   SEARCH HOME Math Central Quandaries & Queries  Question from Adam, a student: Solve the system by graphing: 4y>6x -3x+2y<-6 First change them to standard form: y <=> mx + b.

4y > 6x becomes y > (3/2)x + 0

and

-3x + 2y < -6 becomes y < (3/2)x - 3

Draw two straight lines corresponding to these slopes and y-intercepts. The lines are parallel and divide the plane into three regions. The region above the top line, the region between the lines and the region below the bottom line. Choose a point in each of the three regions and see it is satisfies both inequalities. If it does then any point that lies in the same region also satisfies both inequalities. If none the three points you chose satisfy both inequalities then no points in the plane satisfy both inequalities except perhaps points on the lines. In any case remember to ask yourself if the points on either line are also permitted.

Hope this helps,
Stephen and Penny     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.