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Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!
you graph the boundary lines: x+2y = 12 and 3x-y = 6
pick a point not on the line; (0,0) is an easy testing point
and if those make the equation true; then the side with that point is good, shade it; if false, go with the other side
ah so you can start with any natural number then and plug it into the formula? or no?
you can, but it helps to see that these are line equations; and you just need 2 points to draw a line with right?
the lines themselves are boundaries for the inequalities; so graph them first;
x+2y = 12 and 3x-y = 6; graph these lines so you establish the boundaries for the inequalitites
then pick a point that is NOT on the line to test for a result. that point will be on one side or the other of the line right? The most conviennt point to test is the origin: (0,0) x+2y<=12 and 3x-y>=6 0+2(0)<=12 and 3(0)-(0)>=6 0 <= 12 and 0 >=6 true false
all the points that make a true stament get shaded; all th epoints that make it false... shade the other side of the line
okay i get it use typing it out really help me... :D
lol you* typing it out...
its so weird maybe im asking the wrong question because my book is giving me 2 lines not one for the answer
the answer has 2 lines; I just demonstrated with 1
o sorry i get it now. i was confused for a second so then i can use the same points (0,0) for the second variable correct?
it is the common area that the two have that are solutions
gd i still am confused so now that we have that one line how do we plot the other line? because in the above answer you plug (0,0) into both variables
lemme just write this out and see if i can figure it out before i frustrate you.
you really make 2 solution sets and where they overlap, is the area I shaded
how do i give you a good review?
know a good free graphing programing?
hmmm.... i use wolframalpha.com alot ; but it suits me
youre welcome :)