Let ABC ≠ 0, 3A + 2b-7c = 0, 7a + 4b-15c = 0, find 4a2-5b2-6c2 / A2 + 2B2 + 3c2 Let ABC ≠ 0, 3A + 2b-7c = 0, 7a + 4b-15c = 0, find 4a2-5b2-6c2 / A2 + 2B2 + 3c2 the sooner the better

Let ABC ≠ 0, 3A + 2b-7c = 0, 7a + 4b-15c = 0, find 4a2-5b2-6c2 / A2 + 2B2 + 3c2 Let ABC ≠ 0, 3A + 2b-7c = 0, 7a + 4b-15c = 0, find 4a2-5b2-6c2 / A2 + 2B2 + 3c2 the sooner the better

From 3A + 2B = 7C (1)
7a+4b=15c （2）
（2）-（1）×2,
A = C, B = 2C,
Substituting (4a & sup2; - 5B & sup2; - 6C & sup2;) / (A & sup2; + 2B & sup2; + 3C & sup2;)
=（4c²-20c²-6c²）/（c²+8b²+3c²)
=-11/6.

In △ ABC, a = 2, B = 5, C = 6, then CoSb=______ ．

In ∵ △ ABC, a = 2, B = 5, C = 6, from the cosine theorem, we get CoSb = A2 + C2 − b22ac = 4 + 36 − 252 × 2 × 6 = 58, so the answer is: 58

-3A + 10C = 73,9a + 7b = - 30, 8b + 5C = 11 find out how much ABC equals

-3A + 10C = 73 times 3 - 9A + 30C = 219 plus formula 9A + 7b = - 30
We get 30C + 7b = 189
8b + 5C = 11 times 6 30C + 48b = 66 minus 30C + 7b = 189
B = - 3
Substitute 8b + 5C = 11, C = 7
Substituting - 3A + 10C = 73 gives a = - 1
So ABC = 21

Elementary mathematics. Solve the system of linear equations of three variables. Y + z-3x = 3 and Z + x-3y = 5 and X + y-3z = 6

-3x+y+z=3 ①
x-3y+z=5②
x+y-3z=6③
①X3.④
②X3.⑤
④+③.⑥
⑤+③.⑦
⑥ And (7) do binary linear equations
Calculate the solution of X or Y, and then substitute it with (6) or (7) to calculate the solution of the other
After calculating X and y, substitute them into (1) or (2) or (3) to calculate the solution of Z
{x=...}
{y=...}
{z=...}
I can only tell you the way of thinking. You have to do it yourself,

Given the equations {x + 3Y = 8, y + 3Z = 9, Z + 3x = 11, then x + y + Z=

x+3y=8①
y+3z=9②
z+3x=11③
①+②+③
4x+4y+4z=8+9+11
4(x+y+z)=28
x+y+z=7

The equations are: 1) 3x + y + Z = 6; 2) x + 3Y + Z = 5; 3) x + y-3z = 9

3x+y+z=6 （1）
x+3y+z=5 （2）
x+y-3z=9 （3）
Add it all up and you'll have to
3x+y+z+x+3y+z+x+y-3z=6+5+9
5x+5y-z=20(4)
(3) X5: 5x + 5y-15z = 45 (5)
(4) (5) get: 14z = - 25 z = - 25 / 14
Take z = - 25 / 14 into (1) (3) to get
3X+y=109/14 (6) X+y=9-75/14 X+y=51/14(7)(
6) (7) we get: 2x = 58 / 14, x = 29 / 14, y = 22 / 14 = 11 / 7

Equations x + y-3z = 1 y + z-3x = 2 x + x-3y = 3

According to the three formulas, y = 3z-x, y = 2 + 3x-z, y = (2x-3) / 3
According to the first and the second, we get 3z-x = 2 + 3x-z, 4z-4x = 2 and z-x = 1 / 2
According to the first and the third, we get 3z-x = 2 / 3x-1 and 9z-5x = - 3
According to Formula 1 and formula 2, x = - 15 / 8, z = - 11 / 8
Take x = - 15 / 8 and z = - 11 / 8 into the formula to get y = - 9 / 4

