SOLUTION: how would i graph this linear system, step by step, i dont know how to graph or how to get the coordinates, to graph the system? the system is 2x+y=13 5x-2y=1

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Question 51159This question is from textbook algebra 1, applications, graphs, and equations
: how would i graph this linear system, step by step, i dont know how to graph or how to get the coordinates, to graph the system?
the system is 2x+y=13
5x-2y=1
how would i put 2x+y=13 on a graph and 5x-2y=1 on a graph??
please if you could explain step by step for me??!!
thanks! This question is from textbook algebra 1, applications, graphs, and equations

Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
FIRST GET SOME GRAPH PAPER. NOW DRAW A HORIZONTAL LINE ACROSS THE MIDDLE OF THE PAGE. THIS IS THE X AXIS. NOW DRAW A VERTICLE LINE DOWN THE MIDDLE OF THE PAGE. THIS IS THE Y AXIS. WHERE THESE TWO LINES INTERSECT (CROSS) IS THE (X,Y) POINTS (0,0). NOW MARK EACH SQUARE TO THE RIGHT OF THE (0,0) INTERSECTION ON THE X AXIS 1,2,3,4,5,6,7,8,9 & 10. NOW MARK THE LEFT SIDE OF THIS X LINE STARTING AT (0,0) -1,-2,-3,-4,-5,-6,-7,-8,-9 & -10. NOW MARK THE SQUARES ABOVE THE INTERSECTION OF 0,0 ON THE Y AXIS 1,2,3,4,5,6,7,8,9,10,11,12,13 & 14. NOW MARK ALL THE SQUARES BELOW THE (0,0) INTERSECTION ON THE Y AXIS -1,-2,-3,-4,
-5,-6,-7,-8,-9, & -10.
YOU NOW HAVE A GRAPH FOR PLOTTING THE LINES REPRESENTED BY THE EQUATIONS YOUR TRYING TO GRAPH IN THE FORM Y=mX+b WHERE m=THE SLOPE & b=THE Y INTERCEPT.
2X+Y=13 MUST NOW BE TRANFORMED TO THE STANDARD LINE FORMULA THUS Y=-2X+13.
THIS EQUATION HAS A SLOPE (m) OF -2 & a y intercept of 13 (0,13). NOW TO FIND THE X INTERCEPT TO HAVE TWO POINTS TO DRAW A THE LINE THROUGH WE SET Y=0 THUS
0=-2X+13 OR -2X=-13 OR 2X=13 OR X=13/2 OR X=6.5 (6.5,0).
TO PLOT THE POINT (0,13) START AT THE INTRSECTION (0,0) COUNT UP 13 SQUARES AND MARK THIS POINT(0,13). NOW STARTING AGAIN AT THE INTERSECTION (0,0) AND COUNT 6 AND 1/2 SQUARES TO THE RIGHT OF THIS POINT AND MARK THIS POINT (6.5,0). NOW DRAW A LINE THROUGH THESE 2 POINTS. YOU NOW HAVE A LINE REPRESENTED BY THE EQUATION 2X+Y=13 WITH A SLOPE OF -2 AND A Y INTERCEPT OF 13.
NOW YOU TRY THE NEXT EQUATION USING THE SAME GRAPH.
5X-2Y=1 OR -2Y=-5X+1 OR 2Y=5X-1 OR Y=5/2X-1/2 WHICH HAS A SLOPE OF 5/2 OR 2.5 AND A Y INTERCEPT OF -1/2 OR .5 (0,-.5) NOW SET Y=0 WE GET 5X-2*0=1 OR 5X=1 OR X=1/5 OR .2 (.2,0)