Please use the ternary equation to solve the problem! O O thank you!!

Move the denominator to the addition and subtraction equations, and then add and subtract the same letter to subtract the same letter, and pull it in solving a binary system of equations (the detailed process is not convenient for mobile phone users).

x=(y+z+200)*1/4 1

y=(x+z+200)*2/7 2

z=(x+y+200)*8/7 3

1: Multiply 4, 4x=y+z+200, 4x-y-z=200 2: multiply 7y=2x+2z+400, -2x+7y-2z=400

3: Multiply 7 on both sides 7z=8x+8y+1600, -8x-8y+7z=1600

Multiply 7: 28x-7y-7z=1400

:26x-9z=1800 ⑥

Multiply 8: 32x-8y-8z=1600

Minus: 40x-15z=0 to get x=3 8z substitution, z=substitution and finally get y=600

Solution: x=(y+z+200)*1 4 .ay=(x+z+200)*2/7 ..b

z=(x+y+200)*8/7 ..c

a: multiply both sides by 4, 4x=y+z+200, 4x-y-z=200 b: multiply both sides by 7 7y=2x+2z+400,7y-2x-2z=400 c:

Multiply 7 on both sides 7z=8x+8y+1600, 7z-8x-8y=1600

Multiply 7: 28x-7y-7z=1400

:26x-9z=1800 ⑥

Multiply 8: 32x-8y-8z=1600

Minus: 40x-15z=0 to get x=3 8z substitution, z=substitution and finally get y=600

y+z-4x+200=0 (1)

x+z-7/2y+200=0 (2)

x+y-7 code wax mask code 8z+200=0 (3)1)-(2):9 2y-5x=0 (4).

1)-(3):15/8z=5x=0

Replace y and z with x to represent the expression (1).

10/9x+8/3x-4x+200=0

x=900y=9 and 10x=1000

z=8/3x=2400

4x=y+z+200 ①

7y=2(x+z+200) ②

7z=8(x+y+200) ③

By y=4x-z-200 generation of Xun staring

Finishing yields z=8x 3

then y=4x 3-200

Substituting Keshi Chang jujube to get x=1400

Then y=500 3

z=11200 3

x+y+z=10 (1).

x+1 2y+2z=7 (2).

1 3x+1 2y+z=2 (3).

2)-(3) Deaf.

2/3x+z=5 (4)

2) *2-(1) De.

x+3z=4 (5)

Get x=11 and substitute x=11 into (5) to get.

z==-7/3

Substituting the values of x,z into equation (1).

y=4 3 The key to solving multivariate equations is to eliminate elements, in general, we use the addition and subtraction elimination method, and the key point in the use of this method is to make the absolute value of the coefficients of an unknown number equal.

This is a ternary equation.

First, subtract the two formulas to subtract one element to obtain two new binary one-dimensional equations, and then apply the substitution method to solve the one element, and substitute the previous formula to solve the other two elements is very simple, and it is easy to master the method.

This is a system of ternary equations.

It's simple to use the commutation method.

1 2y-z=3

*3 gives 1 2y+2z=-4 and we get z=-7 3 and then y=4 3

x=11

x+y+z=10 （1）

x+1/2y+2z=7 （2）

1/3x+1/2y+z=2 （3）

1)-(2) 1 2y-z=3 (4)(3)*3-(2) y+z=-1 (5)(4)+(5) gives y=4 3

Substituting y=4 3 into (5) gives z=-7 3 Substituting the values of y and z into (1) gives x=11

i.e. x=11, y=4 3, z=-7 3

x+y+z=10

x+1/2y+2z=7

1/3x+1/2y+z=2

x+y+z=10 (1)

2x+y+4z=14 （2)

2x+3y+6z=12 (3)

2z-y=-6 (4)

y+4z=-8 (5)

4) + (5) formula has.