It is necessary to solve the system of linear equations with three variables
1.x+2y=5
y-3z=-7
4z=x=13
2.x:y:z=4:2:3
2x+3y-5z=-2
3. Three thirds x = four quarters y = five fifths Z
x+y+z=24

The third question
x/3=y/4=z/5=k
x=3k y=4k z=5k
x+y+z=3k+4k+5k=24
k=2
x=6,y=8,z=10
Is there a problem with the first question? 4Z = x = 13?
Second question
x=4k y=2k z=3k
2x+3y-5z=8k+6k-15k=-2
k=2
x=8 y=4 z=6

Solving the system of linear equations with three variables (process required!)
Sorry..... 1. Y = 2x-7 5x + 3Y + 2Z = 2 3x-4z = 4 2.4x + 9y = 12 3y-2z = 1 7x + 5Z = 19 / 4
3.4x-9z = 17 3x + y + 15z = 18 x + 2Y + 3Z = 24.2x + 4Y + 3Z = 9 3x-2y + 5Z = 11 5x-6y + 7z = 13 1!

First question:
（1）.y=2x-7
（2）.5x+3y+2z=2
（3）.3x-4z=4
By substituting Formula 1 into formula 2, it is reduced to: 11x + 2Z = 23 (4)
(3) + (4) * 2 can eliminate Z to get: x = 2
Substituting (4) gives: z = 1 / 2
Substituting the values of X and Z into one of the original equations, we can get y = - 3
The solution is: x = 2, y = - 3, z = 1 / 2
Second question:
(1)4x+9y=12
(2)3y-2z=1
(3)7x+5z=19/4
(2) * 5 + (3) * 2: 14x + 15y = 29 / 2 (4)
(4) * 3 - (1) * 5: x = - 3 / 4
Substituting (4) gives y = 5 / 3
Substituting into the original equation: z = 2
The solution of the original equation is: x = - 3 / 4, y = 5 / 3, z = 2
Third question:
(1)4x-9z=17
(2)3x+y+15z=18
(3)x+2y+3z=2
(2) * 2 - (3): 5x + 27z = 34 (4)
(1) * 3 + (4): x = 5
Substitute (4) to get: z = 1 / 3
Substituting into the original equation, y = - 2
The solution of the original equation is: x = 5, y = - 2, z = 1 / 3
Fourth question:
(1)2x+4y+3z=9
(2)3x-2y+5z=11
(3)5x-6y+7z=13
(2) * 2 + (1): 8x + 13z = 31 (4)
(2) * 3 - (3): x + 2Z = 5 (5)
(5) * 8 - (4): z = 3
Substitute (5) to get: x = - 1
Substituting into the original equation, y = 1 / 2
The solution of the original equation is: x = - 1, y = 1 / 2, z = 3
So tired

To solve a system of linear equations of three variables,
（1）
4x-9z=17
3x+y+15z=18
x+2y-4z=4
（2）
2x+4y+3z=9
3x-2y+5z=11
5x-6y+7z=13

1:4X-9Z=17 (1)
3X+Y+15Z=18 (2)
X+2Y-4Z=4 (3)
(1)-(3)*4 -8Y+7Z=1 (4)
(2)-(3)*3 -5Y+27Z=6 (5)
(4)*5 -40Y+35Z=5 (6)
(5)*8 -40Y+216Z=48 (7)
(7)-(6) 181Z=43
Z=43/181
Substituting (4) y = (7 * 43 / 181-1) / 8 = 120 / 8 / 181 = 15 / 181
Substituting (3) x = 4 + 4 * 43 / 181-2 * 15 / 181 = 582 / 181
2:2X+4Y+3Z=9 (1)
3X-2Y+5Z=11 (2)
5X-6Y+7Z=13 (3)
(1)+(2)*2 8X+13Z=31 (4)
(2)*3-(3) 4X+8Z=20 (5)
(5)*2-(4) 3Z=9
Z=3
Substituting (5) x = (20-8 * 3) / 4 = - 1
X=-1
Substituting (1) y = (9-2 * (- 1) - 3 * 3) / 4 = 1 / 2