LOOKING FOR THE OVERLAP SOLVING SYSTEMS OF INEQUALITIES

A "SYSTEM" OF INEQUALITIES IS A SET OF INEQUALITIES THAT YOU DEAL WITH ALL AT ONCE. USUALLY YOU START OFF WITH TWO OR THREE LINEAR INEQUALITIES. THE TECHNIQUE FOR SOLVING THESE SYSTEMS IS FAIRLY SIMPLE. HERE'S AN EXAMPLE WITH THREE. 2X – 3Y 0 Stapel, Elizabeth. "Systems of Linear Inequalities." Purplemath. Available from

JUST AS WITH SOLVING SINGLE LINEAR INEQUALITIES, IT IS USUALLY BEST TO SOLVE AS MANY OF THE INEQUALITIES AS POSSIBLE FOR "Y" ON ONE SIDE. SOLVING THE FIRST TWO INEQUALITIES, WE GET THE REARRANGED SYSTEM: Y > ( 2 / 3 )X – 4 Y 0

SOLVING SYSTEMS OF INEQUALITIES MEANS GRAPHING EACH INDIVIDUAL INEQUALITY, AND THEN FINDING THE OVERLAPS OF THE VARIOUS SOLUTIONS. SO GRAPH EACH INEQUALITY, AND THEN FIND THE OVERLAPPING PORTIONS OF THE SOLUTION REGIONS. Hint: Using different colored pencils really helps to keep the different inequalities separate and shows the region of the solution more clearly.

THE LINE FOR THE FIRST INEQUALITY IN THE SYSTEM LOOKS LIKE THIS: y > ( 2 / 3 )x – 4

THIS INEQUALITY IS A "GREATER THAN" INEQUALITY, SO WE WANT TO SHADE ABOVE THE LINE. HOWEVER, SINCE THERE WILL BE MORE THAN ONE INEQUALITY ON THIS GRAPH, WE DON'T KNOW (YET) HOW MUCH OF THAT UPPER SIDE WILL ACTUALLY BE IN THE FINAL SOLUTION REGION. y > ( 2 / 3 )x – 4 UNTIL WE KNOW, WE CAN KEEP TRACK OF THE FACT THAT WE NEED THE TOP REGION BY DRAWING A LITTLE FRINGE ALONG THE TOP SIDE OF THE LINE.

NOW, GRAPH THE LINE FOR THE SECOND INEQUALITY y < ( – 1 / 5 )x + 4 y > ( 2 / 3 )x – 4 AND, SINCE THIS IS A "LESS THAN" INEQUALITY, DRAW THE FRINGE ALONG THE BOTTOM OF THE LINE

THE LAST INEQUALITY IS A COMMON "REAL LIFE" CONSTRAINT: ONLY ALLOWING X TO BE POSITIVE. THE LINE "X = 0" IS JUST THE Y-AXIS. I NEED TO REMEMBER TO DASH THE LINE IN, BECAUSE THIS ISN'T AN "OR EQUAL TO" INEQUALITY, SO THE BOUNDARY (THE LINE) ISN'T INCLUDED IN THE SOLUTION: y > ( 2 / 3 )x – 4 y < ( – 1 / 5 )x + 4 x > 0

THE "SOLUTION" OF THE SYSTEM IS THE REGION WHERE ALL THREE INEQUALITIES SHADINGS OVERLAP; THIS AREA CONTAINS ALL THE POINTS THAT MAKE EVERY INEQUALITY IN THE SYSTEM TRUE. x > 0y > ( 2 / 3 )x – 4 y < ( – 1 / 5 )x + 4 This is called a bounded system because it has boundaries on all sides.

THIS TYPE OF SYSTEM IS CALLED "UNBOUNDED", BECAUSE IT CONTINUES FOREVER IN AT LEAST ONE DIRECTION (IN THIS CASE, FOREVER DOWNWARD). y < 2x + 3 y < –4x + 5

OF COURSE, THERE'S ALWAYS THE POSSIBILITY OF GETTING NO SOLUTION AT ALL. y > x + 2 y < x – 2 There is no place where the individual solutions overlap. (Note that the lines y = x + 2 and y = x – 2 are parallel lines.) Since there is no intersection, there is no solution.