math.(三元一次方程)

x + y = 26 - z ─── (1)

x - 1 = y ─── (2)

2x = 18 + y - z ─── (3)



Sub y = x - 1 into (1)

x + (x - 1) = 26 - z

2x = 27 - z

z = 27 - 2x



Sub y = x - 1 and z = 27 - 2x into (3)

2x = 18 + (x - 1) - (27 - 2x)

2x = 18 + x - 1 - 27 + 2x

x = 10


y = 10 - 1 = 9


z = 27 - 2(10) = 7

x + y = 26 - z ─── (1)
x - 1 = y ─── (2)
2x = 18 + y - z ─── (3)

From(2),we' ve got y=x-1

Sub y = x - 1 into (1)
x + (x - 1) = 26 - z
2x = 27 - z
z = 27 - 2x

Sub y = x - 1 and z = 27 - 2x into (3)
2x = 18 + (x - 1) - (27 - 2x)
2x = 18 + x - 1 - 27 + 2x
x = 10

Because y=x-1
Therefore, y = 10 - 1 = 9

Because z=27-2x
Therefore, z = 27 - 2(10) = 7

SUB (x-1=y) IN(x+y=26-z),
x+x-1=26-z
x=(27-z)/2

SUB (x-1=y) AND (x=(27-z)/2) IN (2x=18+y-z),
2[(27-z)/2]=18+x-1-z
2[(27-z)/2]=18+(27-z)/2-1-z
54-2z=36+27-z-2-2z
z=36+27-2-54
z=7

SUB(z=7) IN (x+y=26-z)
x+y=26-7
x+y=19
x=19-y

SUB(x=19-y) IN (x-1=y)
19-y-1=y
2y=18
y=9

SUB(y=9) IN (x=19-y)
x=19-9
x=10
SO, x=10, y=9, z=7.

x + y = 26 - z...............(1)
x - 1 = y.......................(2)
2x = 18 + y - z..............(3)

Sub (2) into (1), we have
x + (x - 1) = 26 - z
2x = 27 - z.....................(4)

Sub (2) into (3), we have
2x = 18 + (x - 1) - z
x = 17 - z.......................(5)

(4) - (5) : x = 10

Sub x = 10 into (5), we have
z = 7

Sub x = 10 into (2), we have
y = 9

Hence, we have x = 10, y = 9, z = 7