x+y+z=30 120x=600z 100y=400z

x+y+z=30 120x=600z 100y=400z

120x=600z
x=5z
100y=400z
y=4z
therefore
5z+4z+z=30
z=3
So y = 4 * 3 = 12
x=5*3=15

How to solve 120x = 80 (x + 20) equation

120x=80(x+20)
120x=80x+80×20
120x-80x=1600
40x=1600
x=1600÷40
x=40

An equation problem, 252x + 224y + 200z = 48100 ≤ y ≤ 20 ≤ Z ≤ 2, to find the integer solution of X
I calculate the money and turn it into an equation problem

0

Solve the equation {x + y = 60 {3 / 100x + 60 / 100y = 10 / 100 * 60}

X + y = 60 (1) 3 / 100x + 60 / 100y = 10 / 100 * 60 (2) (2) * 100, 3x + 60y = 600, x + 20Y = 200 (3) (3) - (1), 19y = 140, y = 140 / 19, substituting y = 140 / 19 into (1), x = 1300 / 19

Given 3a-5b + 19 = 0, a + 8b-1 = 0, find the value of 4a-26b

Subtracting formula two from each other, we get 2a-13b = - 20
Therefore, 4a-26b = - 40

It is known that 3a-5b + 19 = 0, a + 8b-1 = 0. Find the values of the following formulas: (1) - 2a-9b; (2) 4a-26b

This is a system of linear equations with two variables, a = - 147 / 29, B = 22 / 29
So - 2a-9b = - 492 / 29
4a-26b=-40
The slash represents the fraction line, which is to be typed out by formula in word. It's too troublesome, so I won't do it
You can read clams

Given 4A ^ 2 + 4A + B ^ 2-6a + 10 = 0, find a ^ 3b-b ^ 3a

It is known that 3a-5b + 19 = 0, a + 18b-1 = 0, and the value of 4a-4b is obtained

If 3a − 5B = − 19, ① a + 18B = 1, ② × 3 - ①, 59b = 22, i.e. B = 2259, substituting B = 2259 into ②, i.e. a = - 33759, then 4a-4b = - 143659

Given the equations a + 3B + C = 0, 3a-3b-4c = 0, find a: B: C
The answer is a: B: C = 15:7:6, but there is no process

Given the equations a + 3B + C = 0, 3a-3b-4c = 0, find a: B: C
a+3b+c=0①
3a-3b-4c=0②
① + 2: 4a-3c = 0
4A = 3C
a=3 c=4
Substituting a = 3 and C = 4 into (1) gives the following result:
3+3b+4=0
b=-7/3
a：b：c=3:：-7/3：4=9：-7:12

Let ABC not equal to 0, 3A + 2b-7c = 0, 7a + 4b-15c = 0, find the value of 4A ^ 2-5b ^ 2-6c ^ 2 / A ^ 2 + 2B ^ 2 + 3C ^ 2, let a

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.