6z=-14 z=-7/3

Z is substituted into (5).

y=4 3x,y substituted by (1).

x=9 so:

x=9y=4/3

z=-7/3

I'll correct that.

x+y+z=10 ……1) formula.

x+1/2y+2z=7 ……2) formula.

1/3x+1/2y+z=2 ……3) formula.

1）-2） 1/2y-z=3 z=1/2y-31/3* 1）-3） -1/6y-2/3z=4/3-1/6y-2/3(1/2y-3)=4/3y=4/3, z=-7/3,x=11

1. (1)+(2) gives 6x+3y+4z=24 (4)4)-(3) 2y=4

y=2 substituting y=2 into (1) and (2) to obtain.

2x+3z=11 (5)

4x+z=7 (6)

5)*2-(6) gives 5z=22-7

z=3 substituting z=3 into (6) gives 4x+3=7x=1, so x=1, y=2, z=3

2. Let burn (1)+(2)-(3) to get 4y+z=16z=16-4y (4).

Substituting (4) into (3) gives 3x-18y=-48x-6y=-16

x=6y-16 (5)

Substituting (4) and (5) into (3) gives 30y-80-6y+112-28y=20

y=3 substituting y=3 (Tanlu Xu5) to get x=6*3-16=2, substituting y=3 into (4) to get z=16-4*3=4, so x=2, y=3, z=4

。Hope.

It's not a god, but you can, I just became a full member today, and it's exactly 60 points.

Solution: Substituting a*b=2 into a*b+c=2 obtains: 2+c=2 So the solution is Bi Zheng c=0

a=-3-b can be seen from a+b=-3-b

Substituting a=-3-b into a*b=2 obtains: (-3-b)*b=2, i.e., b 2 + 3b + 2 = 0

The solution yields b = -1 or -2

When b=-1, substituting a*b=2 gives a=-2

When b=-2, the hand chakra is substituted for a*b=2 to get a=-1, so the solution of the equation is:

a=-1 b=-2 c=0 or a=-2 b=-1 c=0

2x+y+z=-1 3y-z=-1 3x+2y+3z=-5 From: z=3y+1

Substituting , 2x+4y+1=-1

That is, 2x+4y=-2 3x+11y+3=-5

i.e. 3x+11y=-8 2- 3:

10y=-10

y=-1 substitution: z=-2

Substituting :x=1

The solution of the system of equations is:

x=1，y=-1，z=-2

The sum of the three formulas is 2a+2b+2c=308,,, then a+b+c=154 (4 formulas). Do you want me to talk about the rest of it?

Subtract 123 formulas from 4 formulas.

a＋b＝100 ①

a＋c＝102②

b＋c＝106③

Solution: - get: b-c=-2

Get: 2b = 104

b = 52 Substitute b = 52 to obtain: 52 + c = 106

Substitute c=54 for c=54 to obtain: a+54=102

a=48

Add the 3 equations together.

a+b+c=(100+102+106) 2=102 2 3, so c=8 3

b=2/3a=—10/3

2x+3y-4z=24 ①

x+2z=-4 z-x-2y=6 2 gives 4x+6y-8z=48

3 yields 3z-3x-6y=18

x-5z=66

Get 7z = -70

z=1 gives z=1 to get x+2=-4

x=-6 substitutes x=-6 and z=-1 to get -6+2-2y=6y=-5

2x+3y-4z=24 1

x+2z=-4 2z-x-2y=6 3

Formula 2 x2 gets: 2x+4z=-8 41+4 gets: 4x+3y=16 52+3 gets:

3z-2y=10 63 x4 gets: 4z-4x-8y=24 77 + 1 gets: 2x+5y=-48

i.e.: 4x+10y=-96 8

Subtracting 5 from 8 yields: 4x+10y-4x-3y=-96-16 gets: 7y=-112 y=-16

Substituting y=-16 into equation 5 yields: x=16

Substituting x=16 into equation 2 yields: z=-10