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Math problems in junior one.
If the landlord has studied the function once, he will know that these two equations can be written in the form of y = kx+b, and they actually represent two straight lines.

When k 1=k2 and b 1 is not equal to b2,

These two lines are parallel and have no intersection, that is, the equations have no solution.

When k 1=k2 and b 1=b2,

These two straight lines coincide.

There are countless intersections, so the equations have countless groups of solutions.

When k 1 is not equal to k2,

These two straight lines have only one intersection,

These equations have unique solutions.

2x-3y+m = 0……( 1)-& gt; y=(2/3)x+m/3

(n- 1)x+6y-2 = 0……(2)-& gt; y =-∞(n- 1)/6〕x+ 1/3

Namely: k 1=2/3, k2 =-[(n-1)/6]; b 1=m/3,b2= 1/3 .

According to the above principle, the range of m and n can be found.

If the landlord has not learned this knowledge, you can consider this:

After the two equations are simplified and classified into standard forms,

When the two equations are exactly the same, there is actually only one binary linear equation with countless solutions.

When the coefficients of two equations x and y are exactly the same, but their constants are different, the values of x and y cannot be found, so there is no solution.

When the coefficients of two equations are not proportional, it has a unique set of solutions.

My statement is inaccurate and unscientific, just to be more popular.