SOLUTION: I have been reading almost all the ways to do a graph that has been posted and I am still confused. Here is a problem that I am not clear on.
Solve the following system of linea
Algebra.Com
Question 23198: I have been reading almost all the ways to do a graph that has been posted and I am still confused. Here is a problem that I am not clear on.
Solve the following system of linear inequalities by graphing.
x + 2y < 3
2x – 3y < 6
Thank you
Sherri
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE EXAMPLES BELOW TO PLOT THE GRAPHS FOR FIRST 2 EQUALIT1ES
x + 2y = 3
2x – 3y = 6
AND THEN SHADE THE ZONES PERTAINING TO THE INEQUALITIES
x + 2y < 3
2x – 3y < 6
AFTER SHADING ALL THE ZONES ,SEE THE COMMON ZONE WHICH SATISFIES ALL THE CONDITIONS .THAT IS YOUR SOLUTION ZONE .IF YOU GET INTO ANY DIFFICULTY COME BACK AND I SHALL GIVE YOU MORE DETAILED AND SPECIFIC ANSWER TO YOUR PROBLEM.BUT FIRST TRY TO UNDERSTAND THE METHODOLOGY AND INFORM WHETHER YOU GOT IT OR STILL DOUBTS EXIST.
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Solve and graph the solution set. 8x + 8 > 7x + 9
DONT GET UPSET .TRY TO LEARN WITH A COOL MIND AND YOU WILL BE THE MASTER.I AM GIVING THE SOLUTION FOR YOUR PROBLEM AND FEW OTHER EXAMPLES FOR YOU TO READ AND UNDERSTAND.COME BACK TO ME IF YOU DONOT UNDERSTAND.
HERE WE HAVE 8X+8>7X+9
8X-7X>9-8
X>1
TO DRAW THE GRAPH ,AS A FIRST STEP TAKE
X=1 AND PLOT IT .SINCE X=1 IS CONSTANT ALWAYS ,IT MEANS THAT WHATEVER MAY BE THE VALUE OF Y X IS ALWAYS 1..LET US TABULATE FOR PLOTTING
X...1....1....1.....1.....1.......ALWAYS 1 AS X=1
Y...0....1....2.....-1....-2...ETC...FOR ANY VALUE OF Y
THE GRAPH WILL LOOK LIKE THIS
GREEN LINES ARE X AND Y AXES..THE COLOURED VERTICAL LINE IS THE GRAPH OF X=1
YOU GET A VERTICAL STRAIGHTLINE PERPENDICULAR TO X AXIS AT X=1.WHICH IS THE GRAPH OF X=1.WHATEVER MAY BE THE VALUE OF Y......ON THIS LINE ANY POINT WILL SATISFY THE GIVEN EQUATION X=1.
NOW LET US SEE WHAT THE GRAPH IS FOR X>1
NOW WE FIND THAT ORIGIN IS ON ONE SIDE OF THE LINE.LET US TEST THE ORIGIN IN THE ABOVE EQN.WE FIND THAT AT ORIGIN ,
X=0 AND SO X<1...HENCE THE REGION FROM THE LINE TOWARDS OR IN THE DIRECTION TOWARDS ORIGIN IS FOR X<1.BUT WE WANT THE REGION WHERE X>1....OBVIOUSLY ,IT IS THE REGION FROM THE LINE SPREADING TO THE OTHERSIDE OR POINTING AWAY FROM THE ORIGIN . HENCE THE GRAPH OF X=1 IS THE BORDER LINE WHICH SEPERATES THE X-Y PLANE IN TO 2 PARTS OR 2 ZONES.THE ZONE TOWARDS ORIGIN IS WHERE X<1 AND THE OTHER ZONE IS WHERE X>1.SAME LOGIC APPLIES TO ALL OTHER CASES DISCUSSED BELOW.TO CLEAR YOUR DOUBT WHERE Y> OR < SOME GIVEN FUNCTION OF X , YOU CAN TEST BY EASY SUBSTITUTION OF ORIGIN (X=0 AND Y=0)TO TEST THE ZONE TOWARDS ORIGIN IS THE > OR < SIDE.
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how do you graph this equation y>3x+4
1 solutions
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Answer 11555 by venugopalramana(565) on 2005-12-22 10:59:37 (Show Source):
Y>3X+4...PLOT Y=3X+4 AS USUAL BY GIVING DIFFERENT VALUES TO X AND FINDING Y AS TABULATED BELOW
X.......0........1........2...........-1...........-2
Y.......4........7........10...........1...........-2
THE GRAPH WILL LOOK LIKE THIS
YOU GET A STRAIGHTLINE.....ON THIS LINE ANY POINT WILL SATISFY THE GIVEN EQUATION .NOW WE FIND THAT ORIGIN IS ON ONE SIDE OF THE LINE.LET US TEST THE ORIGIN IN THE ABOVE EQN.WE FIND THAT
Y=0 AND 3X+4=3*0+4=4...SO Y<3X+4...HENCE THE REGION FROM THE LINE TOWARDS OR IN THE DIRECTION TOWARDS ORIGIN IS FOR Y<3X+4.BUT WE WANT THE REGION WHERE Y>3X+4....OBVIOUSLY ,IT IS THE REGION FROM THE LINE SPREADING TO THE OTHERSIDE OR POINTING AWAY FROM THE ORIGIN .
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