Question

me ayudan a resolver estos dos sistemas de ecuaciones.

3/2 a+b=1

a+b/2=7




3x+2y+z=1

5x+3y+4z=2

x+y-z=1

Answer 1

(3/2)a + b = 1

a + (1/2)b = 7

reorganicemos las ecuaciones

3a + 2b = 2
2a + b = 14

ahora, b = 14 - 2a

entonces,
3a+2(14-2a) = 2
3a+28-4a = 2
-a = -26
a = 26

b = 14-2a
b = 14-2(26)
b = 14-52
b = -38

(a,b) = (26,-38)



3x+2y+z = 1
5x+3y+4z = 2
x+y-z = 1

z = 1-3x-2y  ∧  z = x+y-1 

→ z = z
1-3x-2y = x+y-1
4x+3y = 2
4x = 2-3y
x = (1/2) - (3/4)y

z = (1/2) - (3/4)y + y -1
z = (1/4)y - (1/2)

5((1/2) - (3/4)y) + 3y + 4((1/4)y - (1/2)) = 2
(5/2) - (15/4)y + 3y + y - 2 = 2
(1/4)y = 3/2
y = 6

x = (1/2) - (3/4)y
x = (1/2) - (3/4)(6)
x = -4

z = (1/4)(6) - (1/2)
z = 1

(x,y,z) = (-4,6,1)