It is known that the intersection point in the first quadrant of the image of the linear function y = x + m and the inverse scale function y = m + 1 / X (m ≠ - 1) is p (XO, 3) (1) Finding the value of XO (2) Finding the analytic expressions of first order function and inverse proportion function Please answer

It is known that the intersection point in the first quadrant of the image of the linear function y = x + m and the inverse scale function y = m + 1 / X (m ≠ - 1) is p (XO, 3) (1) Finding the value of XO (2) Finding the analytic expressions of first order function and inverse proportion function Please answer

Substituting P into two functions
3=x0+m (1)
3=(m+1)/x0,m+1=3x0 (2)
(1)+(2)
3+m+1=x0+m+3x0
4x0=4
x0=1.m=3-x0=2
m+1=3
So x0 = 1
y=x+2,y=3/x
(1) The point P (x 0,3) is on the graph of the linear function y = x + M
3 = x0 + m, i.e. M = 3-x0
Point P (x0,3) in the inverse scale function y = (M + 1) / X
In this image,
∴3=
(m+1)/x0
M = 3x0-1
∴3-x0=3x0-1，
The solution is x0 = 1;
(2) From (1), we get
m=3-x0=3-1=2，
The expansion of a function of degree
(1) The point P (x 0,3) is on the graph of the linear function y = x + M
3 = x0 + m, i.e. M = 3-x0
Point P (x0,3) in the inverse scale function y = (M + 1) / X
In this image,
∴3=
(m+1)/x0
M = 3x0-1
∴3-x0=3x0-1，
The solution is x0 = 1;
(2) From (1), we get
m=3-x0=3-1=2，
The analytic formula of the first-order function is y = x + 2, and the analytic formula of the inverse proportion function is y=
3 / X. Stow
Let a = {(x, y) | 2x-3y = 1}, B = {(x, y) | 3x + 2Y = 1} be the intersection of a and B
Two nonparallel lines intersect at a point to solve the equations
x=5/13
y=-1/13
So a intersects B = {(5 / 13, - 1 / 13)}
It is known that there are two intersections in the image of the first-order function y = 2x-b and the inverse scale function y = (B + 2) / x, and the abscissa of one focus is 3. Find the value of B and two intersections
Take x = 3, 2x-b = (B + 2) / x, and substitute x = 3 into the equation to get b = 4
Substituting B = 4 into the above two equations and solving the two equations simultaneously, we get X1 = 3, Y1 = 2; x2 = - 1, y2 = - 6
B=4
Given the set a = {(x.y) | 3x-2y = 8}, B = {(x.y) | 2x = 3Y = 1}, find the intersection B of a (expressed by enumeration)
3x-2y=8－－(1)
2x+3y=1－－(2)
(1)×3+(2)×2
9x+4x=24+2
13x=26,x=2
So 3 × 2-2y = 8, 2Y = 2, y = 1
The solution of the equations is x = 2, y = 1
So a ∩ B = {(2,1)}
{(22/5,13/5)}
(it is known that there are two intersections in the image of the first-order function y = 2x-b and the inverse scale function y = (B + 2) / x, in which the abscissa of one focus is 3, (1) find the value of B and
The coordinates of two focal points (2) △ cab area
The coordinates are (- 1, - 6) (3,2) and the important thing is the area
The area of triangle OAB is not cab
1, B = 4, intersection coordinates (- 1, - 6), (3,2)
2,S=8
If a = {x | 1 ≤ x ≤ 3}, B = {y | y = x & # 178; + 2x + A, X ∈ r}
(1) If a ∪ B = B, find the value range of a; (2) if a ∪ B = empty set, find the value range of a; (3) if a ∪ B = {1, + infinity}, find the value range of A
PS: hope to have a detailed process
In set B,
y=(x+1)^2+(a-1)≥(a-1)
B={y|y≥a-1}
B can also be written as:
B={x|x≥a-1}
(1)
∵A∪B=B,
∴A⊆B
be
a-1≤1
∴a≤2
(2)
∵A∪B=∅
∴34
(3)
∵A∪B=[1,+∞)
∴1≤a-1≤3
be
2≤a≤4
Let the plane region represented by the inequality system x + Y-11 ≥ 03x-y + 3 ≥ 05x-3y + 9 ≤ 0 be d. if there are points on region D on the image of exponential function y = ax, then the value range of a is ()
A. （1，3]B. [2，3]C. （1，2]D. [3，+∞]
Make the image of region D, connect the image of exponential function y = ax, and get the point C (2,9) from x + Y-11 = 03x-y + 3 = 0. When the image passes through the boundary point C (2,9), a can take the maximum value 3. Obviously, as long as a is greater than 1, the image must pass through the points in the region
Given the set a = {y | y = x & # 178; + 1}, B = {Y & # 178; = - 2x + 6}, then a intersects B=
The set a and B represent the range, a is 1 - ∞, B is r, so the intersection = a
Inequality system x > = 0, x + 3Y > = 3,3x + y
It's a very common problem in senior high school. Three lines need to be drawn. It can be seen from the figure that the line must intersect with 3x + y = 4. The intersection point can be solved by concatenation. The area of the triangle is generally large and the time is tight. I won't help you
Given a = {(x, y) / 2x + 3Y = 7}, B = {(x, y) / x + y = 5}, then a intersection B =?
U = R, a = {X / x > and equal to 1} then the complement of a in U is
Simultaneous equations
x+y=5
2x+3y=7
We get x = 8, y = - 3
So a intersects B = {(8, - 3)}
It is to solve the system of equations composed of 2x + 3Y = 7 and X + y = 5, where x = 8, y = - 3, a ∩ B = {(x, y) / x = 8, y = - 3}
Set a = {(x, y) | 2x + 3Y = 7}, B = {(x, y) | x + y = 5},
Then a ∩ B = {(x, y) | 2x + 3Y = 7, and X + y = 5}
={（8，-3）}。
U=R,A={x|x>=1}，
Then the complement of a in U is {x | X