Resuelva por cualquiera de los 3 métodos estas ecuaciones
a. 12(x + 2y) – 8(2x +y) = 2(5x –6y) Y 20(x-4y) = --10
b. 5x –3y = 11 Y 4x + y = 2
Respuestas
a.
12(x + 2y) – 8(2x +y) = 2(5x –6y)
12x + 24y - 16x - 8y = 10x - 12y
12x - 16x - 10x + 24y - 8y + 12y = 0
12x - 26x + 36y - 8y = 0
-14x + 28y = 0
20(x-4y) = -10
20x - 80y = -10
Método de sustitución
-14x + 28y = 0 Ec1
20x - 80y = -10 Ec2
x2 = -10 + 80y/20
-14 (-10 + 80y/20) + 28y = 0
-14 (-10 + 80y) = 20 (-28y)
140 - 1120y = -560y
-1120y + 560y = -140
(-) -560y = -140
560y = 140
y = 140/560
y = 1/4
-14x + 28y = 0 Ec1
-14x + 28(1/4) = 0
-14x + 7 = 0
(-) -14x = -7
14x = 7
x = 7/14
x = 1/2
20x - 80y = -10 Ec2
20x - 80 (1/4) = -10
20x - 20 = -10
20x = -10 + 20
20x = 10
x = 10/20
x = 1/2
R// x = 1/2 ; y = 1/4
Comprobación
-14x + 28y = 0 Ec1
-14 (1/2) + 28 (1/4) = 0
-14/2 + 28/4 = 0
-7 + 7 = 0
0 = 0
20x - 80y = -10 Ec2
20 (1/2) - 80 (1/4) = -10
20/2 - 80/4 = -10
10 - 20 = -10
-10 = -10
b.
Método de igualdad
5x –3y = 11 Ec1
4x + y = 2 Ec2
x1 = 11 + 3y/5
x2 = 2 - y/4
11 + 3y 2 - y
_____ = _____
5 4
4 (11 + 3y) = 5 (2 - y)
44 + 12y = 10 - 5y
12y + 5y = 10 - 44
17y = -34
y = -34/17
y = -2
5x –3y = 11 Ec1
5x - 3 (-2) = 11
5x + 6 = 11
5x = 11 - 6
5x = 5
x = 5/5
x = 1
4x + y = 2 Ec2
4x + (-2) = 2
4x - 2 = 2
4x = 2 + 2
4x = 4
x = 4/4
x = 1
x = 1 ; y = -2
Comprobación
5x –3y = 11 Ec1
5(1) - 3(-2) = 11
5 + 6 = 11
11 = 11
4x + y = 2 Ec2
4(1) + (-2) = 2
4 - 2 = 2
2 = 2
Espero haberte ayudado. Suerte en los estudios ≧◠◡◠≦.