It is known that the system of equations x + 2Y = 10, ax + by = 1 has the same solution as the system of equations BX + AX = 6, 2x-y = 5 It is known that the system of equations x + 2Y = 10, ax + by = 1 about X, y has the same solution as the system of equations BX + ay = 6, 2x-y = 5. Find the value of a, B

This means that X and y satisfy the above four equations at the same time, so we first solve two of them x + 2Y = 10,2x-y = 5 to get x, y, that is, x = 4, y = 3. Substitute them into ax + by = 1, BX + ay = 6, that is, 4A + 3B = 1,4b + 3A = 6, and get a = - 2, B = 3

In the equation x + y = 10, it is known that X and y are natural numbers. Try to find the probability that X and y are both positive integers

Just exclude the case where X or Y equals 0
When x is equal to 0 or 10, it is not satisfied
And X has 0 to 10 11 cases
So the probability of dissatisfaction is 2 / 11
The probability of satisfaction is 9 / 11

Given that (x = 2 and (y = 5) and (x = 1 and (y = 10) are two solutions of the equation AX + by = 15, find the values of a and B

Bring in the values of X and y
2a＋5b＝15
1a＋10b＝15
Solution
a＝5
b＝1

Granny Wang's family has 45 ducks and 70 chickens. She has as many geese as ducks. How many chickens, ducks and geese are there?

45+45+70
=90+70
=160

Granny Wang's family has 246 chickens and ducks, three times as many as chickens. How many chickens and ducks are there altogether

246+246*3=984

Granny Wang raised some chickens and ducks in her family. There are 18 ducks. The number of ducks is 3 / 8 of that of chickens. How many chickens do you have to solve the equation

Suppose there are x chickens
3 / 8 * x = 18
Is there a problem with this question?

Granny Wang raised some chickens and ducks. How many chickens and ducks are there? I know that the number of chickens is three times that of ducks. I also know that there are 18 more chickens than ducks
Granny Wang raised some chickens and ducks. How many chickens and ducks are there? I know that the number of chickens is three times that of ducks. I also know that there are 18 more chickens than ducks

If x is set for duck, 3x is set for chicken
The equation system x + 18 = 3x is obtained from the meaning of the problem
If x = 9, then 3x = 27
So there are 27 chickens and 9 ducks

(1) It is known that the sequence {an} is equal difference sequence, and it is proved that the sequence {e ^ an} is equal ratio sequence
(2) It is known that the sequence {an} is an equal ratio sequence, and an > 0. It is proved that the sequence {logean} is an equal difference sequence
(requires complete process Please...)

(1) Let the tolerance of {an} be D, BN = e ^ an, then B (n + 1) / BN = e ^ a (n + 1) / e ^ an = e ^ (a (n + 1) - an) = e ^ D, which is a non-zero constant, so {BN} is an equal ratio sequence. (2) let the common ratio of {an} be q, because an > 0, so Q > 0, then B N = loge (an), then B (n + 1) - BN = loge (a (n + 1)) - loge (an) = loge (a (n + 1) / an

In the arithmetic sequence {an}, A4 = 7, and A2, A5, a7 are equal ratio sequence, find the sum of the first n terms of the sequence {an}
Monthly exam One and a half hours to go

Because it is an arithmetic sequence, A4 + A5 = A2 + A7
And because A2, A5 and A7 are equal ratio series, A5 ^ 2 = A2 * A7
Let the tolerance be d
Then, (7 + D) ^ 2 = (7-2d) (7 + 3D)
d=7/5
Then A1 = 14 / 5
Using the sum formula of the first n terms of arithmetic sequence
Sn=na1+n(n-1)d/2=n^2-2.1n

The sequence an is an equal ratio sequence, Sn is the sum of its first n terms, and a1a7a4 is an equal difference sequence
Prove that 2s3, S6s, s12-s6 are equal ratio sequence
It's not S6s, it's S6

a7=a1*q^6,a4=a1*q^3.
So a1 + A1 * q ^ 3 = 2A1 * q ^ 6
Let Q ^ 6 = x ^ 2, then q ^ 3 = X
solve equations.
The formula of Sn and A1, Q ^ 3, 1-Q are used to represent three numbers respectively
Then, 1-Q will be reduced. The ratio is constant