1. If the two solutions of the equation AX + by = 10 are x = 6, y = - 4 and x = 2, y = 2, then a = () B = () If 3x-6y = 5, then 2y-x + 1 = () Given that the solution of the system of equations 2x + 3Y = 7 is also the solution of the system of equations MX + 2ny = 6, then M = (), n = () 2x-y=3 3nx-my=11

1. If the two solutions of the equation AX + by = 10 are x = 6, y = - 4 and x = 2, y = 2, then a = () B = () If 3x-6y = 5, then 2y-x + 1 = () Given that the solution of the system of equations 2x + 3Y = 7 is also the solution of the system of equations MX + 2ny = 6, then M = (), n = () 2x-y=3 3nx-my=11

1. Substitute x = 6, y = - 4 and x = 2, y = 2 into the original equation
A new system of linear equations with two variables is obtained
6a-4b=10
2a+2b=10
The solution is a = 3
b=2
2. Divide both sides of the equation by - 3, add 1, and you get - 2 / 3

(1+20%)x*90%=270

(1+20%)x*90%=270
1.2x=300
x=250

Solution equation: (x-8000) · (1 + 10%) = (1 + 35%) x-11800
（x-8000 ）·（1+10%）= (1+35%)x-11800
The final result is 12000

Solution
(x-8000)(1+10%)=(1+35%)x-11800
(x-8000)×1.1=1.35x-11800
1.1x-8800=1.35x-11800
1.1x-1.35x=8800-11800
-0.25x=-3000
x=12000

270 / x = 240 / X-10 to solve the equation

270/x=240/x-10
270(x-10)=240x
Transfer of items
270x-240x=2700
30x=2700
x=90
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In non-zero sequence an, if (an + 1-2n-1) (an + 1-an-1) = 0, then an is equal ratio sequence or equal difference sequence, right

That is, a (n + 1) = 2N-1
Or a (n + 1) - an = 1
So it should be an arithmetic sequence
So it's not right

It is known that the K, N, P terms of the arithmetic sequence constitute three consecutive terms of the arithmetic sequence. If the arithmetic sequence is not a constant sequence, what is the common ratio of the arithmetic sequence?

1. To simplify the problem, if the common ratio is Q and the k-th term is a, then the n-th term is AQ and the p-th term AQ ^ 2
A + AQ ^ 2 = 2A. If a is reduced, Q ^ 2-2q + 1 = 0, so q = 1

It is known that the second, third and sixth terms of the arithmetic sequence whose first term is not zero form an arithmetic sequence in turn, and the common ratio of the sequence is obtained
I want to ask why not consider the case of D = 0?

Let the first term be a, the tolerance be D, a! = 0,
Common ratio q = (a + 2D) / (a + D) = (a + 5d) / (a + 2D)
The solution is d = - 2A
So the common ratio q = (a + 2D) / (a + D) = 3
Because the title said, the first item is not 0, a! = 0, so d = - 2A will not be 0

Constant sequence must be equal difference sequence with zero tolerance and equal ratio sequence with one common ratio,
If every term is zero, can it be said that it is an equal ratio sequence with any common ratio?

A constant sequence must be an arithmetic sequence with zero tolerance, but not an arithmetic sequence with one common ratio
A sequence whose every term is zero cannot be said to be an equal ratio sequence whose common ratio is any number
Because if it is, then its common ratio will be 0 divided by 0. We know that 0 can not be used as a divisor, so this common ratio is meaningless. Therefore, it is not an equal ratio sequence

Is the sequence 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,..., an arithmetic sequence or an arithmetic sequence, or both or none?

It's an arithmetic sequence, not an arithmetic sequence
Is an arithmetic sequence with a tolerance of 0
Because the common ratio cannot be zero, this is not an equal ratio sequence