How to move the term of linear equation with one variable

How to move the term of linear equation with one variable

X+8=10 x-(-8)=10
x=10-8 x=10-8
x=2 x=2
As long as you remember one, no matter on the left or on the right, the symbol on the left should be changed. In the second example, the reason why you subtract 8 is that the left negative is still positive, and on the right it becomes negative again

Ask a question of higher number (derivative of implicit function)
If x ^ y = y ^ x, find dy / DX
This problem can be solved directly with the formula dy / DX = - FX / FY, but the result I got is different from the answer in the book. After solving for a long time, I found that the answer in the book is to substitute the conclusion of x ^ y = y ^ x into the result of derivation, which makes the answer more simplified. Do I want to ask whether this substitution is reasonable? Or is the answer solved in another way
Here's another problem for me:
F (x, y) satisfies x * FX (x, y) + y * FY (x, y) = f (x, y), FX (1, - 1) = 3, point P (1, - 1,2) is on the surface z = f (x, y), and the tangent plane equation of point P is obtained

…… This problem can not be solved by dy / DX = - FX / FY, this formula should be learned when learning partial derivatives ~ in fact, x ^ y = y ^ x, taking logarithms on both sides can be changed to xlny = ylnx, that is, xlny ylnx = 0, now it can be solved ~ second, Let f (x, y, z) = z-f (x, y) (I replaced your f with F), then FX = - FX, FY = - fy, FZ = 1 at point P

Factorization - 3x + 2x-1 / 3 =?

-3x+2x-1/3=-1/3(9x-6x+1)= -1/3(3x-1)

Given that x.y satisfies (X-Y + 2 ＞ = 0.x + y-4 ＞ = 0.2x-y-5 ＜ = 0), find the maximum value of Z = [x + 2y-4] Supplement: [] is absolute value sign

According to the system of inequalities, 7 ≥ x ≥ 1,9 ≥ y ≥ 3 can be obtained
So 25 ≥ x + 2Y ≥ 7
So, the maximum value of Z = 21

Let f (x) be a quadratic function, and f (x-1) + F (2x + 1) = 5x ^ 2 + 2x, find f (x)

f(x-1)+f(2x+1)=5x^2+2x
=(x-1)^2+(2x+1)^2-2
=[(x-1)^2-1]+[(2x+1)^2-1]
So f (x) = x ^ 2-1

Given the quadratic function y = x-mx + 1, when x > 2, y increases with the increase of X, the value range of M is obtained

On the right side of the symmetry axis X = m / 2, y increases with the increase of X
So m / 2 ≤ 2
m≤4

The sum of a three digit number, a single digit number and a hundred digit number is equal to a ten digit number. Seven times the sum of a hundred digit number is two times larger than that of a single digit number and a ten digit number. The sum of a single digit number, a ten digit number and a hundred digit number is 14

This three digit number is x on the one digit, y on the ten digit, and z.x + Z = y on the hundred digit. ① 7z = x + y + 2x + y + Z = 14. ③ substitute ① into ③ to get y = 7. Substitute y = 7 into ① to get x + z = 7. ④ substitute ② to get 7z = x + 9. ⑤ ④ - ⑤ to get z = 2, ｜ x = 5. This three digit number is 2 × 100 + 7 × 10 + 5 = 275. Answer: this three digit number is 275

a∧3-6a∧2b＋12ab∧2-8b∧3

Solution equation: 2.5x-x = 1.8

=1.8/1.5=1.2

Through the point P (- 1,1), make a straight line and the ellipse x2 + Y2 + 2 = 1 intersect at two points ab. if the midpoint of the line AB is exactly P point, find the linear equation where AB is

Obviously, AB cannot be parallel to y axis, otherwise a and B are symmetrical about X axis, that is, the midpoint of AB cannot be P
Let the slope of AB be K, then the equation of AB is Y-1 = K (x + 1), that is, y = KX + K + 1
Suppose that the coordinates of a and B are (m, KM + K + 1) and (n, kn + K + 1) respectively
Simultaneous: y = KX + K + 1, x ^ 2 / 4 + y ^ 2 / 2 = 1, eliminate y, get: x ^ 2 / 4 + (KX + K + 1) ^ 2 / 2 = 1,
∴x^2＋2［k^2x^2＋2k（k＋1）x＋（k＋1）^2］－4＝0,
∴（1＋2k^2）x^2＋4k（k＋1）x＋2（k＋1）^2－4＝0.
Obviously, m and N are two parts of the equation (1 + 2K ^ 2) x ^ 2 + 4K (K + 1) x + 2 (K + 1) ^ 2-4 = 0,
By Weida's theorem, there are: M + n = - 4K (K + 1) / (1 + 2K ^ 2),
∴（m＋n）/2＝－2k（k＋1）/（1＋2k^2）.
∵ P is the midpoint of AB, ∵ (M + n) / 2 = - 1, ∵ - 2K (K + 1) / (1 + 2K ^ 2) = - 1,
∴2k（k＋1）＝1＋2k^2,∴2k^2＋2k＝1＋2k^2,∴k＝1/2.
The equation of AB is y = (1 / 2) x + 1 / 2 + 1, that is, x-2y + 3 = 0