How to solve the equation 6x + 8 = 4x-2? (there should be a solution process) thank you!

How to solve the equation 6x + 8 = 4x-2? (there should be a solution process) thank you!

6x+8=4x-2
The result of the transfer is as follows:
6x-4x= -2-8
2x= -10
x= -10÷2
x= -5
The X item moves to the left and the other numbers to the right
6x-4x=-2-8
2x=-10
x=-5
x=-5
6x-4x = - 2-8
Merge, get 2x = - 10
Divide both sides by two to get x = - 5
6x+8=4x-2
6x-4x=-2-8
2x=-10
x=-5
First shift the term, that is, 6x-4x = - 2-8 = - 10, that is, 2x = - 10, so the solution is x = - 5
Solve the system of linear equations of three variables, 14x-4y + 2Z = 9,3x + 2Y + 4Z = - 2,6x-4y-2z = 7, find the value of X, y, Z?
The answer is x = - 1 / 2, y = - 13 / 4, z = 3 / 2
How also can't calculate, simplify out binary linear equation is always the same
(1) + (3) get
20x-8y=16
5x-2y=4 (4)
(2) + (3) × 2
15x-6y=12
5x-2y=4 (5)
The two equations are of the same solution
The system of equations should have innumerable solutions
There's a problem with the functional relationship
(1) 14x-4y + 2Z = 9 and (3) add to get (4)
（2）3x+2y+4z=-2
(3) 6x-4y-2z = 7, both sides simultaneously × 2 get 12x-8y-4z = 14 -- add with (2) to get (5)
（4）20x-8y=16
（5）15x-6y=12
As you said: the equation is the same; did you copy the problem wrong?
Several methods to solve the general term formula of sequence
Sequence is an important part of high school mathematics, which is closely related to functions, equations and inequalities. In the textbook, the definition of sequence is given from two perspectives: one is descriptive definition, that is, "sequence is a sequence of numbers arranged in a certain order"; the other is function definition, that is, "sequence is a special kind of function, that is," sequence is a definition
X = 3Y = − 1 is not only the solution of the equation 4x + my = 9, but also the solution of MX NY = 11=______ ，n=______ .
Substituting x = 3Y = − 1 into the equations 4x + my = 9 and MX NY = 11, we get 12-m = 9, that is, M = 3; substituting M = 3 and x = 3Y = − 1 into MX NY = 11, we get n = 2. So m = 3, n = 2
Given the recurrence formula a (n) = 2 (an-1) + 2 * (- 1) ^ n (n ≥ 2), how to use the iterative method to find the general term?
A1 is 1
a(n) = 2a(n-1) + 2(-1)^n,
a(n)/(-1)^n = 2-2a(n-1)/(-1)^(n-1),
b(n) = a(n)/(-1)^n,
b(n) = 2 - 2b(n-1),
b(n) + x = -2[b(n-1) + x],2 = -3x,x = -2/3,
b(n) -2/3 = -2[b(n-1) - 2/3],
The {B (n) - 2 / 3} is an equal ratio sequence whose first term is B (1) - 2 / 3 = a (1) / (- 1) - 2 / 3 = - 1-2 / 3 = - 5 / 3, and the common ratio is - 2
b(n)-2/3=(-5/3)(-2)^(n-1),
a(n)=b(n)(-1)^n=[-5/3(-2)^(n-1)+2/3](-1)^n=[2(-1)^n + 5*2^(n-1)]/3,n=1,2,...
If x = 0y = 1 and x = 1y = 2 are two solutions of the equation MX + NY = 3, then M=______ ，n=______ .
Substituting x = 0y = 1 and x = 1y = 2 into the equation MX + NY = 3, n = 3M + 2n = 3 and M = - 3N = 3 are obtained
Is the range of application of iterative method and accumulation method in solving the general term formula of sequence the same? It seems that iterative method can solve all the problems of accumulation
Iterative method is only need to know the relationship between the two connected terms can be used, and the cumulative method needs the relationship between the two terms to be a linear relationship with equal coefficients, that is a (n + 1) = a (n) + B. for some relationships, such as a (n + 1) = Ka (n) + B, the cumulative method is more difficult to work. At this time, we need to construct a new sequence, so that a (n + 1) + constant C = K (a (n) +
2 (X-2) - 3 (4x-1) = 9 (1-x) solve equation-2 (x-1) = 4 in different ways
Solve the equation. Thank you
2（x-2）-3（4x-1）=9（1-x）
2x-4-12x+3=9-9x
-1-10x=9-9x
10x-9x=-1-9
x=-10
-2(x-1)=4
-2x+2=4
2x=2-4
2x=-2
x=-1
-2(x-1)=4
x-1=-2
x=-2+1
x=-1
Use the observation method to find the general formula of the following sequence
①1/2,4/9,3/8,8/25,5/18,12/49.
②-3,7,-13,21,-31.
③1,4,9,16.
First question:
Change 1 / 2 to 2 / 4, 3 / 8 to 6 / 16, 5 / 18 to 10 / 36
The original sequence is 2 / 4, 4 / 9, 6 / 16, 8 / 25, 10 / 36, 12 / 49
So the general formula is 2n / (n + 1) & sup2;
Second question:
(-1)^n{(n+1)²-n}
Third question:
n²
1. 2n/(n+1)^2
2. （-1）^n*[(n+1)²-n]
3. n^2
(2) (- 1) & sup2; (2 Λ n + (n-1))
3. n²
①2n/(n+1)^2
②(-1)^n(n^2+n+1)
③n^2
First question
2n/(n+1)^2
Second question
(- 1) ^ n * [n ^ 2 + N + 1] idea: remove the negative sign, because it is one negative and one positive, so use (- 1) ^ n, and then become positive
The rule is that the difference between them is an arithmetic sequence
a1=3
a2-a1=7-3=4
a3-a2=13-7=6
a4-a3=21-13=8
.....
An-an -... Expansion
First question
2n/(n+1)^2
Second question
(- 1) ^ n * [n ^ 2 + N + 1] idea: remove the negative sign, because it is one negative and one positive, so use (- 1) ^ n, and then become positive
The rule is that the difference between them is an arithmetic sequence
a1=3
a2-a1=7-3=4
a3-a2=13-7=6
a4-a3=21-13=8
.....
An-an-1 = self calculation
The third question
N ^ 2
1 2n / (n + 1) square
2 (- 1) n power * (n + 1) * n
3 n square
Solving equation: 2 × (X-2) + 3 × (4x-1) = 9 × (x-1) + 12
I'm going crazy
2×（x-2)+3×（4x-1）=9×（x-1）+12
2x+12x-9x=4+3-9+12
5x=10
X=2
2×（x-2)+3×（4x-1）=9×（x-1）+12
Solution: 2X-4 + 12x-3 = 9x-9 + 12
14X-9X=10
5X=10
X=2
2x-4+12x-3=9x-9+12
10x-7=9x-3
10x-4=9x
X=4