Given the function y = Y1 + Y2, and Y1 = 2x + m, y2 = x / (m-1) + 3, the ordinate of the image focus of the two functions Y1 and Y2 is 4

Given the function y = Y1 + Y2, and Y1 = 2x + m, y2 = x / (m-1) + 3, the ordinate of the image focus of the two functions Y1 and Y2 is 4

Let the intersection of Y1 and Y2 be (s, 4), then the equations can be obtained from the meaning of the problem{
2s+m=4
s/(m-1)+3=4→s=m-1.
Solution
m=2.
Then y = Y1 + y2 = (2x + 2) + X / 1 = 3x + 2

In the same coordinate system, what are the intersection coordinates of the first-order function Y1 = 2x-2, y2 = 0.5x + 1,

Two simultaneous! Y1-y2 = 1.5x-3 = 0
X = 2, because it is two independent first-order functions, we can solve Y1 and Y2 by looking at the same y!

The image of the functions Y1 = x + 1 and y2 = ax + B (a ≠ 0) is shown in the figure. The intersection of these two function images is on the Y axis. Then the value range of X which makes the values of Y1 and Y2 greater than zero is ()
A. x＞-1B. x＞2C. x＜2D. -1＜x＜2

It can be seen from the intersection coordinates of two function images and X-axis that when x ＜ 2, the image of function y2 = ax + B (a ≠ 0) is above x-axis, that is, Y2 ＞ 0; when x ＞ - 1, the image of function y2 = x + 1 is above x-axis, that is, Y1 ＞ 0; therefore, when - 1 ＜ x ＜ 2, the values of Y1 and Y2 are greater than zero