4
de la ec. 3
-x = -14+4y
x = 14-4y ec. 5
sustituyendo la ec. 5 en la ec.4:
5(14-4y) - 2y = 4
70 - 20y - 2y = 4
-22y =4-70
-22y = -66
y = -66/-22
y = 3
Sustituyendo valores en la ec. 5:
x= 14-4*3
x =14-12
x = 2
Comprobémoslo en:
La ec. 1:
3*2 - 4*3 - 2(2*2-7) = 0
6-12 - 2(4-7) = 0
6-12-2*-3 = 0
6-12+6 = 0
0≡0
La ec. 2:
5(2-1)-(2*3-1) = 0
5*1-(6-1) = 0
5-5 = 0
0≡0
Saludos....
Primero las reducimos
metodo suma y resta
2(x+5)=4(y-4x) ec (1)
10(y-x)=11y-12X ec(2)
con ec (1)
2x + 10 =4y -16x
2x+16x-4y = -10
18x -4y = -10 dividimos para 2
9x - 2y = -5 ec (1)
con ecuacion (2)
10(y-x)=11y-12X
10y - 10x = 11y -12 x
10y - 10x -11y+12x = 0
-y+ 2x = 0 ec (2)
9x - 2y = -5 ec (1)
2x - y = 0 ec (2) multiplicamps por 2
9x - 2y = -5
4x -2y = 0
-----------------
5x = -5
x = -1
remplazamos x En ec (1)
2x- y = 0
2(-1 ) -y =0
-2 -y = 0
-y = 2
y= -2
----------------------------------------------------------------------
metodo de sustitucion
3x-4y-2(2x-7)=0 ec (1)
5(x-1)-(2y-1)=0 ec(2)
ec (1)
3x-4y-2(2x-7)=0
3x - 4y-4x +14 = 0
-x-4y = -14 ec (1)
ec (2)
5(x-1)-(2y-1)=0
5x-5 -2y+1 =0
5x -2y = 4 ec (2)
-x-4y = -14 ec (1)
5x -2y = 4 ec (2)
despejamos x en ec (1)
x =14 -4y ec(3)
sustituimos x en ec (2)
5(14-4y) -2y = 4
70 -20y -2y = 4
-22y = 4-70
-22y = -66
y = 66/22
y= 33/11
y = 3
remplazamos y en ec (3)
x =14 -4y ec(3)
x = 14 - 4(3)
x = 14 -12
x =2
saludos