The positive scale function y = 2x and the inverse scale function y = x [K ≠ 0] have only one intersection [2,4] Find another coordinate

The positive scale function y = 2x and the inverse scale function y = x [K ≠ 0] have only one intersection [2,4] Find another coordinate

When x = 2, y = 4
Take x = 2, y = 4 into y = k-x
There are: 4 = K-2
k=8
Take k = 8 into the inverse proportion function to get the analytic formula: y = 8
K > 0, so the inverse scale function image is in the first and third quadrant
If we take y = 4 into y = 2x, we can get x = 2
And because the coordinates are in the third quadrant
So the coordinates are: (- 2, - 4)

Given the solution of binary linear equations 2x + y = 8,3kx + 2Y = 6K, x + y = 10, find the value of K?

1. First, find 2x + y = 8 for 2x + y = 8, x + y = 10 (1) x+y=10…… (2) (1) - (2) get: x = - 2 (2) * 2 - (1) get: y = 12 2, substitute x = - 2; y = 12 into 3kx + 2Y = 6K, get: - 6K + 24 = 6K, get: k = 2

The solution of 3kx + 2Y = 6K 2x + y = 8 is the solution of X + y = 10

2x+y=8
x+y=10
The results show that x = - 2, y = 12;
Then 3K * (- 2) + 2 * 12 = 6K
We get k = 2

The image composed of all points with the solution of equation 2x + y = 5 as the coordinate is the same as that of the linear function y =

The image composed of all points with the solution of equation 2x + y = 5 as coordinates is the same as that of the linear function y = - 2x + 5

Given a function y = 2x + 4, please draw the image of y = 2x + 4 and solve the equations 2x + 4 = 8 and 2x + 4 = 0 according to the image

The image of the first-order function is parallel to the straight line y = 2x-7 and intersects with the straight line y = 0.25x + 3 at a point on the y-axis

Let the analytic formula of this function be y = KX + a
If the image of a linear function is parallel to the line y = 2x-7, then k = 2
25X + 3 intersect with the straight line y = 0.25x + 3 at a point on the y-axis, let x = 0, calculate y = 3, then the intersection coordinate is (0,3), bring in the analytical formula: 3 = 2 × 0 + A, solve a = 3
Therefore, the analytic formula of a function is y = 2x + 3

Given that the image of a linear function passes through points (3,4) and is parallel to the line y = - 2x + 1, the analytic expression of the linear function is obtained

It is parallel to the straight line y = - 2x + 1, so we can set the analytic expression of this first-order function: y = - 2x + B
The image passes through points (3,4),
So, 4 = - 2 * 3 + B
b=10
The analytic expression of this first-order function is y = - 2x + 10

According to the following conditions, it is determined that the image of analytic formula 1 is parallel to the line y = 2x-1, and the image passing through point (1,3) 2 passes through point（
According to the following conditions, it is determined that the image of the analytic formula 1 is parallel to the line y = 2x-1, and passes through the point (1,3) 2. The image passes through the point (2, - 1) and is compared with the line y = - 1 / 2x + 3 at the same point 3 on the y-axis. The area of the triangle formed by the line y = 2x + B and the two coordinate axes is 4

Because the slope of the parallel image is the same as that of the known straight line, it is 2. Let y = 2x + B. let y = 2x + 12 be brought in by passing through the point (1,3). Let x = 0 be brought into the known equation y = 3, so let y = ax + B be brought in by passing through two points

The image of a function passes through points (1,3), and the parallel line y = - 2x + 1

If parallel lines y = - 2x + 1, the slopes are equal, that is, k = - 2
Let the straight line be y = - 2x + B
Substituting (1,3), we get b = 5
The analytic formula is y = - 2x + 5

The image is parallel to the line y = 2x-1 and passes through the point (1,3)

If the image is parallel to the known line y = 2x-1, then the slope of the line k = 2 is known. Let the linear equation be y = 2x + B
If the straight line passes through the point (1,3), then it is substituted into the equation and B = 1
So the line is y = 2x + 1