The ratio of the ternary system of equations is difficult, and the master will !!!!!!!!!

3a-b 3=2a+c 5 ==> a=b 3+c 52a+c 5=2b+c 7 Substituting the above formula a.

2*(b/3+c/5)+c/5=2b+c/7 ==> 16/35*c=4/3*b

c 35=b 12, substituting b 3=4c 35 into a formula.

a=4c 35+c 5=11c 35 == a 11=c 35 so a:b:c=11:12:35

From this equation we get three equations.

3a-b/3=2a+c/5……(1)

2a+c/5=2b+c/7……（2）

3a-b/3=2b+c/7……（3）

The coupling of the three formulas can be solved to give a c=1 7, a b=11 12, then c=7a, b=12a 11

So a:b:c=a:12a 11:7a=11:12:77

From 3a-b 3=2a+c 5, a=b 3+c 5 can be substituted into 2a+c 5=2b+c 7.

2b/3+2c/5+c/5=2b+c/7

12c = 35b

Substituting a=b 3+c 5 gives 12a=11b

then a:b:c=11 12:1:35 12=11:12:35

a：2x+2y+z+8＝0

b：5x+3y+z+34=0

c：3x-y+z+10=0

Step 1: Eliminate an unknown number x to get a system of binary equations for yz. (To see this problem, of course, it is most convenient to eliminate z first, because there is only one z in each of the three equations.) The asterisk * below indicates the multiplier sign:

a：15*(2x+2y+z+8)＝15*0

30x+30y+15z+120=0

b：6*(5x+3y+z+34)=6*0

30x+18y+6z+204=0

c：10*(3x-y+z+10)=10*0

30x-10y+10z+100=0

a－b: (30x+30y+15z+120)－(30x+18y+6z+204)=0

30-30)x+(30-18)y+(15-6)z+(120-204)=0

0x+12y+9z-84=0

12y+11z-84=0

a-c: (30x+30y+15z+120)－(30x-10y+10z+100)=0

30-30)x+(30+10)y+(15-10)z+(120-100)=0

0x+40y+5z-20=0

40y+5z-20=0

The binary system of equations for yz is obtained:

c：12y+9z-84=0

d：40y+5z-20=0

Step 2: Eliminate one more unknown, eliminate Z.

c：12y+9z-84=0

5*（12y+9z-84）＝5*0

60y+45z－420＝0

d：40y+5z-20=0

9*（40y+5z-20）＝5*0

360y+45z－180＝0

c－d：（60y+45z－420）－（360y+45z－1800）＝0

60－360）y+（45－45）z+（－420+180）＝0

300y+0z－600＝0

300y＝600

Y 2 Step 3: Substitute Y = 2 into Group C:

c：12y+9z-84=0

12*（－2）+9z-84=0

24+9z－84＝0

9z－（24+84）＝0

9z＝108

Z 12 Step 4: Substitute (y=2) and (z=12) into group A:

a：2x+2y+z+8＝0

2x+2*(-2)+(12)+8=0

2x=-16

x=-8 and finally get the result:

x=-8y＝-2

z＝12

Eliminate one first, such as z:1 minus 2 and 1 minus 3, so that 2 equations about xy can be obtained, according to the general solution.

1. Remove the denominator from x 2-y 4=1 to get 2x-y=4(1), remove the denominator from x 3+y 5=-1 15 to get 5x+3y=-1 (2), get y=2x-4(3) from (1), substitute (3) into (2) to get x=3(2x-4)=-1 to get x=1, substitute x=1 into (3) to get y=-2, so the solution of the original binary linear equation system is x=1, y=-2.

x-2y+z=3(1) 2x+y-z=4(2)4x+3y+2z=-10(3), z=2x+y-4(4) is obtained from (2), and (4) is substituted into (1)(3) to obtain 3x-2y+(2x+y-4)=3 to obtain 5x-y=7(5), 4x+3y+2(2x+y-4)=-10 is simplified to obtain 8x+5y=-2(6), and x=1 is obtained by substituting (4) into (4) to obtain z=-4 So the solution of the original ternary system of equations is x=1 y=-2 z=-4.

x+6y+3z=6(1) 3x-12y+7z=-3(2) 4x-3y+4z=11(3) x=3-3y-3z 2(4) from (1), and substituting (4) into (2)(3) to obtain 3(3-3y-3z 2)-12y+7z=-3 simplifying to obtain 21y-5z 2=12(5),4(3-3y-3z 2)-3y+4z=11 simplifying to obtain 15y+2z=1(6) from (5)(6) to obtain y=1 3 ,z=- 2 Substituting (4) gives x=5 So the solution of the original ternary system of equations is x=5 , y=1 3 , z=-2 .

;5x+3y=1;

Add the two formulas to get x=13 11; then y=-18 112The process is too lazy to fight, in short, x=1; y=-2;z=-4;y=1/3;z=-2

4x-5y+2z=0 =>4x-5y=-2zx+4y-3z=0 =>x+4y=3z

Addition and subtraction can be obtained.

x=z/3y=2z/3

So x:y:z=1 3:2 3:1=1:2:3

Ways of thinking.

The first elimination of b, and the binary system of equations formed by a and c.

Look at the process experience.

If you are satisfied, please adopt it in time. Thank you!

A system of ternary equations is basically a system of binary equations that uses a formula to replace one of the unknowns with another, and then becomes a system of binary equations.

For example, in this problem, 4a-9c=17, how much a can c be used instead, and then bring the other two formulas to solve the binary system of equations.

a=5,b=-2,c=1 3,The calculation process is shown in the figure below (it should be calculated like this, I haven't done math for a long time, and I have forgotten a lot).

There are an infinite number of solutions to this system of equations, let z=t

Then 2x+4y+10z=50, that is, x+2y=25-5t......x+y+z=16, i.e., x+y=16-t......Yes, y=9-4t

Substituting y=9-4t z=t into x+y+z=16, x=16-y-z=7+3t

x=7+3t y=9-4t z=t

（x+y）-（y+z）=x-z=7

z+x=3 then z and x of the ball, and then find y

The same can be said for the second question.

Let the number of days for the production of parts A, B and C be x, y and z days respectively: x+y+z=63 (1).

600x=300y=500z=k (2) So: x=k 600, y=k 300, z=k 500 Substitute this into (1) respectively: k=9000

i.e.: x=15, y=30, z=18.

It takes 15, 30, and 18 days to produce each.

Solution: If A takes x days, B days, and C z days, then it is derived from the title:

x+y+z=63

600x=300y=500z

6x=3y=5z

y=2x z=6/5x

Substituting x+y+z=63

Solution: x=15 (days).

y=30 (days).

z=18 (days).

Hope it helps!

The idea is, eliminate the element.

Find any one of the unknowns in the two equations, the least common divisor of the coefficients that precede them