SOLUTION: PLEASE HELP...I AM STUCK ON WHAT TO DO HERE...I WOULD REALLY APPRECIATE YOUR HELP THANKS!
sketch the feasible region (quadrant 1 only) for the following system of linear inequa
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-> SOLUTION: PLEASE HELP...I AM STUCK ON WHAT TO DO HERE...I WOULD REALLY APPRECIATE YOUR HELP THANKS!
sketch the feasible region (quadrant 1 only) for the following system of linear inequa
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Question 63957: PLEASE HELP...I AM STUCK ON WHAT TO DO HERE...I WOULD REALLY APPRECIATE YOUR HELP THANKS!
sketch the feasible region (quadrant 1 only) for the following system of linear inequalities
2x-6y< or equal to 0
9x-14y< or equal to 26
1x+3y< or equal to 53
-1+2y< or equal to 17
x> or equal to 0 and y > or equal to 0
SORRY I DON'T KNOW HOW TO WRITE THE EQUAL SIGN UNDER THE GREATER THAN OR LESS THAN SIGNS...I HOPE YOU UNDERSTAND WHAT I MEAN... Answer by tutorcecilia(2152) (Show Source):
You can put this solution on YOUR website! 2x-6y<= 0 [solve for y]
-6y<=-2x
y>=-2x/6 [switch the signs when multiplying or dividing by a negative number]
y>=1/3x [draw the boundary line at y=1/3x. Use a solid line since the line is also 'equal to'.]
.
[shade all points greater than or equal to the line. that is, shade above or greater than the line]