The intersection point of the image of the linear function y = X-2 and y = 2x + 1 is (), that is, x = (), y = () is the solution of the equations X-Y = 2 and 2x-y = - 1

The intersection point of the image of the linear function y = X-2 and y = 2x + 1 is (), that is, x = (), y = () is the solution of the equations X-Y = 2 and 2x-y = - 1

The intersection point of the image of the linear function y = X-2 and y = 2x + 1 is (- 3, - 5), that is, x = (- 3), y = (- 5) is the solution of the equations X-Y = 2 and 2x-y = - 1

What are the connection points and differences between linear function and quadratic equation of two variables, linear function and linear equation of one variable?

1. The relationship between linear equation of one variable and linear function
Any linear equation of one variable can be transformed into the form of AX + B = 0 (a, B are constants, a ≠ 0), so the solution of linear equation of one variable can be transformed into: when the value of a certain linear function is 0, find the value of the corresponding independent variable. From the image, it is equivalent to the known line y = ax + B to determine the abscissa value of its intersection with the X axis
2. The relationship between linear function and linear inequality of one variable
Any linear inequality of one variable can be transformed into ax + b > 0 or ax + B

Linear function and quadratic equation of two variables
X-Y divided by 2 minus x + y divided by 4 = - 1 3 (x + y) - 2 (2x-y) = 8 to find the value of X and y,

Let the intersection point of the image of the first-order function y = 3x-4 and y = - x + 3 be p, and they intersect with the X axis at point a and point B respectively, and try to find the area of the triangle APB
By solving the equations: y = 3x-4 and y = - x + 3, we get x = 1 / 4, y = - 13 / 4, that is, the coordinates of point P are: (1 / 4, - 13 / 4), where the absolute value of ordinate 13 / 4 is the height of AB side of triangle ABP
Let y = 0 in the two equations y = 3x-4 and y = - x + 3, and solve the two values of X: 4 / 3 and - 3 are the abscissa of the two intersections A and B of the function image and X axis,
So the length of the bottom edge ab of the triangle ABP is 4 / 3 - (- 3) = 13 / 3,
So the area of triangle ABP is: 1 / 2 * 13 / 3 * 13 / 4 = 169 / 24
2. In the plane rectangular coordinate system, the line L1 passes through the points (2,3) and (- 1, - 3), the line L2 passes through the origin, and intersects with the line L1 and (- 2, a)
1. Try to find the value of A
2. What kind of solutions of binary linear equations can (- 2, a) be regarded as?
3. Let the intersection point be p, the intersection of line L1 and Y axis and point a, and calculate the area of triangle apo
(1) Let the equation of line L2 be: y = ax + B. by substituting the coordinates of points (2,3) and (- 1, - 3) into the equation, we can get a system of linear equations with respect to a and B, 3 = 2A + B, - 3 = - A + B,
The solution is a = 2, B = - 1,
That is, the equation of line L1 is: y = 2x-1
Because L2 intersects with the line L1 at (- 2, a), so this point is also on the line L2. Substituting it into the equation y = 2x-1, we get a = 2 * (- 2) - 1, a = - 5
(2) (- 2, a) can be regarded as the solution of a system of equations composed of the equations of lines L1 and L2
Because the line L2 passes through the origin and (- 2, a) point, the equation can be obtained by the method in (1): y = - 5 / 2x
Therefore, (- 2, a) can be regarded as the solution of the system of linear equations of two variables: y = 2x - 1 and y = - 5 / 2x
(3) The point of intersection P is (- 2, - 5),
Let: the equation of line L1: y = 0 in y = 2x-1, that is, 2x-1 = 0, and the solution is x = 1 / 2, that is, the abscissa of intersection a of line L1 and Y axis is 1 / 2,
Therefore, the Ao length of the bottom edge of the triangle apo is 1 / 2
The absolute value of the ordinate of point P, 5, is the height of the bottom edge
Therefore, the area of the triangle apo is: 1 / 2 * 1 / 2 * 5 = 5 / 4
I hope my answer can help you