A point whose abscissa and ordinate are all integers is called an integral point. Let K be an integer. When the intersection of the line y = x-3 and y = KX + k is an integral point, how many points does K have From x-3 = KX = k to k = (x-3) / (x + 1), in order to further narrow the scope, how to simplify?

A point whose abscissa and ordinate are all integers is called an integral point. Let K be an integer. When the intersection of the line y = x-3 and y = KX + k is an integral point, how many points does K have From x-3 = KX = k to k = (x-3) / (x + 1), in order to further narrow the scope, how to simplify?

The intersection of y = x-3 and y = KX + k
x-3=kx+k
x=(k+3)/(1-k)=(-1+k+4)/(1-k)=4/(1-k)-1
X is an integer
4 / (1-k) - 1 is an integer
4 / (1-k) is an integer
Because K is an integer
It can be obtained from 4 / (1-k)
1-k=±4 1-k=±2 1-k=±1
k=1±4 k=1±2 k=1±1
k=5、3、2、0、-1、-3
By K = (x-3) / (x + 1)
k=(x+1-4)/(x+1)=1-4/(x+1)
4 / (x + 1) is an integer
Calculate X and get the solvable problem
It's just a little circuitous

In the plane rectangular coordinate system, (x, y) are all integers. We call them integer points. The intersection of y = x + 1 and y = kx-2 (k is an integer) is an integer point. How many values of K are there?

First, find out the intersection coordinates (3 / k-1, K + 2 / k-1), ensure that 3 / k-1 is an integer, K can be - 2,0,2,4, so there are four numbers
(Note: because the y value of the coordinate is brought into the first straight line, when x is an integer point, y must be an integer.)

In Cartesian coordinates, if the vertical and horizontal coordinates of a point are all integers, then the point is called the integral point. Let K be an integer. When the intersection of the line y = X-2 and y = KX + k is the integral point, the value of K can be taken as ()
A. Four B. five C. six D. seven

① When k = 0, y = KX + k = 0, y = KX + KX + k = 0, which is the x-axis, then the intersection point of the straight line y = X-2 and x-axis is (2.0) to satisfy the topic, (2.0) to satisfy the (2.0) to satisfy the (2.0); (k = k = k = 0, \\\\, (k-1) x = (KX + X + X + X + X + X + X + X + K + K + K + K + K + K + 2) and the intersection point of the line y = x = x = x-x-x-2 and the intersection of the straight line, y = 0 or K = 2, k = 0 or K = 2, or K = 4 or K = 4, or K = 4, or K = 2, or K = 2, k = 4, or K = 2, or K = 2, k = 2, or K = k = 4, so K is in common There are four kinds of values

In Cartesian coordinate system, if the ordinates and abscissa of a point are integers, then the point is called integral point. Let K be an integer. When the intersection of the line y = x + 2 and y = kx-4 is integral point, how many values of K can be taken at most?

The intersection of simultaneous straight lines y = x + 2 and y = kx-4: x = 6 / (k-1) x, integer k = - 5, - 2 - 1,0,2,3,4,7
X is an integer and y must be an integer. The maximum number of values is 8