In the known mapping f: a → B, a = b = {(x, y) | x ∈ R, y ∈ r}, the element (x, y) in F: a corresponds to the element (3x + Y-1, x-2y + 1) in B. (1) is there such an element (a, b) that its image is still self? If it exists, find out the element; if it does not exist, explain the reason; (2) judge whether the mapping is one by one?

In the known mapping f: a → B, a = b = {(x, y) | x ∈ R, y ∈ r}, the element (x, y) in F: a corresponds to the element (3x + Y-1, x-2y + 1) in B. (1) is there such an element (a, b) that its image is still self? If it exists, find out the element; if it does not exist, explain the reason; (2) judge whether the mapping is one by one?

(1) Suppose there is such an element (a, b) so that its image is still self, then 3A + B-1 = A and a-2b + 1 = B, the solution is: a = 0, B = 1, that is, there is such an element (0, 1) so that its image is still self, (2) if this mapping is one-to-one mapping, then any element in B has a unique corresponding original image in a, take B

Given the mapping f: a → B = ((x, y), X belongs to R, y belongs to R), the elements (x, y) in F: a correspond to the elements (3x + Y-1, x-2y + 1) in B
Is this a one-to-one mapping?

Yes

Given that the original image of the point (x, y) under the mapping f is (x + 2Y, 2x-y), then the corresponding image of (4,3) under the action of F is (x + 2Y, 2x-y)

（2,1）
x+2y=4
2x-y=3
x=2 y=1

It is known that the mapping from a to B is F1: X → 2x-1, and the mapping from B to C is F2: y → 1 △ (2Y + 1)?

It is known that the mapping from set a to set B is x → 2x-1, and the mapping from set B to set C is y → 1 / (3Y + 1)~

Let y = 2x-1 be substituted into 1 / (3Y + 1)
1 / (6x-2)
So the mapping from a to C is: X → 1 / (6x-2) (x! = 1 / 3)
Pay attention to domain in mapping composition

Let the mapping from set a to B be F1: X → y = 2x + 1, and the mapping from set B to C be F2: y → z = y2-1. Then what is the corresponding rule of mapping f from set a to C? What is the image of element 1 in set a in C? What is the original image of element 0 in set C in a?

Because set a to B is the relationship between X and y, and set B to C is the relationship between Z and y, which is related to y. we can use the relationship between X and y to find out the relationship between X and Z. because z = y ^ 2-1 substitutes y = 2x + 1 into Z = y ^ 2-1, that is, z = (2x + 1) ^ 2-1, which is the relationship between X and Z. the element x = 1 in set a substitutes z = (2 × 1 +)

The solution of the system of equations x + y + 5 = 0 & nbsp; & nbsp; ① XY + 14 = & nbsp; & nbsp; ② is___ ．

The solution of the original equation is: x = 2Y = - 7, x = - 7Y = 2

Solving the equations {x ^ 2-xy = 14 ① X-Y = 14 ②

x^2-xy=14① x-y=14②
① 2
x = 14/14 = 1
y = x -14 = 1-14 = -13

Solve the equations: x + y = 2, XY + Z = 1

x=y=1.z=0

If the values of solution X and y of {3x-2y = 102kx + (K-2) y = 6 are opposite to each other, then the value of K is

3x+2x=10 x=2
4k-2（k-2）=6
2k=2
k=1